🧠:Okay, let's see. I need to prove that there's a positive integer N, bounded by the product of (2|A_i| + 1) for each i, such that for each i, N is not congruent to any element of A_i modulo m_i. The m_i are pairwise relatively prime, which probably means the Chinese Remainder Theorem (CRT) will be useful here. The problem feels like it's related to the pigeonhole principle or some combinatorial argument with the CRT.First, let me restate the problem in simpler terms. We have 2013 moduli m_i, each greater than 1, and pairwise coprime. For each modulus m_i, there's a set A_i of residues modulo m_i that N should avoid. The size of each A_i is |A_i|, and the upper bound for N is the product over all (2|A_i| + 1). So, maybe for each modulus, we have some forbidden residues, and we need to find a number N that avoids all these forbidden residues, with N not too large.Since the moduli are pairwise coprime, the CRT tells us that the system of congruences has a unique solution modulo the product of all m_i. But here, instead of specifying congruences, we need to avoid certain congruences. So perhaps we can model this as a covering problem? Wait, but the forbidden sets A_i are specific to each modulus, and we need a number N that doesn't lie in any of the forbidden residue classes.Alternatively, think of each modulus m_i as a coordinate, and the forbidden residues A_i as "bad" coordinates. Then we want a point N in the product space that doesn't lie in any of the bad coordinates. The bound given is the product of (2|A_i| + 1) over all i. This seems similar to the concept of avoiding forbidden residues in each dimension, with some combinatorial factor.Wait, maybe there's a connection to the inclusion-exclusion principle? But inclusion-exclusion is usually for counting the number of elements not in any of the forbidden sets. However, here we need an existence result with a bound on N.Alternatively, think of each (2|A_i| + 1) as related to the number of possible choices for N modulo m_i. If each modulus m_i has some forbidden residues A_i, then the number of allowed residues is m_i - |A_i|. But since m_i and |A_i| can vary, maybe we need to use some product of these allowed residues? But the given bound is a product over (2|A_i| + 1), which is different.Wait, maybe the (2|A_i| + 1) term is related to the size of some interval or the number of numbers we need to consider for each modulus. Let's take a step back.Suppose for each modulus m_i, the forbidden residues are A_i. Then for each m_i, the forbidden numbers N are those congruent to a in A_i mod m_i. So, for each m_i, the forbidden numbers are the union of arithmetic progressions a + m_i * k for each a in A_i. So, we need to find an N not lying in any of these arithmetic progressions. The problem is similar to the Chinese Remainder Theorem but in reverse; instead of lying in certain residues, we need to avoid them.This seems related to the concept of covering systems or avoiding residue classes. However, in our case, the moduli are pairwise coprime, which might simplify things.Let me think about a simpler case. Suppose there is only one modulus, m_1. Then A_1 is a subset of {1, 2, ..., m_1 - 1}, and we need to find N ≤ (2|A_1| + 1) such that N is not congruent to any element of A_1 modulo m_1. Is this possible?Wait, if we have m_1 and A_1 with |A_1| elements, then the number of forbidden residues is |A_1|. The total residues are m_1, so allowed residues are m_1 - |A_1|. If we pick numbers from 1 up to (2|A_1| + 1), how many residues do they cover? Each number N can be congruent to some residue mod m_1. But by the pigeonhole principle, if we have (2|A_1| + 1) numbers, they might cover up to (2|A_1| + 1) different residues. But since the forbidden residues are |A_1|, maybe there's a way to find an allowed residue within that interval?Wait, actually, if we consider the interval [1, 2|A_1| + 1], since each residue class mod m_1 repeats every m_1 numbers, but since 2|A_1| + 1 might be larger than m_1, this is not straightforward. Maybe the bound is derived from the idea that each forbidden residue can "block" at most two numbers in the interval, so with 2|A_1| + 1 numbers, even if each forbidden residue blocks two numbers, there's still one left. Let me formalize this.Suppose we have an interval of numbers from 1 to S. Each forbidden residue a in A_i corresponds to numbers congruent to a mod m_i. If we look at the numbers 1 to S, how many are congruent to a? It's floor(S/m_i) or floor(S/m_i) + 1. But if S is small, say S ≤ 2|A_i| + 1, but m_i is greater than 1. Wait, this might not be the right approach.Alternatively, think of each forbidden residue a in A_i as excluding numbers N ≡ a mod m_i. For each modulus m_i, the problem is to choose a residue for N mod m_i that is not in A_i. If we can choose such residues independently (since the moduli are coprime), then by the Chinese Remainder Theorem, there exists a unique N modulo M (where M is the product of all m_i) satisfying those residue conditions. But we need N to be positive and bounded by the given product.But the problem states that N is ≤ the product of (2|A_i| + 1) over all i. However, the product of (2|A_i| + 1) might be much smaller than M (the product of m_i). So how can we guarantee that such an N exists within that bound?This seems like a problem that might use the Lovász local lemma or some combinatorial counting argument. Wait, but the problem is asking for existence, not probabilistic existence. The bound given is similar to the one in the pigeonhole principle when you have multiple choices per dimension.Alternatively, think of each modulus m_i as having 2|A_i| + 1 possible "slots" or "candidates" for N. Then, the total number of candidate numbers is the product of these slots. But how does this relate to avoiding the forbidden residues?Wait, maybe for each modulus m_i, we can consider that for each residue not in A_i, there are two residues adjacent to it (modulo m_i) that are also allowed? Not sure. Alternatively, think of each forbidden residue as excluding a congruence class, so for each modulus m_i, the number of allowed residues is m_i - |A_i|. The total number of possible N modulo M (the product of m_i) is the product of (m_i - |A_i|). But the bound given is the product of (2|A_i| + 1). Unless there is some relationship between (m_i - |A_i|) and (2|A_i| + 1), but since m_i can be arbitrary (as long as they are coprime and greater than 1), maybe this is not the case.Wait, but since m_i and m_j are coprime, perhaps the allowed residues for each modulus can be combined independently via CRT, leading to the total number of solutions being the product of (m_i - |A_i|). However, the bound given is the product of (2|A_i| + 1). Since m_i can be as large as needed, the product of (m_i - |A_i|) can be much larger than the product of (2|A_i| + 1). But the problem requires N to be less than the product of (2|A_i| + 1). So, even if there are many solutions modulo M, we need to find one that is small enough.This sounds like a problem that can be approached by the pigeonhole principle combined with the CRT. For example, if we can show that in the interval [1, product(2|A_i| + 1)], there are enough numbers to cover all possible residue combinations, thus forcing at least one of them to avoid all forbidden residues. Wait, but how?Alternatively, maybe use an inductive approach. Let's try to see for a small number of moduli, say 2, and see if the argument can be generalized.Suppose we have two moduli m_1 and m_2, coprime, with forbidden sets A_1 and A_2. The bound is (2|A_1| + 1)(2|A_2| + 1). We need to find N ≤ (2|A_1| + 1)(2|A_2| + 1) such that N ≡ not in A_1 mod m_1 and N ≡ not in A_2 mod m_2.How would we approach this? Let's say we fix residues for N modulo m_1 and m_2. The total number of possible allowed residue pairs is (m_1 - |A_1|)(m_2 - |A_2|). By the Chinese Remainder Theorem, each such pair corresponds to a unique residue modulo M = m_1 m_2. But we need N ≤ (2|A_1| + 1)(2|A_2| + 1). If (2|A_1| + 1)(2|A_2| + 1) ≥ M, then by pigeonhole principle, some N in [1, M] would work, but M could be larger. However, the problem allows N to be as large as the product of (2|A_i| + 1), which might be smaller than M.Wait, perhaps for each modulus m_i, we can consider intervals of numbers where we can "avoid" A_i. For example, in the case of one modulus m, if we have A with |A| elements, then in the first (2|A| + 1) numbers, there must be at least one number not congruent to any element in A modulo m. Is this true?Let me check with an example. Let m = 5, |A| = 2. Then 2|A| + 1 = 5. So numbers 1 to 5. If A = {1, 2}, then numbers 3, 4, 5 are allowed. So in 1-5, there are 3 allowed numbers. So yes, there exists an N ≤ 5. If A = {1, 3}, same thing. If A = {4, 2}, same.But wait, what if m = 3 and |A| = 1. Then 2|A| + 1 = 3. So numbers 1-3. If A = {1}, then allowed residues are 2, 3. So in 1-3, numbers 2 and 3 are allowed. So yes, N can be 2 or 3. If A = {2}, then allowed residues are 1, 3. So N can be 1 or 3. If A = {3}, but A is a subset of {1, 2}, so maybe A can't be {3}. Wait, the problem states A_i ⊆ {1, 2, ..., m_i - 1}. So residues are from 1 to m_i -1. So for m_i = 3, A_i can be subsets of {1, 2}. So if A_i is {1}, then 2|A_i| + 1 = 3, and N can be 2 or 3 in 1-3, but 3 mod 3 is 0, which is allowed since A_i only forbids 1. Wait, N can be 3, which is 0 mod 3, which isn't in A_i. So 3 is allowed. So in that case, N=3 would work.So in the single modulus case, the bound (2|A| + 1) works because in the first (2|A| + 1) numbers, there are |A| forbidden residues, each can block at most two numbers (since residues repeat every m_i), but with (2|A| + 1) numbers, there's at least one number not blocked. Wait, let's formalize this.Suppose we have modulus m, forbidden residues A ⊆ {1, ..., m-1}, |A| = t. Then in the numbers 1 to 2t + 1, how many are forbidden? Each forbidden residue a can appear at most floor((2t + 1)/m) + 1 times. But if m > 2t + 1, then each forbidden residue can appear at most once. So in 1 to 2t + 1, each forbidden residue occurs at most once, so total forbidden numbers are t. So in 2t + 1 numbers, there are t forbidden, so at least (2t + 1 - t) = t + 1 allowed numbers. Wait, but the problem says N must be a positive integer, so there is at least one allowed number. Wait, but if m_i is greater than 2t + 1, then the interval 1 to 2t + 1 contains numbers that are all distinct modulo m_i. So in that case, the forbidden residues block t numbers, so there are (2t + 1) - t = t + 1 allowed numbers.But if m_i is less than or equal to 2t + 1, then numbers start repeating modulo m_i. So in that case, the number of forbidden numbers could be more. Wait, but if m_i is less than 2t + 1, then t = |A_i| ≤ m_i - 1. But if m_i is small, say m_i = 2, then A_i can be {1}, so t = 1. Then 2t + 1 = 3. Numbers 1, 2, 3. Forbidden residue is 1 mod 2. So numbers 1 and 3 are odd, which are forbidden, but 2 is even, which is allowed. So N=2. Wait, but 2t + 1 = 3, which is larger than m_i = 2. So in this case, even though m_i is small, the bound still works.Wait, maybe the general idea is that for each modulus m_i, if we consider an interval of length 2|A_i| + 1, then no matter how the forbidden residues are arranged, there must be at least one number in that interval that is not forbidden modulo m_i. Then, by the Chinese Remainder Theorem, when we take the product over all moduli, we can construct such an N by combining these residues.But how to formalize this? Let's think of each modulus m_i separately. For each m_i, we can find a residue r_i not in A_i, and a number N_i ≤ 2|A_i| + 1 such that N_i ≡ r_i mod m_i. Then, using the CRT, we can find N ≡ r_i mod m_i for all i, and N ≤ product(2|A_i| + 1). But how do we ensure that such N exists within the bound?Wait, maybe use the idea of the simultaneous version of the Chinese Remainder Theorem. If for each modulus m_i, the desired residue r_i is known, then the solution N is unique modulo M = product(m_i). But we need N to be ≤ product(2|A_i| + 1). If product(2|A_i| + 1) ≥ M, then N exists in [1, M], but since M could be larger than the product, this might not hold. However, the problem allows N to be up to the product(2|A_i| + 1). So maybe even if M is larger, there is a solution N ≤ product(2|A_i| + 1).Alternatively, maybe we can model this as a lattice. The set of numbers N ≡ r_i mod m_i for each i forms an arithmetic progression. The problem is to find the smallest N in this progression. But if the progression has a common difference of M, the product of the m_i, then the minimal N could be as large as M, which might be much bigger than the product(2|A_i| + 1). So this approach might not work.Wait, perhaps instead of fixing residues first, we need to construct N in the interval [1, product(2|A_i| + 1)] such that for each i, N mod m_i is not in A_i. To show existence, we can use the probabilistic method or combinatorial counting. Let's see.The total number of possible N in [1, product(2|A_i| + 1)] is exactly the product(2|A_i| + 1). For each modulus m_i, the number of forbidden residues is |A_i|. So for each i, the probability that a random N is forbidden modulo m_i is |A_i| / m_i. But since the moduli are coprime, the events of being forbidden modulo different m_i are independent. Therefore, the probability that a random N is allowed modulo all m_i is product_{i=1}^{2013} (1 - |A_i| / m_i). If we can show that this probability is positive, then there exists such an N.But the problem is that the size of the interval is product(2|A_i| + 1), and we need to show that the number of allowed N is at least 1. So the expected number of allowed N in the interval is product(2|A_i| + 1) * product_{i=1}^{2013} (1 - |A_i| / m_i). If we can show that this expectation is at least 1, then there exists at least one such N.But how can we compute this expectation? Let's see. The expectation is product(2|A_i| + 1) * product(1 - |A_i| / m_i). We need to show that this is ≥ 1.But since m_i > 1 and |A_i| can be up to m_i - 1, but in general, we don't have a relation between |A_i| and m_i. However, the problem statement doesn't give any constraints on the size of A_i relative to m_i. So maybe this approach isn't directly applicable.Alternatively, maybe use a greedy approach. For each modulus m_i, we can choose a residue r_i not in A_i. Then, using the CRT, construct N ≡ r_i mod m_i. However, N could be as large as M, the product of the m_i, which is way larger than the given bound. So this doesn't help.Wait, but maybe for each m_i, instead of choosing any residue, we can choose a residue that corresponds to a small N. For example, using the concept of the "greedy algorithm" where we pick residues that minimize N. But the problem is that the moduli are independent, so minimizing N modulo one might conflict with another.Alternatively, use the idea of the sieve. Each modulus m_i "sieves out" certain numbers. The total number of numbers sieved out by all A_i would be the union over all forbidden arithmetic progressions. The challenge is to show that within the given bound, there's a number not sieved out.But how to bound the number of forbidden numbers in [1, product(2|A_i| + 1)]? For each modulus m_i, the number of forbidden residues is |A_i|. Each forbidden residue a in A_i corresponds to numbers ≡ a mod m_i. The number of such numbers in [1, S] is approximately S / m_i * |A_i|. But if S is product(2|A_j| + 1), then for each modulus m_i, the number of forbidden numbers is roughly |A_i| * product(2|A_j| + 1) / m_i.But the total number of forbidden numbers would be sum_{i=1}^{2013} |A_i| * product(2|A_j| + 1) / m_i. If we can show that this sum is less than product(2|A_j| + 1), then by the union bound, there exists some N not forbidden by any modulus. But this would require sum_{i=1}^{2013} |A_i| / m_i < 1. However, we don't have any constraints on |A_i| relative to m_i. For example, if |A_i| = m_i - 1 for some i, then |A_i| / m_i ≈ 1, and sum would exceed 1, making the union bound insufficient.Therefore, this approach might not work.Wait, but in the problem statement, the m_i are pairwise relatively prime. Maybe there's a way to use the multiplicative structure of the bound. The given bound is a product over (2|A_i| + 1). Let's note that for each modulus, 2|A_i| + 1 is the size of the set { -|A_i|, ..., 0, ..., |A_i| } if we think of residues as centered around 0. But not sure.Alternatively, think of each factor (2|A_i| + 1) as the number of possible "offsets" from some central number. For example, if we fix a number N, then for each modulus m_i, N can be in one of m_i residue classes. The forbidden ones are |A_i|, so allowed are m_i - |A_i|. If we can model the selection of N as choosing for each modulus an allowed residue, then the total number of combinations is product(m_i - |A_i|). But how does this relate to the given bound?Alternatively, think of it as a combinatorial cube. Each dimension corresponds to a modulus m_i, and we have to choose a coordinate in each dimension that's not in A_i. The total volume of the cube is product(m_i - |A_i|). If we can project this cube onto the integers and find a point within the given bound, then we have our N.But how to relate the product(m_i - |A_i|) with the product(2|A_i| + 1). If m_i - |A_i| ≥ 2|A_i| + 1, then the cube's volume is larger, but this is not necessarily the case. For example, if |A_i| = 1, then m_i - |A_i| = m_i - 1, which is at least 2(1) + 1 = 3 only if m_i ≥ 4. But m_i could be 3, so m_i - 1 = 2, which is less than 3. So this inequality doesn't hold in general.Alternatively, maybe the problem is using the concept of the inclusion-exclusion principle but in a more clever way. For each modulus, the forbidden residues block certain numbers, and we need to find a number that's not blocked by any. Since the moduli are coprime, the events of being blocked by different moduli are independent. So the total number of unblocked numbers up to X is roughly X * product(1 - |A_i| / m_i). If X is the product(2|A_i| + 1), then we need X * product(1 - |A_i| / m_i) ≥ 1. But again, this depends on the relation between |A_i| and m_i, which isn't given.Wait, but maybe for each modulus m_i, the term (2|A_i| + 1) is chosen such that even in the worst case where m_i is just slightly larger than 2|A_i| + 1, there's still an overlap. For example, if m_i = 2|A_i| + 2, then the interval 1 to 2|A_i| + 1 would contain exactly one complete residue system mod m_i, so by the pigeonhole principle, there's at least one number in that interval not in A_i. But this is similar to the earlier thought.Wait, let's formalize this. For each modulus m_i, consider the numbers 1 to 2|A_i| + 1. Each residue modulo m_i can appear at most floor((2|A_i| + 1)/m_i) times. If m_i > 2|A_i| + 1, then each residue appears at most once, so the number of forbidden residues in 1 to 2|A_i| + 1 is exactly |A_i|. Hence, there are (2|A_i| + 1) - |A_i| = |A_i| + 1 allowed numbers. So in this case, there exists an N_i ≤ 2|A_i| + 1 such that N_i is allowed modulo m_i.If m_i ≤ 2|A_i| + 1, then the numbers 1 to 2|A_i| + 1 contain multiple full cycles modulo m_i. Let's say the number of full cycles is k = floor((2|A_i| + 1)/m_i), and the remaining numbers are r = (2|A_i| + 1) mod m_i. Each full cycle has m_i numbers, out of which |A_i| are forbidden. So the total forbidden numbers would be k * |A_i| + min(r, |A_i|). The total allowed numbers would be (k * (m_i - |A_i|)) + max(0, r - |A_i|). Since we need at least one allowed number, this would require that even after k full cycles and the remaining r numbers, there's at least one allowed.But since 2|A_i| + 1 ≥ m_i, then k ≥ 1. Let's compute the total allowed numbers:Total numbers: 2|A_i| + 1Forbidden numbers: k * |A_i| + min(r, |A_i|)Allowed numbers: 2|A_i| + 1 - [k * |A_i| + min(r, |A_i|)]But since k = floor((2|A_i| + 1)/m_i), which is at least 1. Let me plug in m_i ≤ 2|A_i| + 1. Let's take m_i = 2|A_i| + 1 - t for some t ≥ 0. Not sure. Alternatively, take m_i ≤ 2|A_i| + 1, so k ≥ 1.Suppose m_i = 2|A_i| + 1. Then k = 1, r = 0. Forbidden numbers: 1 * |A_i| + 0 = |A_i|. Allowed numbers: (2|A_i| + 1) - |A_i| = |A_i| + 1 ≥ 1.If m_i is smaller, say m_i = |A_i| + 1 (the smallest possible m_i, since |A_i| ≤ m_i - 1). Then |A_i| = m_i - 1. Then 2|A_i| + 1 = 2(m_i - 1) + 1 = 2m_i - 1. So k = floor((2m_i - 1)/m_i) = 2, r = (2m_i - 1) - 2 * m_i = -1? Wait, no. Wait, 2m_i - 1 divided by m_i is 1 with remainder m_i - 1. So k = 2, r = m_i - 1. Forbidden numbers = 2 * |A_i| + min(m_i - 1, |A_i|). But |A_i| = m_i - 1, so min(m_i - 1, m_i - 1) = m_i -1. So forbidden numbers = 2(m_i -1) + (m_i -1) = 3(m_i -1). Allowed numbers = (2m_i -1) - 3(m_i -1) = 2m_i -1 -3m_i + 3 = -m_i + 2. But since m_i ≥ 2, this would be negative. Wait, that can't be. There's a mistake here.Wait, when m_i = |A_i| + 1, then |A_i| = m_i -1, so 2|A_i| +1 = 2(m_i -1) +1 = 2m_i -1. So the interval is 1 to 2m_i -1. The number of forbidden residues is m_i -1. Each forbidden residue a in A_i occurs floor((2m_i -1)/m_i) times, which is 2 times, since (2m_i -1)/m_i = 2 - 1/m_i, so floor is 2. Therefore, each forbidden residue occurs twice, except for the last one, which might occur once. Wait, no. The interval 1 to 2m_i -1 includes two full cycles of m_i: 1 to m_i and m_i +1 to 2m_i -1. Each residue from 1 to m_i -1 appears twice, and residue 0 (i.e., m_i) appears once (at m_i). But since A_i is a subset of {1, ..., m_i -1}, the forbidden residues are each appearing twice. So total forbidden numbers are 2(m_i -1). But the interval has 2m_i -1 numbers. So allowed numbers are (2m_i -1) - 2(m_i -1) = 2m_i -1 -2m_i + 2 = 1. So there's exactly one allowed number in this interval. Which is N = m_i, which is 0 mod m_i, but since A_i doesn't include 0, it's allowed. So even in this extreme case, there's exactly one allowed number. So the bound holds.Therefore, in all cases, for each modulus m_i, the interval [1, 2|A_i| +1] contains at least one number that is allowed modulo m_i. Then, using the Chinese Remainder Theorem, we can construct a number N ≤ product(2|A_i| +1) that is allowed modulo each m_i.Wait, how does the Chinese Remainder Theorem help here? If for each modulus m_i, there exists a number N_i ≤ 2|A_i| +1 such that N_i is allowed modulo m_i, then how do we combine these N_i into a single N that works for all moduli?But actually, the Chinese Remainder Theorem says that for any choice of residues r_i modulo m_i (pairwise coprime), there exists a unique N modulo M = product(m_i) such that N ≡ r_i mod m_i. But we need to choose residues r_i such that r_i is allowed (i.e., not in A_i) and such that the corresponding N is ≤ product(2|A_i| +1). However, the problem is that the residues r_i correspond to N_i in [1, 2|A_i| +1], but these N_i are specific to each modulus. How do we ensure that there's a common N that is ≤ product(2|A_i| +1) and satisfies N ≡ r_i mod m_i for all i, where each r_i is allowed?Perhaps use a product argument. For each modulus m_i, we have a set S_i of allowed residues, each corresponding to numbers in [1, 2|A_i| +1]. The size of S_i is at least 1 (from previous analysis). Now, consider all combinations of choosing one residue from each S_i. Each combination corresponds to a unique N modulo M by CRT. The total number of combinations is at least product(1) = 1. However, the number of distinct N in [1, product(2|A_i| +1)] is exactly product(2|A_i| +1). Since M could be much larger, but we need only one N in this interval that satisfies all the residue conditions.This seems like a counting argument. If we have product(2|A_i| +1) numbers in the interval [1, product(2|A_i| +1)], and each number is uniquely determined by its residues modulo each m_i (since the m_i are coprime), then each number corresponds to a tuple of residues (r_1, r_2, ..., r_{2013}), where each r_i is in [1, m_i]. However, we need tuples where each r_i is in S_i (the allowed residues for m_i). The total number of such tuples is product(|S_i|), which is at least product(1) = 1, but potentially more. However, how do we ensure that at least one such tuple corresponds to an N ≤ product(2|A_i| +1)?Wait, maybe use the fact that for each modulus m_i, the allowed residues in S_i can be shifted by multiples of m_i to lie within [1, 2|A_i| +1]. But I'm not sure.Alternatively, think of each allowed residue r_i for modulus m_i as being in [1, 2|A_i| +1]. Then, when constructing N via CRT, the minimal N that satisfies all congruences r_i mod m_i is at most the product of (2|A_i| +1). This is because of the following: in the CRT, the solution N can be written as sum_{i=1}^{n} r_i * M_i * y_i, where M_i = M / m_i, and y_i is the inverse of M_i modulo m_i. But this expression can be as large as M. However, we need N to be bounded by the product(2|A_i| +1).But this approach seems stuck. Maybe need a different perspective.Let me recall a theorem related to covering systems or the Chinese Remainder Theorem with multiplicities. Alternatively, consider that the problem is a direct application of the pigeonhole principle across multiple dimensions.Suppose for each i, we define the interval I_i = [1, 2|A_i| +1]. Then, the product of these intervals is a hypercube in 2013 dimensions, with each side length 2|A_i| +1. The volume of this hypercube is product(2|A_i| +1). Each forbidden residue a in A_i carves out a hyperplane in this hypercube. The total number of such hyperplanes is sum |A_i|, but since they're across different dimensions, their intersections are lower-dimensional.Wait, but how does this help? Perhaps not directly.Alternatively, think of each coordinate i as having 2|A_i| +1 possible positions, each corresponding to a number from 1 to 2|A_i| +1. For each i, the forbidden positions are the numbers in A_i modulo m_i. But since the numbers are up to 2|A_i| +1, which is larger than m_i in some cases, the forbidden positions could wrap around.But this is getting too vague. Let's try a different approach inspired by the one-dimensional case.In one dimension, we have m_i and A_i. We showed that in [1, 2|A_i| +1], there's at least one number not congruent to any a in A_i mod m_i. Suppose we can do this for each dimension, then use the Chinese Remainder Theorem to stitch them together.Wait, but how? For each modulus m_i, select a number N_i in [1, 2|A_i| +1] such that N_i ≡ r_i mod m_i, where r_i ∉ A_i. Then, use the CRT to find N ≡ r_i mod m_i for all i. However, such an N would be unique modulo M = product(m_i). But we need N ≤ product(2|A_i| +1). Since M could be larger than product(2|A_i| +1), how do we ensure that N is within the desired range?Ah, here's a key insight: the number N can be represented in a mixed-radix system where each digit corresponds to a residue modulo m_i, and the digits are chosen within the range [0, 2|A_i|]. This is similar to the standard CRT, but with digits constrained to certain ranges. If each digit (residue) is chosen within [0, 2|A_i|], then the maximal number represented in this system would be sum_{i=1}^{2013} (2|A_i|) * (product_{j=1}^{i-1} m_j). But this is more complicated.Alternatively, use induction on the number of moduli. For the base case, with one modulus, we've established that there's an N ≤ 2|A_1| +1 that works. Assume that for k moduli, there exists an N ≤ product_{i=1}^k (2|A_i| +1) that satisfies the conditions. Now, consider k + 1 moduli. By the induction hypothesis, there's an N' ≤ product_{i=1}^k (2|A_i| +1) that works for the first k moduli. Now, we need to find a number N ≡ N' mod M_k (where M_k is the product of the first k moduli) such that N ≡ r_{k+1} mod m_{k+1}, where r_{k+1} ∉ A_{k+1}, and N ≤ product_{i=1}^{k+1} (2|A_i| +1).Since M_k and m_{k+1} are coprime, the solution N is of the form N = N' + t * M_k for some integer t ≥ 0. We need to choose t such that N' + t * M_k ≡ r_{k+1} mod m_{k+1} and N ≤ product_{i=1}^{k+1} (2|A_i| +1).Solving for t: t ≡ (r_{k+1} - N') * M_k^{-1} mod m_{k+1}. Let t_min be the minimal non-negative solution for t. Then N = N' + t_min * M_k. We need to bound t_min such that N ≤ product_{i=1}^{k+1} (2|A_i| +1).But since product_{i=1}^{k+1} (2|A_i| +1) = product_{i=1}^k (2|A_i| +1) * (2|A_{k+1}| +1) = N' * (2|A_{k+1}| +1). Wait, no. The product up to k+1 is the product up to k multiplied by (2|A_{k+1}| +1). So if we can show that t_min ≤ (2|A_{k+1}| +1), then N = N' + t_min * M_k ≤ product_{i=1}^k (2|A_i| +1) + t_min * M_k. But this doesn't directly give the required bound.Alternatively, notice that t_min can be chosen such that t_min < m_{k+1}, because solutions repeat every m_{k+1}. Therefore, t_min ≤ m_{k+1} -1. So N = N' + t_min * M_k ≤ product_{i=1}^k (2|A_i| +1) + (m_{k+1} -1) * product_{i=1}^k (2|A_i| +1).But this is equal to product_{i=1}^k (2|A_i| +1) * m_{k+1}. However, the desired bound is product_{i=1}^{k+1} (2|A_i| +1). So unless m_{k+1} ≤ 2|A_{k+1}| +1, this doesn't hold. But m_{k+1} could be larger than 2|A_{k+1}| +1.Hmm, this suggests that the inductive approach may not work directly. Maybe there's another way.Wait, perhaps use the fact that for each modulus m_i, we can find a representative N_i in [1, 2|A_i| +1] such that N_i ≡ r_i mod m_i where r_i ∉ A_i. Then, consider the number N = sum_{i=1}^{2013} N_i * (M / m_i) * y_i mod M, where M is the product of all m_i, and y_i is the inverse of (M / m_i) modulo m_i. But this is the standard CRT construction, and N could be as large as M, which is problematic.Alternatively, consider a different numbering system. For example, instead of using the standard CRT, use a greedy algorithm where each digit is chosen within the allowed residues and bounded by 2|A_i| +1. If each digit is chosen within that bound, then the overall number would be bounded by the product.Wait, here's an idea inspired by positional numeral systems. Suppose we construct N such that for each modulus m_i, the coefficient of M_i = product_{j≠i} m_j is chosen from [0, 2|A_i|]. Then, N = sum_{i=1}^{2013} c_i * M_i, where 0 ≤ c_i ≤ 2|A_i|. The maximum N in this case would be sum_{i=1}^{2013} (2|A_i|) * M_i, which is much larger than the desired bound. Not helpful.Wait, but maybe if we use a mixed-radix system where each digit is bounded by (2|A_i| +1). Let me think. In mixed-radix systems, each digit's place value is the product of the previous moduli. If we set each digit d_i to be in [0, 2|A_i|], then the maximum number would be sum_{i=1}^{2013} (2|A_i|) * (product_{j=1}^{i-1} m_j). This might still be too large, but if the m_j are arranged such that each subsequent m_j is larger, maybe this sum is bounded by product(2|A_i| +1). Not sure.Alternatively, think of each modulus as being handled sequentially. Start with N_1 ≡ r_1 mod m_1, where r_1 is allowed and N_1 ≤ 2|A_1| +1. Then, for modulus m_2, find N_2 ≡ r_2 mod m_2 where r_2 is allowed and N_2 ≡ N_1 mod m_1, and N_2 ≤ (2|A_1| +1)(2|A_2| +1). Continue this process inductively.Let me try this with two moduli. Suppose we have m_1, m_2 coprime. Let N_1 ≤ 2|A_1| +1 be such that N_1 ≡ r_1 mod m_1, r_1 ∉ A_1. Then, we need to find N_2 ≡ r_1 mod m_1 and N_2 ≡ r_2 mod m_2, r_2 ∉ A_2, and N_2 ≤ (2|A_1| +1)(2|A_2| +1).By the CRT, the solutions to N_2 ≡ r_1 mod m_1 and N_2 ≡ r_2 mod m_2 are of the form N_2 = r_1 + k * m_1. We need to choose k such that N_2 ≡ r_2 mod m_2. Solving for k: k * m_1 ≡ r_2 - r_1 mod m_2. Since m_1 and m_2 are coprime, k ≡ (r_2 - r_1) * m_1^{-1} mod m_2. Let k = k_0 + t * m_2 for integer t. Then, N_2 = r_1 + (k_0 + t * m_2) * m_1 = r_1 + k_0 * m_1 + t * m_1 m_2. To minimize N_2, take t = 0. Then N_2 = r_1 + k_0 * m_1. Since k_0 is the minimal solution, k_0 ≤ m_2 -1. Therefore, N_2 ≤ r_1 + (m_2 -1) * m_1. But we need N_2 ≤ (2|A_1| +1)(2|A_2| +1). However, m_1 and m_2 can be much larger than 2|A_1| +1 and 2|A_2| +1, so this might not hold.But wait, we have r_1 ≤ 2|A_1| +1, and k_0 ≤ m_2 -1. So N_2 = r_1 + k_0 * m_1 ≤ (2|A_1| +1) + (m_2 -1) * m_1. But unless m_1 and m_2 are small, this could be larger than the product bound. For example, if m_1 = 3, |A_1| = 1, then 2|A_1| +1 = 3, r_1 ≤3. If m_2 = 5, |A_2| =2, then 2|A_2| +1=5. Then N_2 could be up to 3 + (5 -1)*3 = 3 + 12 =15, which is equal to 3*5=15, which matches the product bound. So in this case, it works.Wait, maybe in general, N_2 ≤ (2|A_1| +1) + (m_2 -1)*(2|A_1| +1) = (2|A_1| +1)*m_2. But (2|A_1| +1)(2|A_2| +1) could be less than (2|A_1| +1)*m_2 if m_2 > 2|A_2| +1. But in the earlier example where m_2 =5 and 2|A_2| +1=5, it's equal. So perhaps when m_2 is larger than 2|A_2| +1, the product (2|A_1| +1)(2|A_2| +1) is smaller than (2|A_1| +1)*m_2. Therefore, this approach doesn't ensure N_2 is within the desired bound.But in the example, it worked out. Maybe there's a pattern here. Suppose we have two moduli, m_1 and m_2, and the bound is (2|A_1| +1)(2|A_2| +1). If we find N_1 ≤ 2|A_1| +1 and then solve for N_2, which is N_1 + k * m_1, how large can k be?We need N_2 ≡ r_2 mod m_2, where r_2 ∉ A_2. The minimal k is determined by solving k ≡ (r_2 - N_1) * m_1^{-1} mod m_2. So k can be from 0 to m_2 -1. But N_2 = N_1 + k * m_1 ≤ (2|A_1| +1) + (m_2 -1)*m_1. But in the worst case, this could be (2|A_1| +1) + (m_2 -1)*m_1, which might be larger than (2|A_1| +1)(2|A_2| +1).Wait, unless we can choose r_2 such that k is bounded by (2|A_2| +1). Let me see. For modulus m_2, we need to choose r_2 ∉ A_2 and then find k such that N_2 = N_1 + k * m_1 ≡ r_2 mod m_2. So k ≡ (r_2 - N_1) * m_1^{-1} mod m_2. If we can choose r_2 such that the corresponding k is ≤ 2|A_2| +1, then N_2 = N_1 + k * m_1 ≤ (2|A_1| +1) + (2|A_2| +1)*m_1. But this might not be bounded by (2|A_1| +1)(2|A_2| +1) unless m_1 ≤ 2|A_1| +1. But m_1 can be larger.Wait, this is getting too tangled. Let's try to see if there's an alternative approach.Another angle: think of the problem as a generalization of the Chinese Remainder Theorem with forbidden residues. For each modulus, you have a set of allowed residues. The total number of allowed combinations is the product of (m_i - |A_i|). The question is whether within the first product(2|A_i| +1) numbers, there exists at least one number that is allowed modulo each m_i.This is similar to the hitting set problem, where we want a set that intersects all the forbidden residue classes. The size of the hitting set is product(2|A_i| +1), and we need to show that it contains at least one element not in any forbidden class.But how to prove this? Maybe use inclusion-exclusion. The number of forbidden numbers in [1, S] is sum_{i=1}^{2013} |A_i| * floor(S / m_i) - sum_{i < j} |A_i| |A_j| * floor(S / (m_i m_j)) + ... ± |A_1| ... |A_{2013}| * floor(S / (m_1 ... m_{2013}))|.But since S = product(2|A_i| +1), and the m_i are pairwise coprime, floor(S / (m_{i_1} ... m_{i_k})) ) = product_{i not in {i_1, ..., i_k}} (2|A_i| +1) / product_{i in {i_1, ..., i_k}} m_i. But this is complicated.Alternatively, consider that for each modulus m_i, the forbidden numbers occupy |A_i| residue classes. If we can show that the density of forbidden numbers in the interval [1, S] is less than 1, then there exists an allowed number. The density would be roughly sum_{i=1}^{2013} |A_i| / m_i. But since m_i are pairwise coprime and can be arbitrary, this sum could exceed 1.Alternatively, note that the multiplicative bound given is similar to the Lovász local lemma's condition. In the Lovász local lemma, if the probability of each bad event is bounded and their dependencies are limited, then there's a positive probability that no bad event occurs. However, in this problem, the bad events (being congruent to a forbidden residue) are independent across different moduli because the moduli are coprime. So the probability that a random N is allowed is product_{i=1}^{2013} (1 - |A_i| / m_i). The expected number of allowed N in [1, S] is S * product_{i=1}^{2013} (1 - |A_i| / m_i). If this expectation is at least 1, then there exists such an N.But we need to show that S = product(2|A_i| +1) makes this expectation at least 1. So we need product(2|A_i| +1) * product(1 - |A_i| / m_i) ≥ 1.Taking logarithms, this is equivalent to sum_{i=1}^{2013} [ln(2|A_i| +1) + ln(1 - |A_i| / m_i)] ≥ 0.But how to show this? For each term, ln(2|A_i| +1) + ln(1 - |A_i| / m_i) ≥ ?If we can show that (2|A_i| +1)(1 - |A_i| / m_i) ≥1 for each i, then the product would be ≥1, and hence the logarithm sum ≥0.So let's check if (2|A_i| +1)(1 - |A_i| / m_i) ≥1.Rearranging, (2|A_i| +1)(1 - |A_i| / m_i) ≥1.Let me denote t = |A_i|. Then, (2t +1)(1 - t/m_i) ≥1.We need to verify this inequality.Expand: (2t +1) - (2t +1)(t/m_i) ≥1=> (2t +1) - (2t^2 + t)/m_i ≥1=> (2t +1 -1) - (2t^2 + t)/m_i ≥0=> 2t - (2t^2 + t)/m_i ≥0Multiply both sides by m_i (positive):2t m_i - 2t^2 - t ≥0=> 2t m_i - 2t^2 - t = t(2m_i - 2t -1) ≥0Since t ≥0, we need 2m_i -2t -1 ≥0=> 2m_i ≥2t +1=> m_i ≥t + 0.5But m_i is an integer greater than 1, and t = |A_i| ≤ m_i -1 (since A_i ⊆ {1, ..., m_i -1}).So m_i ≥ t +1. Therefore, 2m_i ≥2t +2 >2t +1. Hence, 2m_i -2t -1 ≥1 >0.Therefore, t(2m_i -2t -1) ≥0, so the inequality holds.Therefore, (2t +1)(1 - t/m_i) ≥1.Hence, product_{i=1}^{2013} (2|A_i| +1)(1 - |A_i| / m_i) ≥ product_{i=1}^{2013} 1 =1.Therefore, the expectation E = product(2|A_i| +1) * product(1 - |A_i| / m_i) ≥1. Therefore, there exists at least one N in [1, product(2|A_i| +1)] that is allowed modulo each m_i.Thus, the proof is complete.Final Answerboxed{left( 2leftlvert A_1 rightrvert + 1 right)left( 2leftlvert A_2 rightrvert + 1 right)cdotsleft( 2leftlvert A_{2013} rightrvert + 1 right)}

✅:To prove the existence of a positive integer ( N ) such that ( N le left( 2leftlvert A_1 rightrvert + 1 right)left( 2leftlvert A_2 rightrvert + 1 right)cdotsleft( 2leftlvert A_{2013} rightrvert + 1 right) ) and for each ( i ), ( N ) is not congruent to any element of ( A_i ) modulo ( m_i ), we can use the following reasoning:1. Single Modulus Case: For each modulus ( m_i ), consider the interval ( [1, 2|A_i| + 1] ). Each forbidden residue ( a in A_i ) can appear at most once in this interval if ( m_i > 2|A_i| + 1 ). If ( m_i le 2|A_i| + 1 ), the interval contains multiple cycles modulo ( m_i ), but there will still be at least one number not in ( A_i ). Thus, there exists an ( N_i le 2|A_i| + 1 ) such that ( N_i notequiv a pmod{m_i} ) for any ( a in A_i ).2. Combining Moduli: Using the Chinese Remainder Theorem (CRT), we can combine the solutions for each modulus. For each ( i ), choose a residue ( r_i ) not in ( A_i ) such that ( r_i le 2|A_i| + 1 ). By CRT, there exists a unique ( N ) modulo ( M = prod_{i=1}^{2013} m_i ) that satisfies all these residues. However, we need ( N ) to be within the bound ( prod_{i=1}^{2013} (2|A_i| + 1) ).3. Expected Value Argument: The expected number of valid ( N ) in the interval ( [1, prod_{i=1}^{2013} (2|A_i| + 1)] ) is given by the product of the probabilities that ( N ) is not in ( A_i ) modulo ( m_i ). This expectation is: [ prod_{i=1}^{2013} (2|A_i| + 1) cdot prod_{i=1}^{2013} left(1 - frac{|A_i|}{m_i}right) ] We show that each term ( (2|A_i| + 1)(1 - frac{|A_i|}{m_i}) ge 1 ), leading to the overall expectation being at least 1, ensuring the existence of such ( N ).4. Inequality Verification: For each ( i ), let ( t = |A_i| ). We verify the inequality ( (2t + 1)(1 - frac{t}{m_i}) ge 1 ). This simplifies to ( 2t m_i - 2t^2 - t ge 0 ), which holds since ( m_i ge t + 1 ).Thus, by the probabilistic method and the Chinese Remainder Theorem, there exists a positive integer ( N ) within the given bound that avoids all forbidden residues.[boxed{left( 2leftlvert A_1 rightrvert + 1 right)left( 2leftlvert A_2 rightrvert + 1 right)cdotsleft( 2leftlvert A_{2013} rightrvert + 1 right)}]

🔑:1. Define the problem and notation: Let ( n = 2013 ) be an odd number. We are given ( 2013 ) pairwise relatively prime positive integers ( m_1, m_2, ldots, m_{2013} ) with ( m_i > 1 ) for all ( i ). Additionally, we have ( 2013 ) sets ( A_1, A_2, ldots, A_{2013} ) where ( A_i subseteq {1, 2, ldots, m_i - 1} ). We need to prove that there exists a positive integer ( N ) such that: [ N le left( 2|A_1| + 1 right)left( 2|A_2| + 1 right)cdotsleft( 2|A_{2013}| + 1 right) ] and for each ( i = 1, 2, ldots, 2013 ), there does not exist ( a in A_i ) such that ( m_i ) divides ( N - a ).2. Define bad numbers: Call a number ( k ) as *( i )-bad* if ( k ) is congruent to some value in ( A_i ) modulo ( m_i ). We aim to find the smallest ( N ) such that ( N + 1 ) is neither ( 1 )-bad, nor ( 2 )-bad, ..., nor ( n )-bad. This means each of ( 1, 2, ldots, N ) is at least one of ( 1 )-bad, or ( 2 )-bad, ..., or ( n )-bad.3. Use the Principle of Inclusion-Exclusion (PIE): Let ( S_i ) be the set of numbers in ( 1, ldots, N ) that are ( i )-bad. By PIE, we have: [ N = |S_1 cup S_2 cup cdots cup S_n| = sum |S_i| - sum |S_i cap S_j| + sum |S_i cap S_j cap S_k| - cdots ]4. Estimate the size of intersections: For any indices ( i_1, ldots, i_c ), we have: [ |S_{i_1} cap cdots cap S_{i_c}| in |A_{i_1}| cdot ldots cdot |A_{i_c}| cdot left[ leftlfloor frac{N}{m_{i_1} cdots m_{i_c}} rightrfloor, leftlfloor frac{N}{m_{i_1} cdots m_{i_c}} rightrfloor + 1 right] ] This follows from the Chinese Remainder Theorem (CRT).5. Upper bound ( N ): Using the bounds from the previous step, we get: [ N le sum left( frac{N}{m_i} + 1 right) |A_i| - sum left( frac{N}{m_i m_j} - 1 right) |A_i| cdot |A_j| + cdots ] Simplifying, we have: [ N le N left[ 1 + left( frac{|A_1|}{m_1} - 1 right) cdots left( frac{|A_n|}{m_n} - 1 right) right] + (|A_1| + 1) cdots (|A_n| + 1) - 1 ] Subtracting ( N ) from both sides and rearranging, we get: [ N le frac{(|A_1| + 1) cdots (|A_n| + 1) - 1}{left( 1 - frac{|A_1|}{m_1} right) cdots left( 1 - frac{|A_n|}{m_n} right)} ]6. Case analysis: - Case 1: ( 2|A_i| + 1 le m_i ) for all ( i ). Then: [ frac{|A_i| + 1}{1 - frac{|A_i|}{m_i}} le 2|A_i| + 1 ] Thus: [ N le (2|A_1| + 1) cdots (2|A_n| + 1) - 1 ] Hence, ( N + 1 le prod (2|A_i| + 1) ). - Case 2: ( 2|A_i| + 1 > m_i ) for some ( i ). Remove this ( i ) and induct down. Essentially, we can ignore the ( mod m_i ) condition and use the CRT to add the restriction that they are all non-( i )-bad modulo ( m_i ). This gives: [ N le N' m_i < N (2|A_i| + 1) ]The final answer is ( boxed{ N le left( 2|A_1| + 1 right)left( 2|A_2| + 1 right)cdotsleft( 2|A_{2013}| + 1 right) } )

🧠:Alright, so I need to prove that OP equals AB in a regular decagon where O is the center, A and B are neighboring vertices, and P is a point on OB such that OP squared equals OB times PB. Hmm, okay. Let me start by visualizing the regular decagon. A decagon has 10 sides, so each central angle should be 360/10 = 36 degrees. That means the angle between OA and OB is 36 degrees. Since it's a regular decagon, all sides are equal, so AB is the length of one side. OB is the radius of the circumscribed circle, right? So OB is equal to the radius, let's denote that as R. Then, OP is a segment from the center to some point P on OB, and we have the equation OP² = OB * PB. Since OB is R, then OP² = R * PB. We need to show that OP is equal to AB. Let me recall that in a regular polygon with n sides, the length of a side can be calculated using the formula 2R sin(π/n). For a decagon, n=10, so AB = 2R sin(π/10). Let me compute sin(π/10). π/10 is 18 degrees, and sin(18°) is (sqrt(5)-1)/4 multiplied by 2, right? Wait, sin(18°) is (sqrt(5)-1)/4 * 2? Let me check. Actually, sin(18°) is (sqrt(5)-1)/4 * 2? Wait, perhaps another approach. The exact value of sin(18°) is (sqrt(5)-1)/4 * 2. Let me verify:Yes, sin(18°) = (√5 - 1)/4 * 2, which simplifies to (√5 - 1)/2 * 1/2? Wait, perhaps I should recall that sin(18°) = (sqrt(5)-1)/4 * 2. Let me compute it numerically. 18 degrees, sine of 18 degrees is approximately 0.3090. If sqrt(5) is approximately 2.236, then sqrt(5)-1 is approximately 1.236. Divided by 4, that's about 0.309, which matches. So sin(18°) = (√5 - 1)/4 * 2? Wait, (√5 - 1)/4 ≈ 0.309, so actually, sin(18°) = (√5 - 1)/4 * 2? Wait, no. Wait, let me check the exact formula. Actually, sin(18°) = (√5 - 1)/4 * 2. Let me look it up. The exact value is (sqrt(5)-1)/4 * 2. Wait, no. Let me recall the formula for sin(18°). It can be derived using the identity that sin(18°) = cos(72°), and using the formula for cos(72°). Alternatively, from the pentagon, since 72° is related to the pentagon. Alternatively, using the formula for the sine of 18 degrees: Let’s set θ = 18°, then 5θ = 90°, so sin(5θ) = 1. Using the multiple-angle formula for sine:sin(5θ) = 16 sin^5θ - 20 sin^3θ + 5 sinθ = 1. Let x = sinθ. Then,16x^5 - 20x^3 + 5x - 1 = 0. Let me factor this equation. Maybe x=1 is a root. Plugging x=1: 16 - 20 + 5 -1 = 0. Yes, x=1 is a root. So factor out (x -1):Using polynomial division or synthetic division:Divide 16x^5 -20x^3 +5x -1 by (x -1). Let's set it up:Coefficients: 16, 0, -20, 0, 5, -1Divide by (x -1):1 | 16   0   -20   0   5   -1   |       16   16   -4   -4   1   -------------------------------     16  16   -4   -4   1   0So the quotient is 16x^4 +16x^3 -4x^2 -4x +1. Therefore,16x^5 -20x^3 +5x -1 = (x -1)(16x^4 +16x^3 -4x^2 -4x +1). Now, we need to factor the quartic. Let's try to factor it further.Let me set y = x². Then the quartic becomes 16x^4 +16x^3 -4x^2 -4x +1. Hmm, not a quadratic in y. Alternatively, perhaps grouping terms:16x^4 +16x^3 -4x^2 -4x +1= (16x^4 +16x^3) + (-4x^2 -4x) +1= 16x^3(x +1) -4x(x +1) +1Hmm, that's 16x^3(x +1) -4x(x +1) +1. Factor out 4x(x +1):= 4x(x +1)(4x^2) -4x(x +1) +1. Wait, no, that might not help. Alternatively, let me try to factor the quartic as a quadratic in x^2. Let me check.Wait, the quartic is 16x^4 +16x^3 -4x^2 -4x +1. Let me see if it can be factored as (ax^2 +bx +c)(dx^2 +ex +f). Let's attempt to factor it.Assume (ax^2 +bx +c)(dx^2 +ex +f) = 16x^4 +16x^3 -4x^2 -4x +1.Multiply out:adx^4 + (ae + bd)x^3 + (af + be + cd)x^2 + (bf + ce)x + cfSet coefficients equal:ad = 16ae + bd = 16af + be + cd = -4bf + ce = -4cf = 1Looking at the last equation, cf =1. So possible c and f are both 1 or -1. Let's try c=1, f=1.Then:ad=16. Let's choose a=4, d=4 (since 4*4=16). Then:ae + bd =16: 4e +4b=16 → e + b=4.Next, af + be + cd= -4: 4*1 + b e +4*1 = -4. So 4 + be +4 = -4 → be= -12.Also, bf + ce= -4: b*1 + e*1= -4 → b + e= -4.But earlier we have e + b=4, which conflicts with b + e= -4. Contradiction. So this choice doesn't work.Try a=8, d=2. Then ad=16. Then:ae + bd=8e +2b=16.af + be + cd=8*1 +b e +2*1=8 + be +2=10 + be= -4 → be= -14.bf + ce= b*1 + e*1= b + e= -4.So we have equations:8e +2b =16 → 4e + b=8be= -14b + e= -4From the third equation, e= -4 -b. Substitute into the first equation:4(-4 -b) +b=8 → -16 -4b +b=8 → -16 -3b=8 → -3b=24 → b= -8Then e= -4 - (-8)=4. Then check be= (-8)(4)= -32 ≠ -14. Not valid.Another possibility for c and f: c= -1, f= -1. Then cf=1. Let's try that.Then ad=16. Again, take a=4, d=4.ae + bd=4e +4b=16 → e + b=4.af + be + cd=4*(-1) +b e +4*(-1)= -4 + be -4= be -8= -4 → be=4.bf + ce= b*(-1) + e*(-1)= -b -e= -4 → b + e=4.So, we have e + b=4 and be=4. So solving, b and e are roots of x² -4x +4=0 → (x-2)^2=0. So b=e=2. So factors (4x² +2x -1)(4x² +2x -1). Wait, let's check:(4x² +2x -1)(4x² +2x -1)=16x^4 +8x^3 -4x² +8x^3 +4x² -2x -4x² -2x +1=16x^4 +16x^3 -4x² -4x +1. Yes! Perfect. So the quartic factors as (4x² +2x -1)^2. Therefore,16x^5 -20x^3 +5x -1 = (x -1)(4x² +2x -1)^2. Therefore, the roots are x=1 and roots of 4x² +2x -1=0. Solving 4x² +2x -1=0:x = [-2 ± sqrt(4 +16)] /8 = [-2 ± sqrt(20)]/8 = [-2 ± 2sqrt(5)]/8 = [-1 ± sqrt(5)]/4.Since sin(18°) is positive and approximately 0.3090, so x= [ -1 + sqrt(5) ] /4 ≈ (-1 + 2.236)/4 ≈1.236/4≈0.309. Yes. So sin(18°)= (sqrt(5)-1)/4 ≈0.309. Therefore, AB=2R sin(18°)=2R*(sqrt(5)-1)/4= R*(sqrt(5)-1)/2.So AB= R*(sqrt(5)-1)/2. So if we can compute OP and show that it equals this value, then we are done.Given that OP²=OB*PB. OB=R. Let’s denote OP=x. Then PB=OB - OP= R - x. Therefore, the equation is x²=R*(R - x). Rearranging: x² + Rx - R²=0. Solving this quadratic equation: x= [ -R ± sqrt(R² +4R²) ] /2= [ -R ± sqrt(5R²) ] /2= [ -R ± R*sqrt(5) ] /2. Since x is positive, we take the positive root: x= [ -R + R*sqrt(5) ] /2= R*(sqrt(5)-1)/2. Which is exactly the expression we have for AB. Therefore, OP=AB.Wait, that seems straightforward. So perhaps the key is to recognize that OP²=OB*PB leads to a quadratic equation, whose positive solution is OP=R*(sqrt(5)-1)/2, which is exactly the length of AB as calculated from the regular decagon.So the conclusion follows directly by solving the quadratic equation given by the condition OP²=OB*PB and recognizing that the positive solution matches the length of the side AB of the decagon.But let me verify once again. Let's recap:Given regular decagon with center O, radius R=OB. AB is a side, so AB=2R sin(π/10)=2R sin(18°)=2R*(sqrt(5)-1)/4=R*(sqrt(5)-1)/2.Point P is on OB such that OP²=OB*PB. Let OP=x, then PB=R -x. Thus x²=R(R -x) →x² +Rx -R²=0 →x=( -R ±sqrt(R² +4R² ))/2=( -R ±R√5)/2. Discarding negative solution, x=R(√5 -1)/2, which is equal to AB. Hence OP=AB.Yes, that seems correct. So the main steps are:1. Express AB in terms of R using the formula for the side length of a regular polygon: AB=2R sin(π/10).2. Use the given condition OP²=OB*PB to set up the quadratic equation in terms of OP.3. Solve the quadratic equation to find OP in terms of R.4. Show that the expression for OP matches that of AB.Therefore, the proof is complete.Final Answerboxed{OP = AB}

✅:Given a regular decagon with center ( O ) and neighboring vertices ( A ) and ( B ). We need to prove that ( OP = AB ) where ( P ) is a point on the line segment ( OB ) such that ( OP^2 = OB cdot PB ).1. Understanding the Regular Decagon: - A regular decagon has 10 sides, so each central angle is ( 36^circ ). - The radius ( OB ) is denoted as ( R ). - The side length ( AB ) of a regular decagon can be calculated using the formula ( 2R sin(18^circ) ).2. Calculating ( AB ): - The exact value of ( sin(18^circ) ) is ( frac{sqrt{5} - 1}{4} ). - Therefore, ( AB = 2R sin(18^circ) = 2R cdot frac{sqrt{5} - 1}{4} = R cdot frac{sqrt{5} - 1}{2} ).3. Setting Up the Equation for ( OP ): - Let ( OP = x ). Then ( PB = OB - OP = R - x ). - Given ( OP^2 = OB cdot PB ), substituting the values we get ( x^2 = R(R - x) ).4. Solving the Quadratic Equation: - Rearranging the equation: ( x^2 + Rx - R^2 = 0 ). - Solving this quadratic equation using the quadratic formula: ( x = frac{-R pm sqrt{R^2 + 4R^2}}{2} = frac{-R pm Rsqrt{5}}{2} ). - Taking the positive root (since ( x ) must be positive): ( x = frac{R(sqrt{5} - 1)}{2} ).5. Conclusion: - The value of ( OP ) is ( frac{R(sqrt{5} - 1)}{2} ), which matches the calculated value of ( AB ).Thus, we have shown that ( OP = AB ).[boxed{OP = AB}]

🔑:1. Define the variables and given conditions: Let ( OA = OB = r ) and ( AB = x ). We are given that ( angle OAB = 36^circ ) and ( angle OBA = 72^circ ). We need to prove that ( OP = AB ) given ( OP^2 = OB cdot PB ).2. Use the Law of Sines in (triangle OAB): [ frac{OA}{sin angle OBA} = frac{AB}{sin angle OAB} ] Substituting the known angles: [ frac{r}{sin 72^circ} = frac{x}{sin 36^circ} ]3. Simplify using known trigonometric values: [ sin 72^circ = 2 sin 36^circ cos 36^circ ] [ cos 36^circ = frac{sqrt{5} + 1}{4} ] [ sin 36^circ = frac{sqrt{10 - 2sqrt{5}}}{4} ] Therefore: [ sin 72^circ = 2 left( frac{sqrt{10 - 2sqrt{5}}}{4} right) left( frac{sqrt{5} + 1}{4} right) = frac{sqrt{10 - 2sqrt{5}} (sqrt{5} + 1)}{8} ]4. Simplify the ratio: [ frac{r}{frac{sqrt{10 - 2sqrt{5}} (sqrt{5} + 1)}{8}} = frac{x}{frac{sqrt{10 - 2sqrt{5}}}{4}} ] [ r = x cdot 2 cos 36^circ ] [ r = x cdot 2 cdot frac{sqrt{5} + 1}{4} = x cdot frac{sqrt{5} + 1}{2} ] This gives us: [ r = x cdot frac{sqrt{5} + 1}{2} quad text{(1)} ]5. Use the given condition ( OP^2 = OB cdot PB ): [ OP^2 = r(r - OP) ] [ OP^2 = r^2 - r cdot OP ] Rearrange into a quadratic equation: [ OP^2 + r cdot OP - r^2 = 0 ]6. Solve the quadratic equation: [ OP = frac{-r pm sqrt{r^2 + 4r^2}}{2} ] [ OP = frac{-r pm sqrt{5r^2}}{2} ] [ OP = frac{-r pm rsqrt{5}}{2} ] [ OP = r cdot frac{-1 pm sqrt{5}}{2} ] Since ( OP ) must be positive: [ OP = r cdot frac{sqrt{5} - 1}{2} ]7. Relate ( OP ) to ( AB ): From (1): [ r = x cdot frac{sqrt{5} + 1}{2} ] Substitute ( r ) into the expression for ( OP ): [ OP = x cdot frac{sqrt{5} + 1}{2} cdot frac{sqrt{5} - 1}{2} ] Simplify: [ OP = x cdot frac{(sqrt{5} + 1)(sqrt{5} - 1)}{4} ] [ OP = x cdot frac{5 - 1}{4} ] [ OP = x cdot 1 ] [ OP = x ]8. Conclusion: [ OP = AB ]The final answer is ( boxed{ OP = AB } )

🧠:Okay, let me try to figure out how to solve this geometry problem. So we have a parallelogram ABCD where AB is 1 unit long. Point K is on side AD such that KD equals 1. Also, angle ABK is 90 degrees, and angle DBK is 30 degrees. We need to find the length of AD. Hmm, let's start by drawing a diagram to visualize the problem.First, in a parallelogram, opposite sides are equal and parallel. So AB is equal to CD, and AD is equal to BC. Given that AB is 1, then CD is also 1. The sides AD and BC are the ones we need to find, right? Because AD is the side we're asked about.Point K is on AD with KD = 1. So since AD is the side we're trying to find, and KD is 1, that means AK must be AD - 1. So if we let AD = x, then AK = x - 1. That might be useful later.We also know that angle ABK is 90 degrees. So triangle ABK is a right-angled triangle at B. Wait, no, angle ABK is 90 degrees, so the right angle is at point B? Wait, hold on. Let me clarify.Angle ABK is 90 degrees. So vertex at B, with segments AB and BK forming the angle. So AB is one side, BK is the other side, and the angle between them is 90 degrees. Therefore, triangle ABK is right-angled at B. Similarly, angle DBK is 30 degrees. So angle at B between DB and BK is 30 degrees. So point D, B, and K are involved here.Let me sketch this mentally. Parallelogram ABCD: AB is 1 unit, so let's place point A at the origin (0,0) for simplicity, then point B would be at (1,0). Since it's a parallelogram, vectors AB and AD define the shape. Let’s denote vector AB as (1,0) and vector AD as (a,b), so point D would be at (a,b), and point C would be at (1 + a, b). Then, the coordinates:A: (0,0)B: (1,0)D: (a,b)C: (1 + a, b)Point K is on AD such that KD = 1. Since AD is from A(0,0) to D(a,b), then the length of AD is sqrt(a² + b²). So KD is the distance from K to D, which is 1. Let’s find coordinates of K. Since K is on AD, we can parametrize AD as t*(a,b), where t ranges from 0 to 1. So K would be at (ta, tb) for some t. Then the distance from K to D is sqrt[(a - ta)² + (b - tb)²] = sqrt[(a(1 - t))² + (b(1 - t))²] = (1 - t)*sqrt(a² + b²). This is equal to 1. So (1 - t)*sqrt(a² + b²) = 1. Therefore, 1 - t = 1/sqrt(a² + b²). So t = 1 - 1/sqrt(a² + b²). Therefore, coordinates of K are:K: (a*(1 - 1/sqrt(a² + b²)), b*(1 - 1/sqrt(a² + b²)))Hmm, that seems a bit complicated. Maybe there's a better coordinate system to use here. Let's think.Alternatively, since AB is 1, let's place AB along the x-axis with A at (0,0) and B at (1,0). Then, since ABCD is a parallelogram, D must be at some point (p, q), and C will be at (1 + p, q). Then AD is the vector from A to D, which is (p, q), so the length of AD is sqrt(p² + q²). Point K is on AD such that KD = 1. So K is 1 unit away from D along AD. Therefore, moving from D towards A by 1 unit. Since AD has length L = sqrt(p² + q²), the coordinates of K can be expressed as D - (unit vector in AD direction)*1.The unit vector in AD direction is (p, q)/L. Therefore, K's coordinates are (p, q) - (p/L, q/L) = (p(1 - 1/L), q(1 - 1/L)).Alternatively, since K is on AD, we can parametrize from A to D. Let’s say K divides AD such that AK = m and KD = 1. Therefore, AD = m + 1. Then, coordinates of K can be found by moving m units from A towards D. Since AD has length sqrt(p² + q²) = m + 1. Wait, but if we parametrize it this way, the coordinates of K would be (p*(m/(m+1)), q*(m/(m+1))). But since KD = 1, then the distance from K to D is sqrt[(p - p*(m/(m+1)))² + (q - q*(m/(m+1)))²] = sqrt[(p/(m+1))² + (q/(m+1))²] = (sqrt(p² + q²))/(m + 1) = (m + 1)/(m + 1) = 1. Wait, that's interesting. So this shows that KD is equal to (sqrt(p² + q²))/(m + 1) = 1. Therefore, sqrt(p² + q²) = m + 1. But sqrt(p² + q²) is AD, which is equal to m + 1. So this is consistent. Therefore, in this parametrization, m = AK, KD = 1, and AD = m + 1. Therefore, coordinates of K are (p*(m/(m + 1)), q*(m/(m + 1))).But maybe this is getting too algebraic. Let's see if we can use coordinate geometry here. Let's assign coordinates:A: (0,0)B: (1,0)D: (p, q)C: (1 + p, q)K: (kp, kq), where 0 ≤ k ≤ 1, since K is on AD.But wait, from earlier, since KD = 1, and AD is length m + 1, but perhaps the coordinates of K can be written as (p - (p/L), q - (q/L)) where L is the length of AD. Wait, but that would be D minus the unit vector in direction AD times 1. So yes, coordinates of K would be (p - p/L, q - q/L). So K is (p(1 - 1/L), q(1 - 1/L)).Alternatively, since moving from D towards A by 1 unit. So vector DA is (-p, -q). The unit vector in direction DA is (-p/L, -q/L). So moving from D by 1 unit in direction DA gives K: (p, q) + (-p/L, -q/L)*1 = (p - p/L, q - q/L). Same as before.Therefore, coordinates of K are (p(1 - 1/L), q(1 - 1/L)), where L = sqrt(p² + q²). So K is ( (p(L - 1))/L , (q(L - 1))/L ).But maybe this is too complicated. Let's see if we can use vectors or coordinate geometry with the given angles.Given that angle ABK is 90 degrees. So vectors BA and BK are perpendicular. Vector BA is A - B = (-1, 0). Vector BK is K - B. If we can write the coordinates of K in terms of p and q, then we can set their dot product to zero.Similarly, angle DBK is 30 degrees. So vectors BD and BK make a 30-degree angle. Vector BD is D - B = (p - 1, q - 0) = (p - 1, q). Vector BK is K - B = (kp - 1, kq - 0) = (kp - 1, kq). Wait, but earlier, we considered K's coordinates as (p(1 - 1/L), q(1 - 1/L)), so k = (1 - 1/L). Therefore, BK would be (p(1 - 1/L) - 1, q(1 - 1/L)).But perhaps this is getting too involved. Let me try to write down the coordinates step by step.Let me set up the coordinate system as follows:- Let point A be at (0,0).- Since AB is 1, point B is at (1,0).- Since ABCD is a parallelogram, the coordinates of D can be (a, b), so point C is (1 + a, b).- Point K is on AD such that KD = 1. Since AD is from (0,0) to (a, b), the length of AD is sqrt(a² + b²). Therefore, the distance from K to D is 1, so K is located 1 unit away from D towards A along AD. Therefore, the coordinates of K can be found by moving 1 unit from D towards A.The unit vector from D to A is (-a, -b)/sqrt(a² + b²). Therefore, moving 1 unit from D towards A gives K's coordinates as:K = D + (-a, -b)/sqrt(a² + b²) * 1 = (a - a/sqrt(a² + b²), b - b/sqrt(a² + b²)).Alternatively, since AD has length L = sqrt(a² + b²), then AK = L - 1, and KD = 1. Therefore, the coordinates of K can be expressed as:K = A + (AD vector normalized) * (L - 1) = (a/L, b/L) * (L - 1) = (a(L - 1)/L, b(L - 1)/L) = (a(1 - 1/L), b(1 - 1/L)).So coordinates of K are (a(1 - 1/L), b(1 - 1/L)) where L = sqrt(a² + b²). So that's K's position.Now, angle ABK is 90 degrees. Let's express vectors BA and BK.Wait, angle at B: angle ABK is 90 degrees. So the angle between BA and BK is 90 degrees. Vector BA is from B to A, which is (-1, 0). Vector BK is from B to K, which is (a(1 - 1/L) - 1, b(1 - 1/L)).Therefore, the dot product of BA and BK should be zero.Dot product BA · BK = (-1)(a(1 - 1/L) - 1) + 0*(b(1 - 1/L)) = -a(1 - 1/L) + 1 = 0.Therefore:-a(1 - 1/L) + 1 = 0.Solving for this:- a + a/L + 1 = 0.=> -a + a/L = -1.=> a(-1 + 1/L) = -1.=> a(1 - 1/L) = 1.So:a = 1 / (1 - 1/L) = L / (L - 1).But L = sqrt(a² + b²). So substituting L:a = sqrt(a² + b²) / (sqrt(a² + b²) - 1).Hmm, this seems complicated. Let's keep this equation as equation (1).Now, let's consider the second condition: angle DBK is 30 degrees.Angle DBK is the angle at B between points D, B, and K. So vectors BD and BK make a 30-degree angle.Vector BD is from B to D: (a - 1, b).Vector BK is from B to K: (a(1 - 1/L) - 1, b(1 - 1/L)).The angle between BD and BK is 30 degrees. Therefore, the cosine of the angle between them is cos(30°) = √3/2.So, the dot product of BD and BK divided by the product of their magnitudes is √3/2.Let's compute BD · BK:= (a - 1)(a(1 - 1/L) - 1) + b * b(1 - 1/L)First, expand the terms:First term: (a - 1)(a(1 - 1/L) - 1)= (a - 1)[a - a/L - 1]= (a - 1)(a - 1 - a/L)Second term: b * b(1 - 1/L) = b²(1 - 1/L)So BD · BK = (a - 1)(a - 1 - a/L) + b²(1 - 1/L)Then, the magnitude of BD is sqrt((a - 1)^2 + b^2).The magnitude of BK is sqrt[(a(1 - 1/L) - 1)^2 + (b(1 - 1/L))^2].Therefore:[ (a - 1)(a - 1 - a/L) + b²(1 - 1/L) ] / [ sqrt((a - 1)^2 + b^2) * sqrt((a(1 - 1/L) - 1)^2 + (b(1 - 1/L))^2) ) ] = √3/2.This seems very complicated. Maybe there's a way to simplify this using equation (1) from earlier.From equation (1):a = L / (L - 1)Where L = sqrt(a² + b²).So let's let L = AD, which is the length we need to find.Therefore, L = sqrt(a² + b²). From equation (1):a = L / (L - 1)So a = L / (L - 1). Let's square both sides:a² = L² / (L - 1)^2Then, since L² = a² + b²,a² + b² = L²Therefore,b² = L² - a² = L² - L²/(L - 1)^2 = L²[1 - 1/(L - 1)^2] = L²[( (L - 1)^2 - 1 ) / (L - 1)^2 ) ]Simplify numerator:(L - 1)^2 - 1 = L² - 2L + 1 - 1 = L² - 2LTherefore,b² = L²*(L² - 2L)/( (L - 1)^2 ) = L³(L - 2)/( (L - 1)^2 )Wait, let me check that again.Wait, numerator: L² - 2L = L(L - 2)Denominator: (L - 1)^2Therefore,b² = L² * [ L(L - 2) / (L - 1)^2 ) ] = L³(L - 2)/(L - 1)^2Hmm, that seems complicated, but maybe manageable.So, we have expressions for a and b in terms of L. Let's try to substitute these into the second equation involving the 30-degree angle.First, let's recall that BD · BK is:(a - 1)(a - 1 - a/L) + b²(1 - 1/L)We can substitute a = L/(L - 1) and b² = L³(L - 2)/(L - 1)^2.First, compute (a - 1):a - 1 = L/(L - 1) - 1 = [L - (L - 1)]/(L - 1) = 1/(L - 1)Next, compute (a - 1 - a/L):a - 1 - a/L = [L/(L - 1) - 1] - [L/(L - 1)]/LFirst term: L/(L - 1) - 1 = 1/(L - 1)Second term: [L/(L - 1)]/L = 1/(L - 1)Therefore,a - 1 - a/L = 1/(L - 1) - 1/(L - 1) = 0Wait, that can't be. Wait, let me check again.Wait:a - 1 - a/L = [L/(L - 1) - 1] - [L/(L - 1)]/LFirst part: L/(L - 1) - 1 = (L - (L - 1))/(L - 1) = 1/(L - 1)Second part: [L/(L - 1)]/L = 1/(L - 1)Therefore,a - 1 - a/L = 1/(L - 1) - 1/(L - 1) = 0So the first term in BD · BK is (a - 1)(a - 1 - a/L) = (1/(L - 1))(0) = 0Therefore, BD · BK = 0 + b²(1 - 1/L) = b²(1 - 1/L)So BD · BK = b²(1 - 1/L)We can substitute b² from earlier:b²(1 - 1/L) = [L³(L - 2)/(L - 1)^2 ]*(1 - 1/L)Simplify 1 - 1/L = (L - 1)/LTherefore,b²(1 - 1/L) = [L³(L - 2)/(L - 1)^2 ] * [ (L - 1)/L ] = [L³(L - 2)(L - 1)] / [L(L - 1)^2] ] = [L²(L - 2)] / (L - 1)So BD · BK = L²(L - 2)/(L - 1)Now, the magnitude of BD is sqrt((a - 1)^2 + b²)From earlier, (a - 1)^2 = (1/(L - 1))² = 1/(L - 1)^2And b² = L³(L - 2)/(L - 1)^2Therefore,|BD| = sqrt[ 1/(L - 1)^2 + L³(L - 2)/(L - 1)^2 ) ]= sqrt[ (1 + L³(L - 2)) / (L - 1)^2 ) ]= sqrt[ (1 + L^4 - 2L³ ) / (L - 1)^2 ) ]Factor numerator:1 + L^4 - 2L³ = L^4 - 2L³ + 1This seems like a quartic. Maybe it factors?Let me try to factor L^4 - 2L³ + 1.Try rational roots. Possible roots are ±1.Testing L=1: 1 - 2 + 1 = 0. So (L - 1) is a factor.Divide L^4 - 2L³ + 1 by (L - 1):Using polynomial long division:L^4 - 2L³ + 0L² + 0L + 1 divided by L - 1.First term: L^4 / L = L^3. Multiply (L - 1) by L^3: L^4 - L^3Subtract from dividend: (-2L³ - (-L³)) = -L³ + 0L² + 0L +1Next term: -L³ / L = -L². Multiply (L -1) by -L²: -L³ + L²Subtract: (-L³ - (-L³)) + 0L² - L² + 0L +1 = -L² +0L +1Next term: -L² / L = -L. Multiply (L -1) by -L: -L² + LSubtract: (-L² - (-L²)) +0L +1 - L = -L +1Next term: -L / L = -1. Multiply (L -1) by -1: -L +1Subtract: (-L +1 - (-L +1)) = 0Therefore, L^4 -2L³ +1 = (L -1)(L³ - L² - L -1)Now, check if L³ - L² - L -1 factors further.Try L=1: 1 -1 -1 -1 = -2 ≠0L= -1: -1 -1 +1 -1 = -2 ≠0So no rational roots. Maybe it factors into quadratics.Assume (L³ - L² - L -1) = (L^2 + aL + b)(L + c)Multiply out:L^3 + (a + c)L² + (b + ac)L + bcSet equal to L³ - L² - L -1So,a + c = -1b + ac = -1bc = -1From bc = -1, possible integer solutions: b=1, c=-1 or b=-1, c=1.Try b=1, c=-1:Then a + (-1) = -1 => a =0Then b + a c =1 +0*(-1)=1≠-1. Doesn't work.Try b=-1, c=1:Then a +1 = -1 => a =-2Then b + a c =-1 + (-2)(1)=-3≠-1. Doesn't work.Therefore, doesn't factor further with integer coefficients. So, the quartic factors as (L -1)(L³ - L² - L -1). So perhaps we can't factor it further, so we need another approach.But let's get back to |BD|:|BD| = sqrt[ (L^4 -2L³ +1)/(L -1)^2 ) ] = sqrt[ (L -1)(L³ - L² - L -1) ) / (L -1)^2 ) ]Simplify:sqrt[ (L³ - L² - L -1) / (L -1) ) ]So |BD| = sqrt( (L³ - L² - L -1)/(L -1) )Hmm, this is getting really messy. Maybe there's a better way.Alternatively, let's compute |BD| using coordinates.BD vector is (a - 1, b). So |BD| = sqrt( (a -1)^2 + b^2 )From earlier, a = L/(L -1), and b² = L³(L - 2)/(L -1)^2. So let's plug these in.(a -1)^2 + b² = [ L/(L -1) -1 ]^2 + L³(L -2)/(L -1)^2= [ (L - (L -1))/(L -1) ) ]^2 + L³(L -2)/(L -1)^2= [ 1/(L -1) ) ]^2 + L³(L -2)/(L -1)^2= [1 + L³(L -2) ] / (L -1)^2Which is the same as before. So back to the same expression.Similarly, |BK| can be computed.Vector BK is (a(1 - 1/L) -1, b(1 -1/L))So |BK| = sqrt[ (a(1 -1/L) -1)^2 + (b(1 -1/L))^2 ]From earlier, a = L/(L -1), so:a(1 -1/L) = L/(L -1)*(1 -1/L) = L/(L -1)*( (L -1)/L ) = 1Therefore, a(1 -1/L) -1 = 1 -1 =0Similarly, b(1 -1/L) = b*( (L -1)/L )But from earlier, b² = L³(L -2)/(L -1)^2So |BK| = sqrt[ 0^2 + (b*(L -1)/L )^2 ] = |b|*(L -1)/LSince lengths are positive, |BK| = b*(L -1)/LBut b² = L³(L -2)/(L -1)^2 => b = sqrt( L³(L -2)/(L -1)^2 ) ) = L*sqrt( L(L -2) )/(L -1 )But since b is a coordinate, it can be positive or negative, but since we're dealing with lengths, we can take positive value.Therefore, |BK| = [ L*sqrt( L(L -2) )/(L -1 ) ] * (L -1)/L ) = sqrt( L(L -2) )Therefore, |BK| = sqrt( L(L -2) )So now, putting it all together:BD · BK = L²(L -2)/(L -1)|BD| = sqrt( (L^4 -2L³ +1)/(L -1)^2 ) ) = sqrt( (L -1)(L³ -L² -L -1) )/(L -1) ) = sqrt( (L³ -L² -L -1)/(L -1) )Wait, no:Wait, earlier, we had:|BD| = sqrt[ (L^4 -2L³ +1)/(L -1)^2 ) ]But L^4 -2L³ +1 = (L -1)(L³ - L² - L -1)So |BD| = sqrt[ (L -1)(L³ - L² - L -1) ) / (L -1)^2 ) ] = sqrt[ (L³ - L² - L -1)/(L -1) ) ]So |BD| = sqrt( (L³ - L² - L -1)/(L -1) )But this seems complicated. However, |BK| we found to be sqrt( L(L -2) )Therefore, the cosine of the angle between BD and BK is:[ BD · BK ] / ( |BD| |BK| ) ) = [ L²(L -2)/(L -1) ) ] / [ sqrt( (L³ - L² - L -1)/(L -1) ) * sqrt( L(L -2) ) ) ]Simplify denominator:sqrt( (L³ - L² - L -1)/(L -1) ) * sqrt( L(L -2) ) = sqrt( [ (L³ - L² - L -1)/(L -1) ) * L(L -2) ) ]Therefore, entire expression becomes:[ L²(L -2)/(L -1) ) ] / sqrt( [ (L³ - L² - L -1) * L(L -2) ) / (L -1) ) ] )= [ L²(L -2)/(L -1) ) ] / [ sqrt( L(L -2)(L³ - L² - L -1)/(L -1) ) ) ]= [ L²(L -2) / (L -1) ) ] / [ sqrt( L(L -2) ) * sqrt( (L³ - L² - L -1)/(L -1) ) ) ]= [ L²(L -2) / (L -1) ) ] / [ sqrt(L(L -2)) * sqrt( (L³ - L² - L -1)/(L -1) ) ) ]= [ L²(L -2) / (L -1) ) ] / [ sqrt(L(L -2)) * sqrt( (L³ - L² - L -1)/(L -1) ) ) ]Let's separate terms:= [ L²(L -2) / (L -1) ) ] / [ sqrt(L(L -2)) * sqrt( (L³ - L² - L -1)/(L -1) ) ) ]= [ L²(L -2) / (L -1) ) ] / [ sqrt(L(L -2)) * sqrt( (L³ - L² - L -1) ) / sqrt(L -1) ) ]= [ L²(L -2) / (L -1) ) ] * [ sqrt(L -1) / ( sqrt(L(L -2)) * sqrt( L³ - L² - L -1 ) ) ) ]= [ L²(L -2) * sqrt(L -1) ) ] / [ (L -1) * sqrt(L(L -2)) * sqrt( L³ - L² - L -1 ) ) ]Simplify numerator and denominator:Numerator: L²(L -2) sqrt(L -1)Denominator: (L -1) sqrt(L(L -2)) sqrt(L³ - L² - L -1)Cancel out sqrt(L -1) in numerator and denominator:Numerator: L²(L -2) / sqrt(L -1)Denominator: sqrt(L(L -2)) sqrt(L³ - L² - L -1 )Combine denominators:sqrt(L(L -2)) * sqrt(L³ - L² - L -1) * sqrt(L -1)But this seems getting too complicated. Maybe square both sides?Since the cosine is √3/2, we can square both sides:[ (BD · BK)^2 ] / [ |BD|^2 |BK|^2 ) ] = 3/4So:[ (L²(L -2)/(L -1))^2 ] / [ |BD|^2 |BK|^2 ) ] = 3/4We have |BD|^2 = (L^4 -2L³ +1)/(L -1)^2 = (L -1)(L³ - L² - L -1)/(L -1)^2 ) = (L³ - L² - L -1)/(L -1)And |BK|^2 = L(L -2)Therefore,[ L^4(L -2)^2/(L -1)^2 ) ] / [ ( (L³ - L² - L -1)/(L -1) ) * L(L -2) ) ] = 3/4Simplify denominator:( (L³ - L² - L -1)/(L -1) ) * L(L -2) ) = L(L -2)(L³ - L² - L -1)/(L -1)Therefore, the entire fraction becomes:[ L^4(L -2)^2/(L -1)^2 ) ] / [ L(L -2)(L³ - L² - L -1)/(L -1) ) ] = [ L^4(L -2)^2/(L -1)^2 ) ] * [ (L -1)/( L(L -2)(L³ - L² - L -1) ) ) ]Simplify:= [ L^4(L -2)^2 (L -1) ) / (L -1)^2 L(L -2)(L³ - L² - L -1) ) ]= [ L^4(L -2)^2 (L -1) ) / ( L(L -2)(L -1)^2 (L³ - L² - L -1) ) ) ]Cancel terms:L^4 / L = L^3(L -2)^2 / (L -2) = (L -2)(L -1) / (L -1)^2 = 1/(L -1)Thus,= L^3 (L -2) / [ (L -1) (L³ - L² - L -1) ) ]Therefore:L^3 (L -2) / [ (L -1)(L³ - L² - L -1) ) ] = 3/4So:4 L^3 (L -2) = 3 (L -1)(L³ - L² - L -1 )Expand both sides:Left side: 4L^4 - 8L^3Right side: 3(L -1)(L³ - L² - L -1)First, expand (L -1)(L³ - L² - L -1):Multiply term by term:L*(L³ - L² - L -1) = L^4 - L³ - L² - L-1*(L³ - L² - L -1) = -L³ + L² + L +1Combine:L^4 - L³ - L² - L - L³ + L² + L +1= L^4 - 2L³ + 0L² +0L +1= L^4 - 2L³ +1Therefore, right side: 3(L^4 - 2L³ +1)Therefore, equation becomes:4L^4 -8L^3 = 3L^4 -6L³ +3Bring all terms to left side:4L^4 -8L^3 -3L^4 +6L³ -3 =0Simplify:(4L^4 -3L^4) + (-8L³ +6L³) -3 =0L^4 -2L³ -3 =0So equation: L^4 -2L³ -3 =0Now, we need to solve L^4 -2L³ -3 =0Looking for real positive roots since L = AD > KD =1, so L >1.Try L=2: 16 -16 -3 = -3 ≠0L=3: 81 -54 -3=24≠0L=1: 1 -2 -3=-4≠0Hmm, maybe rational roots? Possible rational roots are ±1, ±3.Tested L=1: -4, L=3:24, L=-1:1 +2 -3=0. Wait, L=-1: (-1)^4 -2*(-1)^3 -3=1 +2 -3=0. So L=-1 is a root. So factor (L +1).Divide polynomial L^4 -2L³ -3 by (L +1).Using polynomial division:L^4 -2L³ +0L² +0L -3 divided by L +1.First term: L^4 / L = L^3. Multiply (L +1) by L^3: L^4 + L^3Subtract: (-2L³ - L^3) = -3L³ +0L² +0L -3Next term: -3L³ / L = -3L². Multiply (L +1) by -3L²: -3L³ -3L²Subtract: (-3L³ - (-3L³)) + 0L² - (-3L²) +0L -3 = 3L² +0L -3Next term: 3L² / L =3L. Multiply (L +1) by 3L: 3L² +3LSubtract: (3L² -3L²) +0L -3 -3L = -3L -3Next term: -3L / L = -3. Multiply (L +1) by -3: -3L -3Subtract: (-3L - (-3L)) + (-3 - (-3)) =0Therefore, L^4 -2L³ -3 = (L +1)(L³ -3L² +3L -3 )Now, factor the cubic L³ -3L² +3L -3.Try rational roots: possible roots ±1, ±3.Test L=1:1 -3 +3 -3=-2≠0L=3:27 -27 +9 -3=6≠0L= sqrt something? Maybe use rational root theorem didn't work. Let's try to see if it factors.Alternatively, apply the rational root theorem for cubic. If it factors, it would have a linear term times quadratic. Since we couldn't find a rational root, perhaps it's irreducible over rationals. Therefore, we have:L^4 -2L³ -3 = (L +1)(L³ -3L² +3L -3 )Thus, the real roots are L = -1 and roots of L³ -3L² +3L -3 =0.Since L is length AD >1, we discard L=-1. So solve L³ -3L² +3L -3=0.Let’s attempt to find real roots for L³ -3L² +3L -3=0.Let me denote f(L) = L³ -3L² +3L -3Compute f(1):1 -3 +3 -3= -2f(2):8 -12 +6 -3= -1f(3):27 -27 +9 -3=6So there's a root between 2 and3.Use Newton-Raphson method:Let’s start with L=2.5:f(2.5)=15.625 -18.75 +7.5 -3=15.625+7.5=23.125 -18.75=4.375 -3=1.375f(2.5)=1.375f'(L)=3L² -6L +3At L=2.5, f'(2.5)=3*(6.25) -6*(2.5) +3=18.75 -15 +3=6.75Next approximation: L=2.5 - f(2.5)/f'(2.5)=2.5 -1.375/6.75≈2.5 -0.2037≈2.2963Compute f(2.2963):L=2.2963^3 -3*(2.2963)^2 +3*(2.2963) -3≈(2.2963)^3≈12.0(?), let's compute:2.2963^3 = (2 +0.2963)^3≈8 +3*4*0.2963 +3*2*(0.2963)^2 + (0.2963)^3≈8 + 3*4*0.2963=8 +3.5556≈11.5556 + 3*2*0.0878≈11.5556 +0.5268≈12.0824 +0.026≈12.1084Then, -3*(2.2963)^2≈-3*(5.273)≈-15.819+3*2.2963≈6.8889-3≈-3Total≈12.1084 -15.819 +6.8889 -3≈(12.1084 +6.8889) - (15.819 +3)≈18.9973 -18.819≈0.1783f(2.2963)≈0.1783f'(2.2963)=3*(2.2963)^2 -6*(2.2963)+3≈3*(5.273) -13.778 +3≈15.819 -13.778 +3≈5.041Next approximation: 2.2963 -0.1783/5.041≈2.2963 -0.0354≈2.2609Compute f(2.2609):2.2609^3≈2.2609*2.2609*2.2609≈2.2609*5.1127≈11.56-3*(2.2609)^2≈-3*(5.1127)≈-15.338+3*2.2609≈6.7827-3≈-3Total≈11.56 -15.338 +6.7827 -3≈(11.56 +6.7827) - (15.338 +3)≈18.3427 -18.338≈0.0047Close to zero. f(2.2609)=≈0.0047f'(2.2609)=3*(2.2609)^2 -6*2.2609 +3≈3*5.1127 -13.5654 +3≈15.338 -13.5654 +3≈4.7726Next approximation:2.2609 -0.0047/4.7726≈2.2609 -0.001≈2.2599Compute f(2.2599)=≈0. Hence, real root is approximately 2.26.But we need an exact value. Wait, the equation L³ -3L² +3L -3=0. Let’s check if it can be written as (L -1)^3 -2=0.(L -1)^3 = L³ -3L² +3L -1. So if we subtract 2, (L -1)^3 -2 = L³ -3L² +3L -1 -2 = L³ -3L² +3L -3. Yes! So the equation is (L -1)^3 = 2.Therefore, L -1 = cube root of 2 => L =1 + cube root of 2.But cube root of 2 is 2^{1/3}, so L =1 + 2^{1/3}. Is that correct?Wait, let's verify:(L -1)^3 =2 => L -1 = 2^{1/3} => L=1 +2^{1/3}Yes, correct. So the real root is L=1 + ∛2.Therefore, the length AD is 1 + cube root of 2. So the answer is 1 + ∛2, which can be written as boxed{1 + sqrt[3]{2}}.Final AnswerThe length of ( AD ) is (boxed{1 + sqrt[3]{2}}).

✅:Given a parallelogram (ABCD) with (AB = 1). Point (K) is on side (AD) such that (KD = 1), (angle ABK = 90^circ), and (angle DBK = 30^circ). We need to determine the length of (AD).1. Coordinate System Setup: - Place (A) at ((0,0)) and (B) at ((1,0)). - Let (D) be at ((a, b)) and (C) at ((1 + a, b)).2. Point (K) on (AD): - Since (KD = 1), (K) is 1 unit away from (D) along (AD). - The coordinates of (K) are (left(a left(1 - frac{1}{L}right), b left(1 - frac{1}{L}right)right)) where (L = sqrt{a^2 + b^2}) is the length of (AD).3. Right Angle at (B): - (angle ABK = 90^circ) implies vectors (BA) and (BK) are perpendicular. - Dot product of (BA = (-1, 0)) and (BK = left(a left(1 - frac{1}{L}right) - 1, b left(1 - frac{1}{L}right)right)) must be zero. - Solving this gives (a = frac{L}{L - 1}).4. 30-Degree Angle at (B): - (angle DBK = 30^circ) implies the cosine of the angle between vectors (BD) and (BK) is (frac{sqrt{3}}{2}). - Using the dot product and magnitudes of vectors (BD) and (BK), we derive the equation (L^4 - 2L^3 - 3 = 0).5. Solving the Quartic Equation: - Factoring the equation (L^4 - 2L^3 - 3 = 0) gives roots (L = -1) and (L = 1 + sqrt[3]{2}). - Since (L) must be positive and greater than 1, the valid solution is (L = 1 + sqrt[3]{2}).Thus, the length of (AD) is (boxed{1 + sqrt[3]{2}}).

🔑:1. Define Variables and Setup: Let ( AK = x ), ( BD = y ), and ( BK = z ). We know that ( AB = 1 ), ( KD = 1 ), and ( angle ABK = 90^circ ) and ( angle DBK = 30^circ ).2. Identify Key Points and Relationships: Since ( angle DBK = 30^circ ) and ( angle ABK = 90^circ ), we can use trigonometric relationships and properties of the parallelogram to find the lengths of the sides.3. Use Perpendicularity and Similar Triangles: Let ( E ) be the point where ( CD ) intersects ( BK ). Since ( angle DBE = 30^circ ) and ( BE perp CD ), we have ( DE = frac{y}{2} ).4. Apply Similar Triangles: From the similarity of triangles ( Delta DEK sim Delta ABK ), we have: [ frac{DE}{AB} = frac{DK}{AK} ] Substituting the known values: [ frac{frac{y}{2}}{1} = frac{1}{x} ] Solving for ( y ): [ y = frac{2}{x} ]5. Apply Stewart's Theorem: Using Stewart's Theorem on ( Delta ABD ) with cevian ( BK ): [ AB^2 cdot DK + BD^2 cdot AK = BK^2 cdot AD + AD cdot AK cdot BD ] Substituting the known values and relationships: [ 1^2 cdot 1 + left(frac{2}{x}right)^2 cdot x = z^2 cdot (x + 1) + (x + 1) cdot x cdot frac{2}{x} ] Simplifying: [ 1 + frac{4}{x} = z^2 cdot (x + 1) + 2(x + 1) ]6. Solve for ( x ): From the equation above, we need to solve for ( x ). Simplifying further: [ 1 + frac{4}{x} = z^2 cdot (x + 1) + 2(x + 1) ] Since ( z = BK ) and using the right triangle ( Delta ABK ): [ z = sqrt{AB^2 + BK^2} = sqrt{1^2 + x^2} = sqrt{1 + x^2} ] Substituting ( z ) back into the equation: [ 1 + frac{4}{x} = (1 + x^2)(x + 1) + 2(x + 1) ] Simplifying: [ 1 + frac{4}{x} = x^3 + x^2 + x + 1 + 2x + 2 ] [ 1 + frac{4}{x} = x^3 + x^2 + 3x + 3 ] Multiplying through by ( x ) to clear the fraction: [ x + 4 = x^4 + x^3 + 3x^2 + 3x ] Rearranging: [ x^4 + x^3 + 3x^2 + 2x - 4 = 0 ] Solving this polynomial equation, we find: [ x = sqrt[3]{2} ]7. Determine ( AD ): Since ( AD = AK + KD ): [ AD = x + 1 = sqrt[3]{2} + 1 ]The final answer is ( boxed{1 + sqrt[3]{2}} ).

🧠:Okay, so I need to find all second-degree polynomials with integer coefficients of the form P(X) = aX² + bX + c, where a ≠ b, and they satisfy the conditions P(a) = b and P(b) = a. Hmm, let's break this down step by step.First, since P(X) is a quadratic polynomial, the coefficients a, b, c must all be integers. The conditions given are P(a) = b and P(b) = a, and importantly, a ≠ b. So, the leading coefficient is a, the linear term is b, and the constant term is c. Let's write down the equations based on the given conditions.Starting with P(a) = b. Substituting X = a into the polynomial gives:P(a) = a*(a)² + b*(a) + c = a³ + a*b + c. And this is supposed to equal b. So, equation 1 is:a³ + a*b + c = b.Similarly, substituting X = b into P(X) gives:P(b) = a*(b)² + b*(b) + c = a*b² + b² + c. And this should equal a. So, equation 2 is:a*b² + b² + c = a.Now, we have two equations:1. a³ + a*b + c = b2. a*b² + b² + c = aSince both equations have c, maybe we can subtract them or eliminate c to find a relationship between a and b. Let's subtract equation 1 from equation 2:(a*b² + b² + c) - (a³ + a*b + c) = a - bSimplifying the left side:a*b² + b² + c - a³ - a*b - c = a*b² + b² - a³ - a*bSo, that's equal to a - b.Therefore:a*b² + b² - a³ - a*b = a - bLet me factor some terms here. Let's see:Left side: a*b² - a³ + b² - a*b = a(b² - a²) + b(b - a)Wait, notice that b² - a² factors as (b - a)(b + a). So:a*(b - a)(b + a) + b*(b - a) = (b - a)(a(b + a) + b) = (b - a)(a² + a*b + b)So, the left side becomes (b - a)(a² + a*b + b). Therefore, we have:(b - a)(a² + a*b + b) = a - bBut note that a - b = -(b - a), so we can write:(b - a)(a² + a*b + b) = -(b - a)Assuming that b ≠ a (which is given), we can divide both sides by (b - a), which gives:a² + a*b + b = -1So, now we have the equation:a² + a*b + b = -1Let me write that as:a² + a*b + b + 1 = 0Hmm, maybe factor this somehow? Let's see. Let's group terms:a² + a*b + b + 1 = a² + a*b + b + 1Factor out b from the middle terms:a² + b(a + 1) + 1 = 0Not sure if that helps. Alternatively, maybe factor in terms of a:a² + b*a + (b + 1) = 0This is a quadratic equation in terms of a. Maybe we can solve for a in terms of b?Using the quadratic formula:a = [-b ± sqrt(b² - 4*1*(b + 1))]/2Compute the discriminant:D = b² - 4*(b + 1) = b² - 4b - 4Since a must be an integer, the discriminant D must be a perfect square. Let's set D = k² where k is an integer:b² - 4b - 4 = k²Let me rearrange this equation:b² - 4b - 4 - k² = 0Hmm, maybe complete the square for the b terms:b² - 4b = (b - 2)^2 - 4So, substituting back:(b - 2)^2 - 4 - 4 - k² = (b - 2)^2 - 8 - k² = 0Thus:(b - 2)^2 - k² = 8This is a difference of squares:[(b - 2) - k][(b - 2) + k] = 8Let me denote m = (b - 2) - k and n = (b - 2) + k, so that m*n = 8. Also, since m and n are integers (because b and k are integers), we need to find all pairs of integers (m, n) such that m*n = 8 and n >= m (since k is non-negative? Wait, not necessarily. But m and n can be positive or negative. However, m and n must satisfy that (b - 2) = (m + n)/2 and k = (n - m)/2. Since k must be an integer, m and n must be both even or both odd. But 8 is even, so m and n must both be even.Let's list all factor pairs of 8 (positive and negative):1. (1, 8)2. (2, 4)3. (-1, -8)4. (-2, -4)5. (4, 2)6. (8, 1)7. (-4, -2)8. (-8, -1)But since m and n are ordered such that m <= n (to avoid duplicates), but actually, since m*n =8, we can consider all possible pairs regardless of order, but with both positive and negative factors.Wait, but actually, since m and n can be any integers as long as m*n = 8. So let's list all possible pairs (m, n):(1,8), (2,4), (-1,-8), (-2,-4), (4,2), (8,1), (-4,-2), (-8,-1). However, since m and n are factors such that m <= n if considering positive, but considering negatives, we can have different orders. However, for each pair, we can compute (b - 2) and k.But since m and n are factors of 8, let's go through each pair:1. (m, n) = (1, 8): Then: (b - 2) = (1 + 8)/2 = 9/2 = 4.5 → Not integer. Discard.2. (m, n) = (2, 4): (b - 2) = (2 + 4)/2 = 6/2 = 3 ⇒ b = 5 k = (4 - 2)/2 = 1 So, k = 1, check discriminant: D = k² = 1. Then, original quadratic for a is: a = [-b ± k]/2 = [-5 ± 1]/2 ⇒ (-5 +1)/2 = -4/2 = -2; (-5 -1)/2 = -6/2 = -3 So possible a = -2 or a = -3. Let's check these.3. (m, n) = (-1, -8): (b - 2) = (-1 + (-8))/2 = (-9)/2 = -4.5 → Not integer. Discard.4. (m, n) = (-2, -4): (b - 2) = (-2 + (-4))/2 = (-6)/2 = -3 ⇒ b = -1 k = (-4 - (-2))/2 = (-2)/2 = -1 ⇒ k² = 1 Then, a = [-b ± k]/2 = [-(-1) ± (-1)]/2 = [1 ± (-1)]/2 So, [1 -1]/2 = 0/2 = 0; [1 + (-1)]/2 = 0/2 = 0. Wait, k was -1, but in discriminant, k² is 1. Wait, maybe the sign of k doesn't matter because it's squared. So perhaps a = [ -b ± |k| ]/2? Wait, let's re-examine. If D = k², then sqrt(D) = |k|. So, in the quadratic formula, sqrt(D) is non-negative. So, regardless of k's sign, sqrt(D) is positive. So in this case, even though k = -1, sqrt(D) is 1, so: a = [-b ± 1]/2. For b = -1: a = [ -(-1) ± 1 ]/2 = [1 ± 1]/2 ⇒ (1 +1)/2 = 1, (1 -1)/2 = 0. So a = 1 or a = 0. But since in the original problem, a is the leading coefficient of a quadratic, so a ≠ 0. So a = 1 is valid, but a = 0 would make it a linear polynomial, which is not allowed. So only a = 1.5. (m, n) = (4, 2): Same as (2,4), which was already considered. So (b - 2) = 3, b =5.6. (8,1): Same as (1,8), non-integer.7. (-4, -2): Same as (-2, -4), already considered. (b -2) = -3 ⇒ b = -1.8. (-8, -1): Same as (-1, -8), non-integer.So the valid factor pairs are (2,4) leading to b=5, and (-2,-4) leading to b=-1. For each of these, we can find possible a's.So for b=5, possible a's are -2 and -3.For b=-1, possible a's are 1 (since a=0 is invalid).Now, let's check these possibilities.First, case 1: b=5, a=-2 or a=-3.Then, we can find c using one of the original equations. Let's use equation 1: a³ + a*b + c = b.For a=-2, b=5:Left side: (-2)^3 + (-2)*5 + c = -8 -10 + c = -18 + c. This should equal b=5, so:-18 + c = 5 ⇒ c = 23.Check with equation 2: a*b² + b² + c = a.Compute left side: (-2)*(5)^2 + (5)^2 + 23 = (-2)*25 + 25 +23 = -50 +25 +23 = (-50 +25) +23 = -25 +23 = -2. Which should equal a=-2. Yes, that works. So P(X) = -2X² +5X +23. Let's check if a ≠ b: a=-2, b=5. Yes, -2 ≠5. Good.Next, a=-3, b=5:Equation 1: (-3)^3 + (-3)*5 + c = -27 -15 + c = -42 + c = 5 ⇒ c=47.Check equation 2: a*b² + b² + c = (-3)*25 +25 +47 = -75 +25 +47 = (-75+25)+47 = -50 +47 = -3. Which equals a=-3. That works. So P(X)=-3X² +5X +47. Check a ≠b: -3 ≠5. Good.Case 2: b=-1, a=1.Compute c using equation 1: a³ +a*b +c = b.So, 1^3 +1*(-1) + c =1 -1 +c =0 +c =c. This should equal b=-1. So c=-1.Check equation 2: a*b² +b² +c =1*(-1)^2 + (-1)^2 + (-1) =1*1 +1 -1 =1 +1 -1 =1. Which should equal a=1. Yes. So P(X)=1X² + (-1)X + (-1) = X² -X -1. Check a ≠b: 1 ≠ -1. Good.So, possible polynomials are:1. -2X² +5X +232. -3X² +5X +473. X² -X -1Wait, but we need to check if these polynomials indeed satisfy the original conditions. Let me verify each one.First, P(X)=X² -X -1. Check P(a)=b where a=1, b=-1.P(1)=1² -1 -1 =1 -1 -1=-1, which is equal to b=-1. Then P(b)=P(-1)=(-1)^2 -(-1) -1=1 +1 -1=1, which is equal to a=1. So yes, this works.Next, P(X)=-2X² +5X +23. Check P(a)=b where a=-2, b=5.P(-2)= -2*(-2)^2 +5*(-2) +23 = -2*4 + (-10) +23= -8 -10 +23=5. Which is b=5. Then P(b)=P(5)= -2*25 +5*5 +23= -50 +25 +23= -50 +48= -2, which is a=-2. Correct.Similarly, P(X)=-3X² +5X +47. Check P(-3)= -3*(-3)^2 +5*(-3) +47= -3*9 -15 +47= -27 -15 +47= (-42)+47=5=b. Then P(5)= -3*25 +5*5 +47= -75 +25 +47= (-75+25)+47= -50+47= -3=a. Correct.So these three polynomials satisfy the conditions. Are there any other possibilities?Wait, let's check if there are other factor pairs we might have missed. Earlier, when listing the factor pairs of 8, we considered (1,8), (2,4), etc., but maybe there are other pairs like ( -8, -1 ), but we considered them. Alternatively, could there be fractions? But since m and n need to be integers (because b and k are integers), so only integer factor pairs.So all possibilities have been considered, leading to three polynomials. But let's confirm whether there are other solutions. Let's consider the equation a² +ab +b = -1. Let me see if there are other integer solutions (a, b) where a ≠ b.Suppose we try small integer values for a and b. Let's see:Case 1: a=1. Then the equation becomes 1 + b + b = -1 ⇒1 +2b = -1 ⇒2b= -2 ⇒b=-1. Which is the solution we already have.Case 2: a=0. But a=0 would make it a linear polynomial, which is not allowed.Case 3: a=-1. Then equation: (-1)^2 + (-1)*b +b =1 -b +b=1= -1? 1= -1? No. Contradiction. So no solution here.Case 4: a=2. Equation:4 +2b +b=4 +3b =-1 ⇒3b= -5 ⇒b= -5/3. Not integer.a=3: 9 +3b +b=9 +4b=-1 ⇒4b=-10 ⇒b=-10/4=-5/2. Not integer.a=-2: Equation:4 + (-2)b +b =4 -b =-1 ⇒-b= -5 ⇒b=5. Which is the solution already found.a=-3: Equation:9 + (-3)b +b=9 -2b=-1 ⇒-2b= -10 ⇒b=5. Which is the other solution.a=4: 16 +4b +b=16 +5b=-1 ⇒5b=-17 ⇒b=-17/5. Not integer.a=-4: 16 + (-4)b +b=16 -3b=-1 ⇒-3b= -17 ⇒b=17/3. Not integer.Similarly, trying other a's, we get non-integer b's. So the only integer solutions are (a,b) = (1, -1), (-2,5), (-3,5). Therefore, these correspond to the three polynomials found earlier.Therefore, the solutions are:1. P(X) = X² - X -12. P(X) = -2X² +5X +233. P(X) = -3X² +5X +47But let me check if there's another approach where maybe a and b are swapped? Wait, the problem states a ≠ b, but maybe there's a symmetry. However, since a and b are coefficients in the polynomial, swapping them would change the polynomial, unless the polynomial is symmetric in some way, which doesn't seem necessarily the case here.But in our solutions, for a=-2, b=5 and a=-3, b=5, so different a's but same b. So these are distinct.Therefore, these three polynomials are the only solutions.Final AnswerThe polynomials are boxed{X^2 - X - 1}, boxed{-2X^2 + 5X + 23}, and boxed{-3X^2 + 5X + 47}.

✅:Given the problem of finding all second-degree polynomials ( P(X) = aX^2 + bX + c ) with integer coefficients such that ( P(a) = b ), ( P(b) = a ), and ( a neq b ), we start by setting up the equations derived from the given conditions:1. ( a^3 + ab + c = b )2. ( ab^2 + b^2 + c = a )Subtracting these equations to eliminate ( c ):[ab^2 + b^2 - a^3 - ab = a - b]Factoring and simplifying, we get:[(b - a)(a^2 + ab + b) = -(b - a)]Since ( a neq b ), we divide both sides by ( b - a ):[a^2 + ab + b = -1]This equation is a quadratic in ( a ):[a^2 + ab + (b + 1) = 0]Using the quadratic formula to solve for ( a ):[a = frac{-b pm sqrt{b^2 - 4(b + 1)}}{2}]The discriminant ( b^2 - 4b - 4 ) must be a perfect square. Let ( k^2 = b^2 - 4b - 4 ). Rewriting this as a difference of squares:[(b - 2 - k)(b - 2 + k) = 8]Considering all factor pairs of 8, we find the integer solutions for ( b ):- For factor pairs (2, 4) and (-2, -4), we find ( b = 5 ) and ( b = -1 ).For ( b = 5 ), solving the quadratic equation for ( a ):- ( a = -2 ) or ( a = -3 )For ( b = -1 ), solving the quadratic equation for ( a ):- ( a = 1 )Substituting these values back into the original equations to find ( c ):1. For ( a = -2 ), ( b = 5 ): - ( c = 23 ) - Polynomial: ( -2X^2 + 5X + 23 )2. For ( a = -3 ), ( b = 5 ): - ( c = 47 ) - Polynomial: ( -3X^2 + 5X + 47 )3. For ( a = 1 ), ( b = -1 ): - ( c = -1 ) - Polynomial: ( X^2 - X - 1 )Verifying these polynomials confirms they satisfy the given conditions. Thus, the solutions are:[boxed{X^2 - X - 1}, boxed{-2X^2 + 5X + 23}, boxed{-3X^2 + 5X + 47}]

🔑:1. Given the polynomial ( P(X) = aX^2 + bX + c ), we need to satisfy the conditions ( P(a) = b ) and ( P(b) = a ) with ( a neq b ).2. First, we substitute ( X = a ) into the polynomial: [ P(a) = a cdot a^2 + b cdot a + c = a^3 + ab + c = b ] This gives us the equation: [ a^3 + ab + c = b quad text{(1)} ]3. Next, we substitute ( X = b ) into the polynomial: [ P(b) = a cdot b^2 + b cdot b + c = ab^2 + b^2 + c = a ] This gives us the equation: [ ab^2 + b^2 + c = a quad text{(2)} ]4. We now have two equations: [ a^3 + ab + c = b quad text{(1)} ] [ ab^2 + b^2 + c = a quad text{(2)} ]5. Subtract equation (1) from equation (2): [ (ab^2 + b^2 + c) - (a^3 + ab + c) = a - b ] Simplifying, we get: [ ab^2 + b^2 - a^3 - ab = a - b ] Factor out common terms: [ a(b^2 - a^2) + b(b - a) = a - b ] [ a(b - a)(b + a) + b(b - a) = a - b ] Factor out ( (b - a) ): [ (b - a)(a(b + a) + b) = a - b ] Since ( a neq b ), we can divide both sides by ( (a - b) ): [ a(b + a) + b = -1 ] Simplifying further: [ a^2 + ab + b + 1 = 0 quad text{(3)} ]6. We now solve equation (3) for ( b ): [ a^2 + ab + b + 1 = 0 ] Rearrange to solve for ( b ): [ b(a + 1) = -a^2 - 1 ] [ b = frac{-a^2 - 1}{a + 1} ]7. Since ( b ) must be an integer, ( frac{-a^2 - 1}{a + 1} ) must be an integer. We test integer values for ( a ) to find suitable ( b ): - For ( a = 1 ): [ b = frac{-1^2 - 1}{1 + 1} = frac{-2}{2} = -1 ] This gives ( (a, b) = (1, -1) ). - For ( a = -2 ): [ b = frac{-(-2)^2 - 1}{-2 + 1} = frac{-4 - 1}{-1} = 5 ] This gives ( (a, b) = (-2, 5) ). - For ( a = -3 ): [ b = frac{-(-3)^2 - 1}{-3 + 1} = frac{-9 - 1}{-2} = 5 ] This gives ( (a, b) = (-3, 5) ).8. We discard the case ( (a, b) = (0, -1) ) since ( a neq b ).9. We now find ( c ) for each pair ( (a, b) ): - For ( (a, b) = (1, -1) ): [ a^3 + ab + c = b implies 1^3 + 1(-1) + c = -1 implies 1 - 1 + c = -1 implies c = -1 ] Thus, ( (a, b, c) = (1, -1, -1) ). - For ( (a, b) = (-2, 5) ): [ a^3 + ab + c = b implies (-2)^3 + (-2)(5) + c = 5 implies -8 - 10 + c = 5 implies c = 23 ] Thus, ( (a, b, c) = (-2, 5, 23) ). - For ( (a, b) = (-3, 5) ): [ a^3 + ab + c = b implies (-3)^3 + (-3)(5) + c = 5 implies -27 - 15 + c = 5 implies c = 47 ] Thus, ( (a, b, c) = (-3, 5, 47) ).The final answer is ( boxed{ (a, b, c) = (1, -1, -1), (-2, 5, 23), (-3, 5, 47) } ).

🧠:Okay, so I need to solve this system of equations where each variable is defined in terms of another through a fifth-degree polynomial and a cubic polynomial. Let's write down the equations again to make sure I have them right:1. ( x^5 = 2y^3 + y - 2 )2. ( y^5 = 2z^3 + z - 2 )3. ( z^5 = 2x^3 + x - 2 )Hmm, these equations are cyclic. Each variable depends on the next one. That makes me think that maybe there's a symmetric solution where all variables are equal. Let me check that first.Suppose ( x = y = z = k ). Then substituting into each equation, we'd get:( k^5 = 2k^3 + k - 2 )So, rearranging:( k^5 - 2k^3 - k + 2 = 0 )I need to solve this equation for real ( k ). Let's try to factor this polynomial. Maybe there are integer roots. Let's use Rational Root Theorem. Possible rational roots are factors of 2 over factors of 1, so ±1, ±2.Testing k=1:1 - 2 -1 +2 = 0. Oh, that works! So (k - 1) is a factor.Let's perform polynomial division or use synthetic division.Divide ( k^5 - 2k^3 - k + 2 ) by (k - 1).Using synthetic division:Coefficients: 1 (k^5), 0 (k^4), -2 (k^3), 0 (k^2), -1 (k), 2 (constant)Bring down 1. Multiply by 1: 1. Add to next coefficient: 0 +1=1. Multiply by 1:1. Add to -2: -1. Multiply by1: -1. Add to 0: -1. Multiply by1: -1. Add to -1: -2. Multiply by1: -2. Add to 2: 0. Perfect.So the polynomial factors as (k - 1)(k^4 + k^3 - k^2 - k - 2).Now we have to factor the quartic: ( k^4 + k^3 - k^2 - k - 2 ).Let me try possible roots here again. Possible roots are ±1, ±2.Testing k=1:1 +1 -1 -1 -2 = -2 ≠0k= -1:1 -1 -1 +1 -2 = -2 ≠0k=2:16 +8 -4 -2 -2=16+8=24; 24-4=20; 20-2=18; 18-2=16 ≠0k=-2:16 -8 -4 +2 -2=16-8=8; 8-4=4; 4+2=6; 6-2=4 ≠0Hmm, no rational roots. Maybe factor into quadratics.Suppose ( k^4 + k^3 - k^2 - k - 2 = (k^2 + ak + b)(k^2 + ck + d) )Expanding:( k^4 + (a + c)k^3 + (ac + b + d)k^2 + (ad + bc)k + bd )Set equal to coefficients:1. ( a + c = 1 )2. ( ac + b + d = -1 )3. ( ad + bc = -1 )4. ( bd = -2 )We need integers a, b, c, d. Since bd = -2, possible pairs (b, d): (1,-2), (-1,2), (2,-1), (-2,1)Let's try b=2, d=-1:Then equations:1. a + c =12. ac +2 -1 = ac +1 = -1 ⇒ ac= -23. a*(-1) + c*2 = -a + 2c = -1From 1: c =1 -aPlug into ac= -2:a(1 - a) = -2 ⇒ a - a² = -2 ⇒ a² -a -2=0 ⇒ a=(1±√(1+8))/2=(1±3)/2 ⇒ a=2 or -1If a=2, then c=1 -2= -1Check equation 3: -2 + 2*(-1)= -2 -2= -4 ≠ -1. Wait, no:Wait equation 3: -a +2c = -2 + 2*(-1)= -2 -2= -4 ≠ -1. Not good.If a=-1, then c=1 -(-1)=2Check equation 3: -(-1) + 2*(2)=1 +4=5 ≠-1. Also not good.So b=2, d=-1 doesn't work.Try b=1, d=-2:Equations:1. a + c =12. ac +1 -2 = ac -1 = -1 ⇒ ac=03. a*(-2) + c*1 = -2a +c = -1From ac=0, either a=0 or c=0.If a=0, then c=1 (from 1). Then equation3: -0 +1=1≠-1. Not good.If c=0, then a=1. Then equation3: -2*1 +0= -2≠-1. Not good.Next pair: b=-1, d=2.Equations:1. a +c=12. ac -1 +2=ac +1= -1 ⇒ ac= -23. a*2 +c*(-1)=2a -c= -1From 1: c=1 -aPlug into ac=-2:a(1 -a)= -2 ⇒ a -a²= -2 ⇒ a² -a -2=0 same as before. Solutions a=2 or -1.If a=2, c= -1. Then equation3: 2*2 - (-1)=4 +1=5≠-1.If a=-1, c=2. Then equation3: 2*(-1) -2= -2 -2= -4≠-1.Nope.Last pair: b=-2, d=1.Equations:1. a +c=12. ac -2 +1=ac -1= -1 ⇒ ac=03. a*1 +c*(-2)=a -2c= -1From 1: c=1 -aFrom ac=0, either a=0 or c=0.If a=0, c=1. Then equation3:0 -2*1= -2≠-1.If c=0, a=1. Then equation3:1 -0=1≠-1.So none of these factorings work. Maybe the quartic is irreducible over integers. So, perhaps we need to factor it another way or use the rational root theorem with fractions? But since it's a quartic, maybe it has real roots. Let's check for real roots.Let’s denote ( f(k) = k^4 + k^3 - k^2 - k - 2 )Compute f(1): 1 +1 -1 -1 -2 = -2f(2): 16 +8 -4 -2 -2=16+8=24-8=16f(-1):1 -1 -1 +1 -2= -2f(0):0 +0 -0 -0 -2= -2f(-2):16 -8 -4 +2 -2=4So, between k=1 and k=2, f(k) goes from -2 to 16, so crosses zero. Similarly, between k=-2 and k=-1, f(k) goes from 4 to -2, so crosses zero. Hence, there are two real roots: one in (1,2) and another in (-2,-1). The other roots might be complex.Therefore, the original quintic equation has roots at k=1, and four other roots from the quartic, two real and two complex. But since we are interested in real solutions, the real roots are k=1, and the two from the quartic.But wait, the quartic has two real roots, so total three real roots for the quintic equation. But maybe?Wait, the quartic is degree 4. If it has two real roots, then the other two are complex conjugates. So total for quintic, 1 (from k=1) plus two from quartic, but quartic has two real roots and two complex. So quintic has 1 + 2 = 3 real roots. Wait, quintic is degree 5, so total five roots, with two complex (since coefficients are real, complex roots come in pairs). So quintic has three real roots and two complex roots.Therefore, when we assumed x=y=z=k, we get three real solutions: k=1, and two others. Let me check approximate values.First, let's find the root between 1 and 2. Let's try k=1.5:f(1.5)= (1.5)^4 + (1.5)^3 - (1.5)^2 -1.5 -21.5^2=2.25; 1.5^3=3.375; 1.5^4=5.0625So 5.0625 +3.375 -2.25 -1.5 -2 = 5.0625 +3.375=8.4375; 8.4375 -2.25=6.1875; 6.1875 -1.5=4.6875; 4.6875 -2=2.6875 >0. Still positive. So between 1 and 1.5, f(k) goes from -2 to 2.6875. So crosses zero somewhere.At k=1.2:1.2^4=2.0736; 1.2^3=1.728; 1.2^2=1.44So f(1.2)=2.0736 +1.728 -1.44 -1.2 -2 ≈2.0736+1.728=3.8016 -1.44=2.3616 -1.2=1.1616 -2= -0.8384So f(1.2)≈-0.8384At k=1.3:1.3^4=2.8561; 1.3^3=2.197; 1.3^2=1.69f(1.3)=2.8561 +2.197 -1.69 -1.3 -2 ≈2.8561+2.197=5.0531 -1.69=3.3631 -1.3=2.0631 -2=0.0631 >0So between 1.2 and 1.3, f(k) crosses zero. Similarly, negative at k=1.2, positive at 1.3. Let's approximate with linear approx:From k=1.2 to 1.3, delta k=0.1, delta f≈0.0631 -(-0.8384)=0.9015We need f(k)=0. Assume linear: 0 = f(1.2) + (0.9015/0.1)(k -1.2)0 = -0.8384 +9.015(k -1.2)9.015(k -1.2)=0.8384k -1.2≈0.8384/9.015≈0.093k≈1.2 +0.093≈1.293So approximately 1.293. Let's check f(1.293):Compute 1.293^4, etc. Maybe this is overkill. Let's note that the approximate root is around 1.293. Let's call it r1≈1.293.Similarly, the other real root is between -2 and -1. Let's check k=-1.5:f(-1.5)= (-1.5)^4 + (-1.5)^3 - (-1.5)^2 - (-1.5) -2=5.0625 -3.375 -2.25 +1.5 -2=5.0625 -3.375=1.6875; 1.6875 -2.25= -0.5625; -0.5625 +1.5=0.9375; 0.9375 -2= -1.0625 <0k=-1.5 gives f=-1.0625k=-1:f(-1)=1 -1 -1 +1 -2= -2k=-2:f(-2)=16 -8 -4 +2 -2=4So between k=-2 and -1.5, f(k) goes from 4 to -1.0625. Crossing zero there.At k=-1.75:(-1.75)^4=9.37890625; (-1.75)^3=-5.359375; (-1.75)^2=3.0625f(-1.75)=9.37890625 -5.359375 -3.0625 -(-1.75) -2=9.37890625 -5.359375=4.01953125; 4.01953125 -3.0625=0.95703125; 0.95703125 +1.75=2.70703125; 2.70703125 -2=0.70703125>0So f(-1.75)≈0.707At k=-1.6:(-1.6)^4=6.5536; (-1.6)^3=-4.096; (-1.6)^2=2.56f(-1.6)=6.5536 -4.096 -2.56 -(-1.6) -2=6.5536 -4.096=2.4576; 2.4576 -2.56= -0.1024; -0.1024 +1.6=1.4976; 1.4976 -2= -0.5024 <0So between k=-1.75 and -1.6, f(k) goes from positive to negative. Let's check k=-1.7:(-1.7)^4=8.3521; (-1.7)^3=-4.913; (-1.7)^2=2.89f(-1.7)=8.3521 -4.913 -2.89 +1.7 -2 ≈8.3521 -4.913=3.4391; 3.4391 -2.89=0.5491; 0.5491 +1.7=2.2491; 2.2491 -2=0.2491>0k=-1.7: f≈0.2491k=-1.65:(-1.65)^4=(1.65)^4≈ (1.65^2)^2≈(2.7225)^2≈7.412; (-1.65)^3≈-4.489; (-1.65)^2≈2.7225f(-1.65)=7.412 -4.489 -2.7225 +1.65 -2≈7.412-4.489=2.923; 2.923-2.7225≈0.2005; 0.2005+1.65≈1.8505; 1.8505-2≈-0.1495 <0So between -1.7 and -1.65, f(k) crosses zero. Let's approximate:At k=-1.7: f=0.2491At k=-1.65: f≈-0.1495Delta k=0.05, delta f≈-0.1495 -0.2491≈-0.3986Zero crossing at k= -1.7 - (0.2491)*(0.05)/0.3986≈ -1.7 - (0.012455/0.3986)≈-1.7 -0.031≈-1.731Wait, linear approximation:Let’s model f(k) between k=-1.7 (0.2491) and k=-1.65 (-0.1495). The change is -0.3986 over 0.05 increase in k. To find when f(k)=0:From k=-1.7 to k=-1.65: slope is -0.3986/0.05≈-7.972 per unit k.Starting at k=-1.7, f=0.2491. To reach 0:0 =0.2491 -7.972*(k +1.7)So, 7.972*(k +1.7)=0.2491 ⇒ k +1.7≈0.2491/7.972≈0.0312 ⇒k≈-1.7 +0.0312≈-1.6688So approximate root at k≈-1.6688. Let's call this r2≈-1.669.Therefore, the quintic equation has real roots at k=1, k≈1.293, and k≈-1.669. So for the symmetric case x=y=z=k, there are three real solutions: (1,1,1), (r1,r1,r1), and (r2,r2,r2).But wait, we need to confirm if these approximate roots actually satisfy the original quintic equation. Also, note that the equations are cyclic but not necessarily symmetric. So there might be solutions where x, y, z are not all equal.However, solving a cyclic system usually starts by checking symmetric solutions first. Let's verify if these symmetric solutions work.First, k=1: substituting into each equation:1. (1^5 = 2*1^3 +1 -2 →1=2+1-2=1) ✓Similarly for y and z. So (1,1,1) is indeed a solution.For the other approximate roots, let's check if they satisfy the quintic equation.Take k≈1.293:Compute LHS: k^5 ≈1.293^5. Let me compute step by step:1.293^2≈1.6721.293^3≈1.672*1.293≈2.1631.293^4≈2.163*1.293≈2.7981.293^5≈2.798*1.293≈3.618RHS: 2k^3 +k -2≈2*(2.163) +1.293 -2≈4.326 +1.293 -2≈5.619 -2≈3.619. So LHS≈3.618, RHS≈3.619. Close enough, considering rounding errors. So this is a valid solution.Similarly, for k≈-1.669:LHS: (-1.669)^5≈-13.06 (since (-1.669)^2≈2.785, ^3≈-4.65, ^4≈7.76, ^5≈-12.95)RHS: 2*(-1.669)^3 + (-1.669) -2≈2*(-4.65) -1.669 -2≈-9.3 -1.669 -2≈-12.969. Close to LHS≈-12.95. So approximately valid.Thus, the symmetric solutions are (1,1,1), (r1,r1,r1), and (r2,r2,r2) where r1≈1.293 and r2≈-1.669.But are there non-symmetric solutions? That's a more complex question. To check, we need to see if there are solutions where x, y, z are not all equal.Suppose that x, y, z are not all equal. Let's consider if the system has cycles of higher length. For example, x→y→z→x, forming a cycle. If they are not equal, but follow a cyclic permutation.But given the complexity of the equations, it might be challenging to find such solutions. Let's try to analyze the functions involved.Define functions f(a) = 2a^3 + a - 2. Then the equations are:x^5 = f(y)y^5 = f(z)z^5 = f(x)So, composing these functions: x^5 = f(y), y^5 = f(z), z^5 = f(x). Therefore, x^(5^3) = f(f(f(x))). Wait, composing three times.But this might not be helpful. Alternatively, substituting:From first equation: y = [ (x^5 + 2 - y)/2 ]^(1/3). Wait, not straightforward.Alternatively, suppose we consider the function g(a) = (f(a))^(1/5). Then the system can be written as:x = g(y)y = g(z)z = g(x)Therefore, x = g(g(g(x))). So x is a fixed point of the function h(x) = g(g(g(x))).Finding fixed points of such a composite function might be difficult. However, fixed points of the individual functions would correspond to symmetric solutions where x=y=z=k, since h(k) = g(g(g(k))) = g(g(k^{5})) = g( (2k^3 +k -2)^{1/5} ) ... which is complicated.Alternatively, let's analyze the behavior of the functions to see if non-symmetric solutions can exist.First, consider the function f(a) = 2a^3 +a -2. Let's analyze its monotonicity.Derivative f’(a) =6a^2 +1, which is always positive. So f(a) is strictly increasing. Therefore, f is invertible.Similarly, the function g(a) = (f(a))^(1/5) = (2a^3 +a -2)^{1/5}. Since f(a) is strictly increasing, and the fifth root is also strictly increasing, g(a) is strictly increasing wherever it is defined. The domain of g(a) requires that 2a^3 +a -2 ≥0 when considering real numbers, since we are taking the fifth root, which is defined for all real numbers (since fifth root of a negative number is negative). Wait, fifth root is defined for all real numbers, so g(a) is defined for all real a.But even though f(a) is strictly increasing, g(a) is also strictly increasing. Therefore, the composition g(g(g(x))) is strictly increasing. Therefore, the equation x = g(g(g(x))) can have only one solution for each fixed point. But we already have three symmetric solutions. Wait, but if h(x) = g(g(g(x))) is strictly increasing, then h(x) can intersect the line y=x at multiple points, depending on its shape. However, since g is strictly increasing, h(x) is also strictly increasing. Therefore, h(x) - x is strictly increasing? Wait, if h(x) is strictly increasing and x is strictly increasing, then h(x) -x could be increasing or decreasing depending on the derivatives.Wait, derivative of h(x) is h’(x) = g’(g(g(x))) * g’(g(x)) * g’(x). Since g is strictly increasing, g’ is positive wherever it exists (which is wherever f(a) ≠0, since derivative of g(a) = (1/5)(2a^3 +a -2)^{-4/5}*(6a^2 +1). Since 6a^2 +1 is always positive, and (2a^3 +a -2)^{-4/5} is real for all a (since fifth roots can handle negative bases), but (2a^3 +a -2)^{-4/5} is positive when 2a^3 +a -2 ≠0. Therefore, h’(x) is positive wherever the denominator doesn't cause issues. So h(x) is strictly increasing wherever it's differentiable, which is everywhere except where 2a^3 +a -2 =0, which is at the roots of 2a^3 +a -2 =0. Let's see, 2a^3 +a -2 =0. Let's check if this has real roots.At a=1: 2 +1 -2=1≠0a=0:0 +0 -2=-2a=1. Let's try a=0.8:2*(0.512) +0.8 -2≈1.024 +0.8 -2≈-0.176a=0.85:2*(0.614125) +0.85 -2≈1.22825 +0.85 -2≈0.07825>0So there's a root between 0.8 and 0.85. Let's call it s≈0.83.Therefore, at a=s≈0.83, 2a^3 +a -2=0. At this point, g(a) =0, since (0)^{1/5}=0. So the function g(a) is smooth everywhere except at a=s, where it has a vertical tangent? Because derivative of g(a) at a=s would involve division by zero. However, even with that, h(x) is still strictly increasing except at points where x, g(x), or g(g(x)) equals s. But since s is a specific value, unless the system cycles through s, which is complicated, h(x) is piecewise strictly increasing. Therefore, the equation x = h(x) can have multiple fixed points if h(x) crosses y=x multiple times. But given that each composition is increasing, it's possible to have multiple intersections.But how does this relate to our problem? If h(x) is strictly increasing, it can cross y=x at multiple points, which would correspond to multiple fixed points, i.e., multiple solutions where x = h(x). However, each such fixed point would correspond to a symmetric solution x=y=z=k. Wait, because x = g(g(g(x))) implies that if we set x=y=z=k, then each equation is satisfied. Therefore, all fixed points of h(x) are symmetric solutions. Therefore, the only real solutions are the symmetric ones. Hence, there are no non-symmetric real solutions.Wait, but this is under the assumption that h(x) is strictly increasing. However, even if h(x) is strictly increasing, it can still intersect y=x multiple times, as long as its slope is less than 1 in certain regions. For example, h(x) could cross y=x from below to above, creating multiple fixed points. However, with h(x) being strictly increasing, the number of fixed points is determined by how many times h(x) crosses y=x. Since we already found three real solutions from the symmetric case, and given the complexity of the equations, it's plausible that these are the only real solutions.To confirm, let's analyze the function h(x) = g(g(g(x))).Given that each g is strictly increasing, h(x) is strictly increasing. Suppose h(x) -x has multiple zeros. However, the derivative of h(x) -x is h’(x) -1. If h’(x) is always greater than 1, then h(x) -x is strictly increasing, so only one zero. If h’(x) is sometimes less than 1, there could be multiple zeros. But computing h’(x) is complicated.Alternatively, consider that the symmetric solutions correspond to fixed points of the function h(x). Since we found three real fixed points, and h(x) is strictly increasing, it's possible that there are exactly three intersections between h(x) and y=x. Hence, three symmetric solutions.However, another way to see is to note that if we assume the system has only symmetric solutions, then we can limit ourselves to those. But how to be sure?Suppose there exists a non-symmetric solution. Then, since the system is cyclic, we could permute the variables and get other solutions. For example, if (x, y, z) is a solution, then so are (y, z, x) and (z, x, y). However, given the high degree of the equations, it's not clear if such solutions exist apart from the symmetric ones.Alternatively, suppose that x, y, z are distinct. Let's analyze the possibility.Assume x > y > z. Then, from the first equation: x^5 = 2y^3 + y -2. If x > y, then x^5 > 2y^3 + y -2. Similarly, y^5 = 2z^3 + z -2. If y > z, then y^5 > 2z^3 + z -2. And z^5 = 2x^3 +x -2. If z < x, then z^5 < 2x^3 +x -2. But combining these inequalities:x^5 > 2y^3 + y -2y^5 > 2z^3 + z -2z^5 < 2x^3 +x -2This seems contradictory. Let's see:From first inequality: x^5 > 2y^3 + y -2But since x > y, and if x and y are positive, x^5 is much larger than 2y^3 + y.But then from the third inequality: z^5 < 2x^3 +x -2If z is the smallest, and if z is negative, then z^5 is negative (since odd power), while 2x^3 +x -2. If x is positive, 2x^3 +x -2 could be positive or negative.Wait, this is getting too vague. Maybe a better approach is to consider the function's growth rates.Note that for large positive x, x^5 grows much faster than 2x^3 +x -2. Similarly, for large negative x, x^5 is negative and 2x^3 +x -2 is also negative, but which one is more negative?For x → ∞: x^5 dominates 2x^3, so x^5 > 2x^3 +x -2.For x → -∞: x^5 → -∞, and 2x^3 +x -2 → -∞. Which grows faster? x^5 vs 2x^3. x^5 grows faster in magnitude, so for large negative x, x^5 < 2x^3 +x -2.Therefore, the equation x^5 = 2x^3 +x -2 has solutions where x^5 intersects 2x^3 +x -2, which we found earlier at x=1, x≈1.293, and x≈-1.669.Now, suppose we have a non-symmetric solution. Let's say x > y. Then, from x^5 = 2y^3 + y -2. If x > y, then 2y^3 + y -2 must be less than x^5. But y must satisfy y^5 = 2z^3 + z -2. If y > z, then similarly. But if z is the smallest, then z^5 = 2x^3 +x -2. If x is the largest, then 2x^3 +x -2 would be large positive, so z^5 must also be large positive, implying z is positive. But if z is positive and the smallest, then x > y > z >0.But in that case, z^5 =2x^3 +x -2. If x is large, z would have to be roughly (2x^3)^{1/5} = 2^{1/5} x^{3/5}. But as x increases, z grows much slower than x. So z is much smaller than x. But then y is between x and z. This seems possible, but the equations are highly intertwined.Alternatively, maybe suppose that x, y, z are in a cycle where each is larger than the next, but given the cyclic nature, this would require x > y > z >x, which is impossible. Therefore, such a cycle cannot exist. Similarly for other orderings.Therefore, the only possible real solutions are the symmetric ones. Hence, the real triplets are (1,1,1), (r1,r1,r1), and (r2,r2,r2) where r1≈1.293 and r2≈-1.669.But to confirm, let's check if substituting these approximate roots into the system works.Take r1≈1.293:Compute x^5 ≈1.293^5≈3.618Compute 2y^3 +y -2 with y=1.293:2*(1.293)^3 +1.293 -2≈2*(2.163) +1.293 -2≈4.326 +1.293 -2≈3.619. Close to x^5≈3.618. So consistent.Similarly, y^5=z^5=x^5≈3.618, so each equation holds.For r2≈-1.669:x^5≈(-1.669)^5≈-12.95Compute 2y^3 +y -2 with y=-1.669:2*(-1.669)^3 + (-1.669) -2≈2*(-4.648) -1.669 -2≈-9.296 -1.669 -2≈-12.965≈-12.95. Close enough.Therefore, these approximate symmetric solutions work.Therefore, the conclusion is that the only real solutions are the symmetric ones: (1,1,1), and two others where each variable is approximately 1.293 and -1.669. However, the problem asks for exact solutions, not approximations. So we need to express them exactly.We already found that k=1 is an exact solution. The other solutions come from the quartic equation k^4 +k^3 -k^2 -k -2=0. This quartic might factor into quadratics with real coefficients, but as we saw earlier, it doesn't factor with integer coefficients. However, maybe it can be factored using real numbers.Alternatively, perhaps we can express the roots using the quartic formula, but that's very complicated. Alternatively, notice that the quartic equation can be rewritten as:k^4 +k^3 -k^2 -k -2=0Let me try grouping terms:(k^4 -k^2) + (k^3 -k) -2=0k^2(k^2 -1) +k(k^2 -1) -2=0(k^2 -1)(k^2 +k) -2=0Which is (k-1)(k+1)(k(k +1)) -2=0Hmm, not sure if helpful. Alternatively, let’s attempt substitution. Let m=k^2 + ak +b. Not sure.Alternatively, let's use the substitution k = t - b to eliminate the cubic term. But quartic equations are generally tough.Alternatively, check if the quartic is biquadratic, but it's not. Alternatively, maybe a substitution t =k +1/k or similar. Not obvious.Alternatively, since it didn't factor with integer coefficients, and we couldn't find rational roots, perhaps the real roots can only be expressed via radicals but it's very messy. Given that the original problem is from a competition or similar, it's likely that the only exact solution is k=1, and the others are irrational and must be left as roots of the quartic. However, the problem says "find all triplets of real numbers", so we might need to present all real solutions, including the symmetric ones, expressed in terms of roots of the quintic equation.But the quintic equation factors as (k -1)(k^4 +k^3 -k^2 -k -2)=0, so the real solutions are k=1, and the real roots of the quartic. Since the quartic has two real roots, as we saw, there are three real solutions for k. Hence, three real triplets (k,k,k) where k is each real root.Therefore, the answer should be three triplets: (1,1,1), and two others which can be written as (r,r,r) where r is a real root of k^4 +k^3 -k^2 -k -2=0. Since the problem might accept this form, or require exact expressions, but since the quartic is not solvable nicely, the boxed answer would be:boxed{(1, 1, 1)}, and the other two solutions can be expressed using the roots of the quartic equation, but since the problem asks for all triplets, and we can’t express them in radicals neatly, we can present them as (r, r, r) where r satisfies r^5 = 2r^3 + r - 2. However, typically in such problems, if symmetric solutions are the only real ones and others can't be expressed simply, the answer is just the boxed (1,1,1). But wait, the original quintic has three real roots, so we need to mention all three.But maybe the problem expects only (1,1,1) because it's the only rational solution, and the others are irrational. But the question says "real numbers", not rational. So we need to include all real solutions.But how to present them? Since we can't write the irrational ones in exact form without radicals, which are too complicated, perhaps we can state all real solutions are the triplets where x=y=z and each is a real root of the equation k^5 -2k^3 -k +2=0. But since the problem might expect us to recognize that the only real solution is (1,1,1), but our analysis shows there are three.Wait, let's double-check the original quintic equation: k^5 -2k^3 -k +2=0. We factored out (k-1), leaving quartic k^4 +k^3 -k^2 -k -2=0. We determined that the quartic has two real roots and two complex roots. Therefore, the quintic has three real roots: 1, r1, r2. Hence, three real solutions.But in competition math, often such systems have only the symmetric solution as the real one, but here we have more. However, given the analysis, unless there's a mistake in assuming that non-symmetric solutions don't exist, the answer should include all three symmetric solutions.However, perhaps I made a mistake in the quartic analysis. Let me confirm the number of real roots of the quartic k^4 +k^3 -k^2 -k -2.Using Descartes' Rule of Signs for positive roots: coefficients are +1, +1, -1, -1, -2. Sign changes: +1 to +1 (none), +1 to -1 (1), -1 to -1 (none), -1 to -2 (none). So 1 sign change ⇒1 positive real root.For negative roots, substitute k=-m:(-m)^4 + (-m)^3 - (-m)^2 - (-m) -2 = m^4 -m^3 -m^2 +m -2.Coefficients: +1, -1, -1, +1, -2.Sign changes: +1 to -1 (1), -1 to -1 (0), -1 to +1 (1), +1 to -2 (1). Total 3 sign changes ⇒ 3 or 1 negative real roots.But the quartic is degree 4, so possible 1 positive, 3 negative (but total 4), but earlier analysis showed there's a root at ≈-1.669 and another positive root. Wait, but Descartes' Rule says up to 3 negative roots, but we found only one negative root. There's a discrepancy.Wait, when I tested f(-1.5)= -1.0625, f(-2)=4, so one real root between -2 and -1.5. But Descartes' Rule allows up to 3 negative roots, but we only found one. Therefore, there must be two more negative roots or complex roots. Wait, but the original quartic must have 4 roots: we found one positive, one negative, and two complex. That matches Descartes' Rule (1 positive, 3 or 1 negative). Since we found 1 negative, the other two must be complex. Therefore, quartic has two real roots (1 positive, 1 negative) and two complex. Therefore, quintic has three real roots:1, r1 (positive), r2 (negative). Therefore, three real solutions.Hence, the answer should be three triplets: (1,1,1), (r1,r1,r1), (r2,r2,r2) where r1≈1.293 and r2≈-1.669. Since the problem asks for all real triplets, and we can’t express r1 and r2 in exact form, we have to present them as roots of the equation.But in mathematical problems, especially Olympiad-style, if an equation can't be solved exactly, we still need to present the solutions in terms of the equation. So the final answer would be the three symmetric solutions where each variable is a real root of the equation k^5 -2k^3 -k +2=0. But since k=1 is one of them, we can specify them as:The triplets are (1,1,1) and (r,r,r) where r is a real root of r^4 + r^3 - r^2 - r -2=0.But to confirm, since the quintic factors as (k-1)(k^4 +k^3 -k^2 -k -2)=0, the other real roots are from the quartic. Therefore, the solutions are:1. (1,1,1)2. (r,r,r) where r is a real root of k^4 +k^3 -k^2 -k -2=0.Since the quartic has two real roots, there are two such triplets. However, since the problem asks for triplets of real numbers, we need to mention all three.But perhaps the problem expects only (1,1,1) as the solution, assuming that the other roots are extraneous or that the system only has the symmetric solution. But given our analysis, that's not the case.Wait, let's verify with substitution. Suppose we have a solution where x, y, z are not all equal. Let's assume x=1. Then from the first equation:1=2y^3 +y -2→2y^3 +y -3=0. Let's solve this for y.Try y=1:2+1-3=0. So y=1 is a root. Then factor: (y-1)(2y^2 +2y +3)=0. The quadratic has discriminant 4 -24= -20 <0, so only real root y=1. Then z would be from y=1:1=2z^3 +z -2→2z^3 +z -3=0. Again, z=1 is a root:2+1-3=0. Factor: (z-1)(2z^2 +2z +3)=0. So z=1. Then x=1. So even if we assume one variable is 1, the others must be 1. Hence, the only solution with any variable 1 is (1,1,1).Suppose we take x=r1≈1.293, then y would be such that y^5=2z^3 +z -2, and z^5=2x^3 +x -2. But since we are in the symmetric case, this holds. However, if we perturb one variable, it's not clear if the system can still hold. Given the complexity, and the fact that the system is cyclic and each equation is strictly increasing, it's likely that only the symmetric solutions exist.Therefore, the conclusion is that the only real solutions are the symmetric ones: (1,1,1), (r1,r1,r1), and (r2,r2,r2), where r1 and r2 are the other two real roots of the quintic equation. Therefore, the answer should include all three.But the problem asks to "find all triplets", so we need to present all real solutions. However, since r1 and r2 cannot be expressed in a simple radical form, we have to describe them in terms of the equation. In some contexts, it's acceptable to present them as (k,k,k) where k satisfies k^5=2k^3 +k -2, which is exactly the equation from the symmetric case. Hence, the solutions are all triplets where x=y=z=k and k is a real solution to k^5=2k^3 +k -2.Therefore, the final answer is all triplets (k,k,k) where k is a real root of the equation k^5 -2k^3 -k +2=0. Since this equation factors as (k-1)(k^4 +k^3 -k^2 -k -2)=0, the real roots are k=1 and the real roots of the quartic. Therefore, the triplets are (1,1,1) and two others where each component is a real root of the quartic.However, in the absence of exact forms for r1 and r2, we can write the solutions as:boxed{(1, 1, 1)}, and the two other solutions are real triplets of the form boxed{(r, r, r)} where r is a real root of the equation r^5 - 2r^3 - r + 2 = 0.But typically, in competitions, if an equation is solvable and has rational roots, they expect the explicit solutions, and for the others, they might accept indicating their existence. However, since the problem doesn't specify the format, but asks for precise solutions, it's safer to box all three solutions, but since we can't write r1 and r2 explicitly, we can state the symmetric solutions as above.Alternatively, the problem might only expect (1,1,1) as the solution, but given the analysis, that's incomplete. To check, let's verify if substituting r1 and r2 back into the original system works.We saw that approximately, they do. Hence, they are valid solutions. Therefore, the correct answer is that all real triplets are the symmetric ones, which are three in total: (1,1,1), and two others where each variable is a real root of the quartic.But since the problem is in Chinese or from an unspecified source, but given the international standards, the answer should include all real solutions, expressed as triplets where all components are equal to a real root of the quintic equation.Therefore, the final answer is:All real triplets are given by boxed{(1, 1, 1)} and the two real solutions of the form boxed{(r, r, r)} where ( r ) satisfies ( r^5 = 2r^3 + r - 2 ).But to adhere to the instruction of putting the final answer within a single box, perhaps list all solutions, but since two can't be expressed neatly, the standard approach is to present all solutions as (k,k,k) where k satisfies the equation. But the problem might expect only (1,1,1). However, given the analysis, three solutions exist. Therefore, the answer is the three symmetric solutions.But in the absence of exact forms for the other two roots, and following mathematical conventions, the answer is:The real triplets are all solutions with ( x = y = z ), which are ( boxed{(1, 1, 1)} ), and two other triplets where each component is a real root of ( k^5 - 2k^3 - k + 2 = 0 ).But since the problem might expect only the boxed (1,1,1), but given the instructions to provide accurate solutions, and knowing there are three real solutions, the correct answer includes all three. However, without explicit forms, we can only explicitly write (1,1,1) and reference the others.But considering the problem's origin and common practices, it's possible that (1,1,1) is the only expected answer, with the others considered extraneous or complex. But no, we saw they are real. Hence, the answer is three triplets, which we can write as:The real solutions are the triplets where all components are equal and satisfy ( k^5 = 2k^3 + k - 2 ). These are ( boxed{(1, 1, 1)} ), and two other triplets with each component being a real root of the equation ( k^5 - 2k^3 - k + 2 = 0 ).But since the user might expect a boxed answer, and multiple answers are allowed in boxes, perhaps:The real triplets are ( boxed{(1, 1, 1)} ), ( boxed{(r, r, r)} ), and ( boxed{(s, s, s)} ), where ( r ) and ( s ) are the other real roots of ( k^5 - 2k^3 - k + 2 = 0 ).But without knowing r and s explicitly, this is acceptable. However, in many math competitions, if a system reduces to a single equation with multiple real roots, all solutions are mentioned in this way. Therefore, this is the most precise answer possible.

✅:To solve the system of equations:[begin{cases}x^5 = 2y^3 + y - 2 y^5 = 2z^3 + z - 2 z^5 = 2x^3 + x - 2end{cases}]we first consider the possibility of symmetric solutions where ( x = y = z = k ). Substituting ( x = y = z = k ) into the equations, we get:[k^5 = 2k^3 + k - 2]Rearranging this equation, we have:[k^5 - 2k^3 - k + 2 = 0]Using the Rational Root Theorem, we find that ( k = 1 ) is a root. Factoring out ( (k - 1) ) from the polynomial, we get:[(k - 1)(k^4 + k^3 - k^2 - k - 2) = 0]The quartic polynomial ( k^4 + k^3 - k^2 - k - 2 ) is analyzed for real roots. Using Descartes' Rule of Signs and intermediate value theorem, we determine that there are two additional real roots, one positive and one negative. These roots are approximately ( k approx 1.293 ) and ( k approx -1.669 ).To check for non-symmetric solutions, we consider the cyclic nature of the system and the monotonicity of the functions involved. Since each function is strictly increasing, the system can only have symmetric solutions. Non-symmetric solutions would require cyclic permutations leading to contradictions in growth rates and inequalities.Thus, the only real solutions are the symmetric ones where ( x = y = z = k ). These solutions correspond to the real roots of the polynomial ( k^5 - 2k^3 - k + 2 = 0 ).The real triplets are:[boxed{(1, 1, 1)}]and the two other real solutions of the form:[boxed{(r, r, r)}]where ( r ) is a real root of the equation ( k^5 - 2k^3 - k + 2 = 0 ). These roots can be approximated numerically as ( r approx 1.293 ) and ( r approx -1.669 ).

🔑:1. Define the function and its properties: Let ( varphi(t) = 2t^3 + t - 2 ). We need to show that ( varphi(t) ) is strictly increasing. To do this, we compute its derivative: [ varphi'(t) = 6t^2 + 1 ] Since ( 6t^2 + 1 > 0 ) for all ( t in mathbb{R} ), ( varphi(t) ) is strictly increasing.2. Equating the functions: Given the system of equations: [ x^5 = 2y^3 + y - 2 ] [ y^5 = 2z^3 + z - 2 ] [ z^5 = 2x^3 + x - 2 ] We observe that if ( x = y = z ), then each equation reduces to: [ t^5 = 2t^3 + t - 2 ] Therefore, any real root ( r ) of the equation ( t^5 = 2t^3 + t - 2 ) will satisfy ( (r, r, r) ).3. Uniqueness of the solution: Assume ( x = max{x, y, z} ). Then ( x ge z ) implies ( x^5 ge z^5 ). Since ( t^5 ) is strictly increasing, we have: [ 2y^3 + y - 2 ge 2x^3 + x - 2 ] Since ( varphi(t) = 2t^3 + t - 2 ) is strictly increasing, it follows that: [ y ge x ] Thus, ( x = y ). Similarly, by considering ( y = max{x, y, z} ) and ( z = max{x, y, z} ), we can show that ( y = z ) and ( z = x ). Therefore, ( x = y = z ).4. Conclusion: The only solutions to the system are of the form ( (r, r, r) ), where ( r ) is a real root of the polynomial equation ( t^5 = 2t^3 + t - 2 ).The final answer is ( boxed{ (r, r, r) } ) where ( r ) is a real root of ( t^5 = 2t^3 + t - 2 ).

🧠:Okay, let's tackle this problem step by step. So, we have two polynomials P and Q with real coefficients, their degrees don't exceed n. They satisfy the identity P(x) x^{n+1} + Q(x) (x+1)^{n+1} = 1. We need to find all possible values of Q(-1/2). First, let me make sure I understand the problem correctly. We have an equation involving two polynomials, P and Q, each of degree at most n. The equation is P(x) multiplied by x^{n+1} plus Q(x) multiplied by (x+1)^{n+1} equals 1. And we need to figure out what values Q(-1/2) can take. Hmm, so this is an identity, meaning it's true for all real numbers x. So, the left-hand side must simplify to the constant polynomial 1. Therefore, all the coefficients of x^k for k ≥ 1 must be zero, and the constant term must be 1. But maybe there's a better way to approach this. Since we have an identity that holds for all x, perhaps we can use some properties of polynomials or some substitution to find Q(-1/2). Wait, the polynomials P and Q are of degree at most n, but they are multiplied by x^{n+1} and (x+1)^{n+1}, respectively. Let me think about the degrees here. If P is degree at most n, then P(x) x^{n+1} would be degree at most n + (n+1) = 2n + 1. Similarly, Q(x) (x+1)^{n+1} would be degree at most n + (n+1) = 2n + 1. So the sum is degree at most 2n + 1. But the sum equals 1, which is a constant. Therefore, all coefficients of x^k for k ≥ 1 must cancel out, leaving only the constant term 1. So, this equation is essentially saying that the two polynomials P(x) x^{n+1} and Q(x) (x+1)^{n+1} are additive inverses for all terms except the constant term, which sums to 1. But how do we find Q(-1/2)? Maybe we can plug in specific values of x into the equation to get some equations involving P(x) and Q(x). Let's see.If we plug x = 0 into the equation, we get P(0) * 0^{n+1} + Q(0) * 1^{n+1} = 1. Since 0^{n+1} is 0, this simplifies to Q(0) = 1. So, Q(0) must equal 1. That's one value we know. Similarly, if we plug x = -1, we get P(-1) * (-1)^{n+1} + Q(-1) * 0^{n+1} = 1. The second term is 0, so we get P(-1) * (-1)^{n+1} = 1. Therefore, P(-1) = (-1)^{n+1}. So that's another condition. But we need Q(-1/2). Maybe we can plug x = -1/2 into the equation? Let's try that. Plugging x = -1/2 into the equation:P(-1/2) * (-1/2)^{n+1} + Q(-1/2) * (-1/2 + 1)^{n+1} = 1.Simplify the second term: (-1/2 + 1) = 1/2, so (1/2)^{n+1}. Therefore, the equation becomes:P(-1/2) * (-1/2)^{n+1} + Q(-1/2) * (1/2)^{n+1} = 1.Hmm, so that's an equation involving P(-1/2) and Q(-1/2). If we can relate P(-1/2) to Q(-1/2), maybe we can solve for Q(-1/2). But how?Alternatively, perhaps we can consider the original equation as a Bézout's identity. The equation P(x) x^{n+1} + Q(x) (x+1)^{n+1} = 1 suggests that the greatest common divisor (gcd) of x^{n+1} and (x+1)^{n+1} is 1, which is true since x and x+1 are coprime. Therefore, there exist polynomials P(x) and Q(x) such that this identity holds. But the problem states that the degrees of P and Q are at most n. So, we need to find such P and Q, and then evaluate Q at -1/2.Wait, Bézout's identity in polynomial rings tells us that if two polynomials are coprime, there exist polynomials such that a linear combination equals their gcd. Here, since x^{n+1} and (x+1)^{n+1} are coprime (as x and x+1 are coprime), their gcd is 1, so such P and Q exist. But the question is about all possible values of Q(-1/2). Hmm, so there might be multiple pairs (P, Q) satisfying the equation, each giving a different value for Q(-1/2). Wait, but Bézout coefficients are unique modulo the ideal generated by the other polynomial. Wait, maybe not. Wait, in Bézout's identity, the coefficients are not unique. So, there might be infinitely many pairs (P, Q) satisfying the equation, but with the constraint that the degrees of P and Q are at most n. Hmm, but how many such pairs are there?Wait, let's think in terms of linear algebra. The equation P(x) x^{n+1} + Q(x) (x+1)^{n+1} = 1 must hold for all x. Since polynomials are determined by their coefficients, this equation gives a system of equations for the coefficients of P and Q. Let me denote P(x) = a_0 + a_1 x + ... + a_n x^n and Q(x) = b_0 + b_1 x + ... + b_n x^n. Then, when we multiply P(x) by x^{n+1}, the resulting polynomial has terms from x^{n+1} up to x^{2n+1}, and similarly for Q(x) multiplied by (x+1)^{n+1}. Then, the sum of these two products must equal 1, so all coefficients of x^k for k ≥ 1 must be zero, and the constant term must be 1. But this seems complicated because expanding (x+1)^{n+1} would give cross terms. However, since the left-hand side must equal 1, a constant polynomial, all coefficients of x^k for k ≥ 1 in the expansion must be zero. Therefore, this gives a system of equations for the coefficients of P and Q. But since the problem asks for Q(-1/2), maybe there's a smarter way to approach this rather than solving the entire system. Let me think.Another approach: Since the equation is P(x) x^{n+1} + Q(x) (x+1)^{n+1} = 1, perhaps we can consider this equation modulo (x+1)^{n+1} or modulo x^{n+1}. Wait, that's similar to modular inverses. Let me explain. If we take the equation modulo x^{n+1}, then the term P(x) x^{n+1} is congruent to 0 modulo x^{n+1}, so we have Q(x) (x+1)^{n+1} ≡ 1 mod x^{n+1}. Therefore, Q(x) must be the inverse of (x+1)^{n+1} modulo x^{n+1}. Similarly, if we take modulo (x+1)^{n+1}, then Q(x) (x+1)^{n+1} is 0 modulo that, so P(x) x^{n+1} ≡ 1 mod (x+1)^{n+1}, meaning P(x) is the inverse of x^{n+1} modulo (x+1)^{n+1}. But how does this help us find Q(-1/2)? Maybe we can use polynomial division or the extended Euclidean algorithm? But since we have modulus x^{n+1} and (x+1)^{n+1}, which are both high degree polynomials, this might be tricky.Alternatively, since Q(x) is the inverse of (x+1)^{n+1} modulo x^{n+1}, we can write Q(x) (x+1)^{n+1} ≡ 1 mod x^{n+1}. Therefore, Q(x) ≡ [(x+1)^{n+1}]^{-1} mod x^{n+1}. To compute the inverse of (x+1)^{n+1} modulo x^{n+1}, we can use the expansion of (x+1)^{n+1}. Let's compute (x+1)^{n+1} = sum_{k=0}^{n+1} binom{n+1}{k} x^k. However, modulo x^{n+1}, this is just the same as the polynomial itself, but when we take the inverse, we need a polynomial Q(x) of degree at most n such that Q(x) * (x+1)^{n+1} ≡ 1 mod x^{n+1}. But how do we find such an inverse? Perhaps we can use the fact that (x+1)^{n+1} = x^{n+1} + binom{n+1}{1} x^n + ... + 1. But modulo x^{n+1}, this becomes the polynomial sum_{k=0}^n binom{n+1}{k} x^k. Let's denote this as S(x) = sum_{k=0}^n binom{n+1}{k} x^k. Then, we need to find Q(x) such that Q(x) S(x) ≡ 1 mod x^{n+1}. Since S(x) is a polynomial of degree n, we can find its inverse modulo x^{n+1} using the extended Euclidean algorithm, but maybe there's a pattern here. Alternatively, consider that S(x) = (x + 1)^{n+1} - x^{n+1}, but actually, S(x) is exactly (x + 1)^{n+1} truncated at degree n. Hmm. Maybe we can use the inverse of (1 + x) modulo x^{n+1}. Wait, but (1 + x) is a degree 1 polynomial. However, here we have (1 + x)^{n+1}, which is a higher power.Alternatively, note that (1 + x)^{n+1} ≡ S(x) mod x^{n+1}, so to find the inverse of S(x), perhaps we can write it as (1 + x)^{n+1} divided by something? Wait, no. Let me think differently.If we let’s consider the inverse of (1 + x) modulo x^{n+1}. The inverse of (1 + x) is 1 - x + x^2 - x^3 + ... ± x^n, since (1 + x)(1 - x + x^2 - ... + (-1)^n x^n) = 1 + (-1)^{n} x^{n+1} ≡ 1 mod x^{n+1}. So, the inverse of (1 + x) modulo x^{n+1} is the alternating sum up to x^n. But here, we need the inverse of (1 + x)^{n+1} modulo x^{n+1}. So, perhaps we can take the inverse of (1 + x) and raise it to the (n+1)th power modulo x^{n+1}. Since inverses modulo x^{n+1} can be multiplied, so if we let T(x) = 1 - x + x^2 - x^3 + ... + (-1)^n x^n, then T(x) is the inverse of (1 + x) modulo x^{n+1}. Therefore, T(x)^{n+1} would be the inverse of (1 + x)^{n+1} modulo x^{n+1}. But computing T(x)^{n+1} modulo x^{n+1} seems complicated. Maybe there's a better way. Alternatively, since (1 + x)^{n+1} * Q(x) ≡ 1 mod x^{n+1}, then Q(x) ≡ (1 + x)^{-(n+1)} mod x^{n+1}. The expansion of (1 + x)^{-(n+1)} as a power series is given by the binomial theorem for negative exponents, which would be an infinite series, but since we are working modulo x^{n+1}, we can truncate it at x^n. The expansion of (1 + x)^{-(n+1)} is sum_{k=0}^{infty} (-1)^k binom{n + k}{k} x^k. Truncating this at x^n gives Q(x) = sum_{k=0}^n (-1)^k binom{n + k}{k} x^k. Therefore, Q(x) is this polynomial. Then, once we have Q(x), we can evaluate it at x = -1/2. Wait, but is this correct? Let me check for a small n. Let’s take n = 1. Then, (1 + x)^2 = 1 + 2x + x^2. Modulo x^2, this is 1 + 2x. The inverse of 1 + 2x modulo x^2 should be 1 - 2x, since (1 + 2x)(1 - 2x) = 1 - 4x^2 ≡ 1 mod x^2. But according to the formula above, (1 + x)^{-2} expansion would be sum_{k=0}^{infty} (-1)^k binom{2 + k - 1}{k} x^k? Wait, actually, the expansion for (1 + x)^{-m} is sum_{k=0}^{infty} (-1)^k binom{m + k - 1}{k} x^k. So for m = n + 1 = 2, it's sum_{k=0}^{infty} (-1)^k binom{2 + k - 1}{k} x^k = sum_{k=0}^{infty} (-1)^k binom{k + 1}{k} x^k = sum_{k=0}^{infty} (-1)^k (k + 1) x^k. Truncating at k = 1 (since n=1), we get 1 - 2x. Which matches the inverse we found earlier. So, yes, this seems correct. Therefore, in general, Q(x) is the truncation of the series expansion of (1 + x)^{-(n+1)} up to degree n. Therefore, Q(x) = sum_{k=0}^n (-1)^k binom{n + k}{k} x^k. Then, to find Q(-1/2), we substitute x = -1/2 into this polynomial. Let's compute that:Q(-1/2) = sum_{k=0}^n (-1)^k binom{n + k}{k} (-1/2)^k = sum_{k=0}^n binom{n + k}{k} (1/2)^k.Because (-1)^k * (-1/2)^k = (1/2)^k. Therefore, Q(-1/2) is the sum from k=0 to n of binom{n + k}{k} (1/2)^k. Hmm, so the question is asking for all possible values of Q(-1/2), but according to this, Q(x) is uniquely determined as the inverse of (1 + x)^{n+1} modulo x^{n+1}, so Q(-1/2) is uniquely determined. Therefore, there is only one possible value. But the problem says "determine all possible values," which suggests that there might be multiple, but according to this reasoning, it's unique. Wait, but earlier I thought that Bézout coefficients are not unique, but perhaps under the degree constraints (degrees of P and Q at most n), the coefficients are unique. Let me check. Suppose we have two different pairs (P1, Q1) and (P2, Q2) satisfying the equation. Then, (P1 - P2) x^{n+1} + (Q1 - Q2) (x + 1)^{n+1} = 0. Therefore, (P1 - P2) x^{n+1} = - (Q1 - Q2) (x + 1)^{n+1}. But the left-hand side is a polynomial multiple of x^{n+1}, and the right-hand side is a polynomial multiple of (x + 1)^{n+1}. Since x^{n+1} and (x + 1)^{n+1} are coprime (their gcd is 1), the only way this equality holds is if both sides are zero. Therefore, P1 - P2 = 0 and Q1 - Q2 = 0. Hence, the solution is unique. Therefore, Q is uniquely determined, so Q(-1/2) is uniquely determined. Therefore, there's only one possible value. Therefore, the answer is the sum from k=0 to n of binom{n + k}{k} (1/2)^k. But let me verify this for a small n. Let's take n = 1. Then, Q(x) = 1 - 2x (from before). Then Q(-1/2) = 1 - 2*(-1/2) = 1 + 1 = 2. Alternatively, compute the sum: sum_{k=0}^1 binom{1 + k}{k} (1/2)^k. For k=0: binom{1}{0} (1/2)^0 = 1. For k=1: binom{2}{1} (1/2)^1 = 2*(1/2) = 1. Sum is 1 + 1 = 2. Which matches. Another example, n = 0. Then, the original equation is P(x) x^{1} + Q(x) (x + 1)^1 = 1. But since n=0, P and Q are constants. Therefore, P x + Q (x + 1) = 1. Since this must hold for all x, the coefficients of x and the constant term must match. Coefficient of x: P + Q = 0. Constant term: Q = 1. Therefore, Q = 1, then P = -1. Then Q(-1/2) = 1. According to the formula: sum_{k=0}^0 binom{0 + 0}{0} (1/2)^0 = 1. So, correct. Another test, n = 2. Then Q(x) is the inverse of (1 + x)^3 mod x^3. Let's compute (1 + x)^3 = 1 + 3x + 3x^2 + x^3. Modulo x^3, this is 1 + 3x + 3x^2. The inverse of this modulo x^3 can be found by solving (1 + 3x + 3x^2)(a + bx + cx^2) ≡ 1 mod x^3. Multiplying out:a(1) + (3a + b)x + (3a + 3b + c)x^2 + (3b + 3c)x^3 + (3c)x^4. But modulo x^3, the x^3 and x^4 terms disappear. So:Constant term: a = 1Coefficient of x: 3a + b = 0 ⇒ 3*1 + b = 0 ⇒ b = -3Coefficient of x^2: 3a + 3b + c = 0 ⇒ 3*1 + 3*(-3) + c = 0 ⇒ 3 - 9 + c = 0 ⇒ c = 6Therefore, inverse is 1 - 3x + 6x^2. Therefore, Q(x) = 1 - 3x + 6x^2. Then Q(-1/2) = 1 - 3*(-1/2) + 6*(1/4) = 1 + 3/2 + 6/4 = 1 + 1.5 + 1.5 = 4. Using the formula: sum_{k=0}^2 binom{2 + k}{k} (1/2)^k.k=0: binom{2}{0} (1/2)^0 = 1k=1: binom{3}{1} (1/2)^1 = 3*(1/2) = 1.5k=2: binom{4}{2} (1/2)^2 = 6*(1/4) = 1.5Sum: 1 + 1.5 + 1.5 = 4. Correct. Therefore, the formula seems to hold. Therefore, Q(-1/2) = sum_{k=0}^n binom{n + k}{k} (1/2)^k. But can we write this sum in a closed-form expression? Let's recall that the sum sum_{k=0}^n binom{n + k}{k} x^k is equal to frac{1}{(1 - x)^{n + 1}} } minus the sum from k=n+1 to infinity binom{n + k}{k} x^k. But since we are truncating at k=n, it's not the full generating function. However, perhaps there's a generating function identity here. Wait, the generating function for sum_{k=0}^infty binom{n + k}{k} x^k = frac{1}{(1 - x)^{n + 1}}}. So, the infinite sum is 1/(1 - x)^{n + 1}. But we need the sum up to k=n. However, Q(x) is the truncation of the inverse of (1 + x)^{n + 1} modulo x^{n + 1}, which corresponds to the first n + 1 terms of the generating function. However, evaluating Q(-1/2) gives sum_{k=0}^n binom{n + k}{k} (1/2)^k. Is there a closed-form expression for this finite sum? Hmm. Alternatively, since the infinite sum is 1/(1 - 1/2)^{n + 1} } = 1/(1/2)^{n + 1} } = 2^{n + 1}. But the finite sum up to k=n is 2^{n + 1} minus the sum from k=n+1 to infinity binom{n + k}{k} (1/2)^k. So, maybe Q(-1/2) = 2^{n + 1} - something. But I don't know if that's helpful. Alternatively, note that when we plug x = 1/2 into the generating function, we get sum_{k=0}^infty binom{n + k}{k} (1/2)^k = 2^{n + 1}. But our sum is only up to k=n, so the difference is the tail starting from k=n+1. But how much is that? Alternatively, maybe there's a combinatorial interpretation. Wait, binom{n + k}{k} is the number of ways to distribute k indistinguishable items into n + 1 distinguishable boxes, which is also the number of multisets of size k from a set of size n + 1. But I don't see an immediate connection. Alternatively, using generating functions for finite sums. Let me consider that Q(-1/2) = sum_{k=0}^n binom{n + k}{k} (1/2)^k. Let’s see for n=0: 1. For n=1: 1 + 2*(1/2) = 2. For n=2: 1 + 3*(1/2) + 6*(1/4) = 1 + 1.5 + 1.5 = 4. For n=3: Let's compute the sum. binom{3+0}{0}(1/2)^0 = 1, binom{4}{1}(1/2)^1 = 4*(1/2) = 2, binom{5}{2}(1/2)^2 = 10*(1/4) = 2.5, binom{6}{3}(1/2)^3 = 20*(1/8) = 2.5. Sum is 1 + 2 + 2.5 + 2.5 = 8. So, n=0:1=2^0, n=1:2=2^1, n=2:4=2^2, n=3:8=2^3. So, the pattern seems to be Q(-1/2) = 2^n. Wait, but for n=0, Q(-1/2)=1=2^0, n=1:2=2^1, n=2:4=2^2, n=3:8=2^3. So, this suggests that Q(-1/2) = 2^n. But wait, according to our previous examples, for n=2, we got 4=2^2, for n=3, 8=2^3. But according to the sum formula, for n=3, sum_{k=0}^3 binom{3 + k}{k} (1/2)^k = 1 + 4*(1/2) + 10*(1/4) + 20*(1/8) = 1 + 2 + 2.5 + 2.5 = 8. So, yes, 2^3. Similarly, for n=0: 1=2^0. So, this suggests that in general, Q(-1/2) = 2^n. Wait, this contradicts the earlier thought that the sum equals 2^{n+1} minus something, but in reality, the finite sum equals 2^n. Wait, but for n=0, 2^{0} =1, which matches. For n=1, 2^{1}=2, which matches. For n=2, 4=2^2, and for n=3, 8=2^3. So, this suggests that Q(-1/2) = 2^n. But wait, according to the infinite generating function, sum_{k=0}^infty binom{n + k}{k} (1/2)^k = 2^{n + 1}. But our finite sum is sum_{k=0}^n binom{n + k}{k} (1/2)^k = 2^n. Therefore, the tail sum_{k=n+1}^infty binom{n + k}{k} (1/2)^k = 2^{n + 1} - 2^n = 2^n. Interesting. So, if we can show that sum_{k=0}^n binom{n + k}{k} (1/2)^k = 2^n, then Q(-1/2) = 2^n. Let me try to prove this identity. Let’s consider generating functions. Let’s define S(n) = sum_{k=0}^n binom{n + k}{k} (1/2)^k. We need to show that S(n) = 2^n. We can use induction. For n=0, S(0)=1=2^0. Assume it holds for n=m: S(m)=2^m. Now, for n=m+1:S(m+1) = sum_{k=0}^{m+1} binom{(m+1) + k}{k} (1/2)^k. We need to relate this to S(m). Let’s see:Note that binom{(m+1) + k}{k} = binom{m + k + 1}{k} = binom{m + k + 1}{m + 1}. There's an identity for binomial coefficients: binom{m + k + 1}{m + 1} = binom{m + k}{m + 1} + binom{m + k}{m}. But I don't know if that helps. Alternatively, consider the generating function for S(n). Since S(n) = sum_{k=0}^n binom{n + k}{k} (1/2)^k, let’s write the generating function G(x) = sum_{n=0}^infty S(n) x^n. But maybe another approach. Let’s consider that S(n) satisfies a recurrence relation. For example, note that S(n+1) = sum_{k=0}^{n+1} binom{n +1 + k}{k} (1/2)^k. Let’s split the sum into k=0 to k=n+1. Notice that binom{n + 1 + k}{k} = binom{n + k}{k} + binom{n + k}{k - 1} (using the identity binom{m + 1}{k} = binom{m}{k} + binom{m}{k - 1}). Therefore,S(n+1) = sum_{k=0}^{n+1} [ binom{n + k}{k} + binom{n + k}{k - 1} ] (1/2)^k= sum_{k=0}^{n+1} binom{n + k}{k} (1/2)^k + sum_{k=0}^{n+1} binom{n + k}{k - 1} (1/2)^k= sum_{k=0}^{n+1} binom{n + k}{k} (1/2)^k + sum_{k=1}^{n+1} binom{n + k}{k - 1} (1/2)^k= sum_{k=0}^{n+1} binom{n + k}{k} (1/2)^k + sum_{j=0}^{n} binom{n + j + 1}{j} (1/2)^{j + 1} (letting j = k - 1)= [sum_{k=0}^n binom{n + k}{k} (1/2)^k + binom{2n + 1}{n + 1} (1/2)^{n + 1}] + (1/2) sum_{j=0}^n binom{n + j + 1}{j} (1/2)^jBut S(n) = sum_{k=0}^n binom{n + k}{k} (1/2)^k = 2^n (by the induction hypothesis). Therefore,S(n+1) = 2^n + binom{2n + 1}{n + 1} (1/2)^{n + 1} + (1/2) sum_{j=0}^n binom{n + j + 1}{j} (1/2)^jBut the sum sum_{j=0}^n binom{n + j + 1}{j} (1/2)^j is similar to S(n+1), but not exactly. Wait, perhaps there's a better way. Let's consider the original recurrence. Alternatively, note that S(n+1) = sum_{k=0}^{n+1} binom{n+1+k}{k} (1/2)^k. Let’s shift the index: let m = k. Then,S(n+1) = sum_{m=0}^{n+1} binom{n+1 + m}{m} (1/2)^m. This looks similar to the original definition. But how does it relate to S(n)? Alternatively, use generating functions. Let’s consider that the generating function for S(n) is sum_{n=0}^infty S(n) x^n = sum_{n=0}^infty [sum_{k=0}^n binom{n + k}{k} (1/2)^k] x^n. But swapping the sums:= sum_{k=0}^infty (1/2)^k sum_{n=k}^infty binom{n + k}{k} x^n. Let’s set m = n - k, so n = m + k. Then, = sum_{k=0}^infty (1/2)^k sum_{m=0}^infty binom{m + 2k}{k} x^{m + k}= sum_{k=0}^infty (1/2 x)^k sum_{m=0}^infty binom{m + 2k}{k} x^m. But the inner sum is sum_{m=0}^infty binom{m + 2k}{k} x^m. Wait, binom{m + 2k}{k} is the number of ways to choose k items from m + 2k, which doesn't immediately suggest a generating function. Alternatively, we can use generating functions for binomial coefficients. Recall that sum_{m=0}^infty binom{m + r}{r} x^m = 1/(1 - x)^{r + 1}}. But here, we have sum_{m=0}^infty binom{m + 2k}{k} x^m. Let’s let r = 2k, then sum_{m=0}^infty binom{m + r}{r} x^m = 1/(1 - x)^{r + 1} = 1/(1 - x)^{2k + 1}}. But in our case, the binomial coefficient is binom{m + 2k}{k}, not binom{m + 2k}{2k}. Wait, binom{m + 2k}{k} is different. For example, if m=0, it's binom{2k}{k}; for m=1, binom{2k +1}{k}; etc. This is the central binomial coefficient with some offset. I don't recall a standard generating function for this. Alternatively, perhaps there's a generating function identity for sum_{m=0}^infty binom{m + 2k}{k} x^m. Let’s try to find it. Let’s denote this sum as C_k(x). Consider that binom{m + 2k}{k} is the number of lattice paths from (0,0) to (m + k, k) without crossing the line y = x, but I might be mixing up things. Alternatively, using generating functions for such coefficients. Alternatively, perhaps use generating functions in terms of hypergeometric functions, but that might be complicated. Alternatively, notice that S(n) = 2^n fits with the examples we checked. If we assume S(n) = 2^n, then S(n+1) = 2^{n+1}, which is twice S(n). Let’s check if the recurrence holds. If S(n+1) = 2 S(n), then since S(n) = 2^n, S(n+1) = 2^{n+1} = 2 * 2^n. Let’s see if our previous computation for n=1 to n=2 holds. S(1) = 2, S(2) = 4 = 2*2. Yes. Similarly, S(2) = 4, S(3)=8=2*4. So, the recurrence seems to hold. Therefore, if we can show that S(n+1) = 2 S(n), then by induction, S(n) = 2^n. Let’s try to show that. From the definition:S(n+1) = sum_{k=0}^{n+1} binom{n+1 + k}{k} (1/2)^kWe want to relate this to S(n) = sum_{k=0}^n binom{n + k}{k} (1/2)^k.Let’s write S(n+1) as sum_{k=0}^{n+1} binom{n+1 +k}{k} (1/2)^k. Using the identity binom{n+1 +k}{k} = binom{n +k}{k} + binom{n +k}{k -1}, we have:S(n+1) = sum_{k=0}^{n+1} [binom{n +k}{k} + binom{n +k}{k -1}] (1/2)^k= sum_{k=0}^{n+1} binom{n +k}{k} (1/2)^k + sum_{k=0}^{n+1} binom{n +k}{k -1} (1/2)^kThe first sum is sum_{k=0}^{n+1} binom{n +k}{k} (1/2)^k. Note that when k = n+1, binom{n + (n+1)}{n+1} = binom{2n +1}{n+1}. So this sum is S(n) + binom{2n +1}{n+1} (1/2)^{n+1}.The second sum is sum_{k=0}^{n+1} binom{n +k}{k -1} (1/2)^k. When k=0, binom{n +0}{-1}=0, so we can start the sum from k=1. Then, let j = k -1:sum_{j=0}^n binom{n + j +1}{j} (1/2)^{j +1} = (1/2) sum_{j=0}^n binom{n +j +1}{j} (1/2)^jBut binom{n +j +1}{j} = binom{n +j +1}{n +1}. This is the same as the coefficients in S(n+1), but shifted. However, if we consider that sum_{j=0}^n binom{n +j +1}{j} (1/2)^j = sum_{j=0}^n binom{n +j +1}{n +1} (1/2)^j = sum_{j=0}^n binom{n +j +1}{n +1} (1/2)^jBut this is the same as sum_{m=n+1}^{2n+1} binom{m}{n+1} (1/2)^{m -n -1 + j}... Hmm, not sure. Alternatively, notice that sum_{j=0}^n binom{n +j +1}{n +1} (1/2)^j = sum_{j=0}^n binom{n +j +1}{j} (1/2)^j = ?Wait, but if we look back at our previous examples, when we computed S(n+1), the second sum turned out to be equal to S(n). For example, when n=1, S(n+1) = S(2) = 4. The second sum was (1/2) sum_{j=0}^1 binom{1 +j +1}{j} (1/2)^j = (1/2)[ binom{2}{0}(1/2)^0 + binom{3}{1}(1/2)^1 ] = (1/2)[1 + 3*(1/2)] = (1/2)(1 + 1.5) = (1/2)(2.5) = 1.25. Then S(n+1) = S(1) + binom{3}{2}(1/2)^2 + 1.25 = 2 + 3*(1/4) + 1.25 = 2 + 0.75 + 1.25 = 4. Which matches 2*2=4. But how does this generalize? Let’s try to see if the second sum is equal to S(n). If that's the case, then:S(n+1) = S(n) + binom{2n +1}{n+1} (1/2)^{n+1} + (1/2) sum_{j=0}^n binom{n +j +1}{j} (1/2)^jBut if sum_{j=0}^n binom{n +j +1}{j} (1/2)^j = 2^{n+1} (which would be S(n+1) if it were up to j=n+1), but we only go up to j=n. Wait, this might not be straightforward. Alternatively, if we accept that the pattern holds for the examples and assume by induction that S(n) = 2^n, then S(n+1) = 2^{n+1}. Therefore, the answer is Q(-1/2) = 2^n. Therefore, the possible value of Q(-1/2) is uniquely determined as 2^n. Hence, the only possible value is 2^n, so the answer is boxed{2^n}.

✅:To determine all possible values of ( Qleft(-frac{1}{2}right) ) given the identity ( P(x) x^{n + 1} + Q(x) (x+1)^{n + 1} = 1 ) where ( P ) and ( Q ) are polynomials of degree at most ( n ), we proceed as follows:1. Understanding the Identity: The equation must hold for all ( x ), implying that the left-hand side must simplify to the constant polynomial 1. This requires the coefficients of all non-constant terms to cancel out.2. Modular Inverses: By considering the equation modulo ( x^{n+1} ), we find that ( Q(x) ) must be the inverse of ( (x+1)^{n+1} ) modulo ( x^{n+1} ). This inverse can be expressed as a polynomial truncated from the series expansion of ( (1+x)^{-(n+1)} ).3. Series Expansion: The inverse polynomial ( Q(x) ) is given by the truncation of the series expansion of ( (1+x)^{-(n+1)} ) up to degree ( n ). This results in ( Q(x) = sum_{k=0}^n (-1)^k binom{n + k}{k} x^k ).4. Evaluating at ( x = -frac{1}{2} ): [ Qleft(-frac{1}{2}right) = sum_{k=0}^n (-1)^k binom{n + k}{k} left(-frac{1}{2}right)^k = sum_{k=0}^n binom{n + k}{k} left(frac{1}{2}right)^k ] This sum simplifies to ( 2^n ) through induction and pattern recognition from small cases, confirming that the finite sum matches the closed-form expression.5. Uniqueness: The constraints on the degrees of ( P ) and ( Q ) ensure the solution is unique, thus ( Qleft(-frac{1}{2}right) ) is uniquely determined.Final Answer: (boxed{2^n})

🔑:1. Uniqueness of Polynomials ( P ) and ( Q ): - Suppose there exist two pairs of polynomials ((P_1, Q_1)) and ((P_2, Q_2)) such that: [ P_1(x)x^{n+1} + Q_1(x)(x+1)^{n+1} = 1 ] and [ P_2(x)x^{n+1} + Q_2(x)(x+1)^{n+1} = 1. ] - Subtracting these two equations, we get: [ (P_1(x) - P_2(x))x^{n+1} + (Q_1(x) - Q_2(x))(x+1)^{n+1} = 0. ] - Since ((x+1)^{n+1}) divides ((Q_1(x) - Q_2(x))(x+1)^{n+1}), it follows that: [ (x+1)^{n+1} mid (P_1(x) - P_2(x))x^{n+1}. ] - Because (x) and (x+1) are coprime, we have: [ (x+1)^{n+1} mid P_1(x) - P_2(x). ] - Since the degree of (P_1(x) - P_2(x)) is at most (n), the only way this can hold is if (P_1(x) - P_2(x) = 0), i.e., (P_1(x) = P_2(x)). - Similarly, we can show that (Q_1(x) = Q_2(x)). - Therefore, the polynomials (P) and (Q) are uniquely determined.2. Swapping Terms: - To find a relationship between (P) and (Q), let (x = -y - 1). Then: [ Q(-y-1)(-1)^{n+1}y^{n+1} + P(-y-1)(-y-1)^{n+1} = 1. ] - Simplifying, we get: [ Q(-y-1)(-1)^{n+1}y^{n+1} + P(-y-1)(-1)^{n+1}(y+1)^{n+1} = 1. ] - By the lemma, we have: [ P(x) = (-1)^{n+1}Q(-x-1) ] and [ Q(x) = (-1)^{n+1}P(-x-1). ]3. Evaluating at (x = -frac{1}{2}): - Setting (x = -frac{1}{2}), we get: [ Pleft(-frac{1}{2}right) = (-1)^{n+1}Qleft(-frac{1}{2}right). ] - Substituting into the original identity: [ Pleft(-frac{1}{2}right) left(-frac{1}{2}right)^{n+1} + Qleft(-frac{1}{2}right) left(frac{1}{2}right)^{n+1} = 1. ] - Using (Pleft(-frac{1}{2}right) = (-1)^{n+1}Qleft(-frac{1}{2}right)), we get: [ (-1)^{n+1}Qleft(-frac{1}{2}right) left(-frac{1}{2}right)^{n+1} + Qleft(-frac{1}{2}right) left(frac{1}{2}right)^{n+1} = 1. ] - Simplifying, we have: [ (-1)^{n+1}Qleft(-frac{1}{2}right) left(-frac{1}{2}right)^{n+1} + Qleft(-frac{1}{2}right) left(frac{1}{2}right)^{n+1} = 1. ] [ Qleft(-frac{1}{2}right) left[(-1)^{n+1} left(-frac{1}{2}right)^{n+1} + left(frac{1}{2}right)^{n+1}right] = 1. ] - Since ((-1)^{n+1} left(-frac{1}{2}right)^{n+1} = left(frac{1}{2}right)^{n+1}), we get: [ Qleft(-frac{1}{2}right) left[ left(frac{1}{2}right)^{n+1} + left(frac{1}{2}right)^{n+1} right] = 1. ] [ Qleft(-frac{1}{2}right) cdot 2 left(frac{1}{2}right)^{n+1} = 1. ] [ Qleft(-frac{1}{2}right) cdot 2^{-(n+1)} cdot 2 = 1. ] [ Qleft(-frac{1}{2}right) cdot 2^{-n} = 1. ] [ Qleft(-frac{1}{2}right) = 2^n. ]The final answer is ( boxed{2^n} ).