🧠:Okay, so I have this problem: a polynomial P(x) with integer coefficients takes the value 2 for four different integer values of x. I need to prove that P(x) can never be 1, 3, 5, 7, or 9 for any integer x. Hmm, let me think about how to approach this.First, I remember that if a polynomial with integer coefficients takes the same value for several integer arguments, there's a way to factor it. Specifically, if P(a) = P(b) for integers a and b, then (x - a)(x - b) divides P(x) - P(a). But since there are four different integers where P(x) = 2, maybe I can write P(x) - 2 as a product of four linear factors multiplied by another polynomial with integer coefficients. Let me write that down.Suppose the four integer roots where P(x) = 2 are a, b, c, d. Then P(x) - 2 = (x - a)(x - b)(x - c)(x - d)Q(x), where Q(x) is a polynomial with integer coefficients. Right, so P(x) = 2 + (x - a)(x - b)(x - c)(x - d)Q(x). Since all coefficients are integers, Q(x) must also have integer coefficients.Now, the problem states that we need to show that P(x) cannot take the values 1, 3, 5, 7, or 9 for any integer x. So, suppose for contradiction that there exists some integer k such that P(k) is one of these values. Let's take a general case: P(k) = m, where m is in {1, 3, 5, 7, 9}. Then substituting into the equation above:m = 2 + (k - a)(k - b)(k - c)(k - d)Q(k)Rearranging, we get:(k - a)(k - b)(k - c)(k - d)Q(k) = m - 2Since m is one of 1, 3, 5, 7, 9, then m - 2 is in {-1, 1, 3, 5, 7}. So the product of four integers (k - a), (k - b), etc., multiplied by Q(k), which is also an integer (since Q has integer coefficients and k is integer), equals to one of these numbers: -1, 1, 3, 5, 7.But here's the thing: the left-hand side is a product of four integers (from the differences) times another integer (Q(k)). So the entire left-hand side is an integer. The right-hand side is one of -1, 1, 3, 5, 7. So we need the product (k - a)(k - b)(k - c)(k - d)Q(k) to equal one of these numbers. Let me denote the product (k - a)(k - b)(k - c)(k - d) as N. Then N * Q(k) = m - 2.But since m - 2 is a small number in absolute value (the largest being 7), the product N * Q(k) must be small. However, N is a product of four integers. So unless each of the factors is ±1, the product N could be larger. But given that Q(k) is also an integer, the product N * Q(k) is constrained.Wait, perhaps we can use the fact that if we have four integers multiplied together, the possible divisors of m - 2 would be limited. Since m - 2 is one of -1, 1, 3, 5, 7, which are all primes except -1 and 1. So maybe each of these m - 2 values can't be expressed as a product of four integers multiplied by another integer unless some of the factors are ±1.Alternatively, think about the factors (k - a), (k - b), etc. Each of these is an integer. If we have four such factors multiplied together, their product is N. Then Q(k) is also an integer. So N * Q(k) is m - 2. So N divides m - 2. But m - 2 is a prime number or ±1. Therefore, the product N must divide m - 2. But since m - 2 is a prime or ±1, the possible divisors are limited. For example, if m - 2 = 3, then N could be ±1, ±3. But N is the product of four integers. So 3 can be written as the product of four integers? Let's see: 3 = 1*1*1*3, or -1*-1*1*3, etc. But the problem is that if N is 3, then Q(k) would have to be 1, but 3*1=3. Similarly, if N is 1, then Q(k) would be 3. Wait, but Q(k) is an integer. So in either case, possible.But maybe the key is that since we have four factors multiplied together, and m - 2 is either ±1 or a prime. Let's check each case:Case 1: m - 2 = -1. Then N * Q(k) = -1. So possible N and Q(k) are either (-1,1) or (1,-1). But N is a product of four integers. Let's see: if N = -1, then the product of four integers is -1. Each of those four integers must be ±1, but the product is -1. So that would require an odd number of -1s. For four factors, possible if 1, 1, 1, -1 or three -1s and one 1. Then Q(k) would be 1 or -1. But then, if N is -1, then (k - a)(k - b)(k - c)(k - d) = -1. Each (k - a), (k - b), etc., is ±1. So that would mean that k is adjacent to each of a, b, c, d. But since a, b, c, d are distinct integers, how can k be adjacent (i.e., differ by 1) to four distinct integers? For example, if k is 0, then the four numbers a, b, c, d would have to be 1 and -1, but there's only two possibilities. Wait, this seems impossible. Because you can't have four distinct integers all differing from k by 1 or -1. Because in integers, adjacent numbers differ by 1. So if k is an integer, the numbers adjacent to k are k + 1 and k - 1. But there are only two such numbers. So how can there be four distinct integers a, b, c, d such that each (k - a) is ±1? It's impossible. Therefore, N cannot be ±1 because that would require four distinct integers a, b, c, d all within ±1 of k, which is only possible if there are two values. Hence, N can't be ±1. Therefore, N * Q(k) = -1 is impossible, so m cannot be 1.Similarly, for m - 2 = 1, then N * Q(k) = 1. So N and Q(k) are both 1 or both -1. But similar reasoning applies: N would have to be 1 or -1. If N = 1, then (k - a)...(k - d) = 1. So each factor is 1 or -1, and even number of -1s. But again, to have four factors of ±1, each (k - a) is ±1. But as before, k can only be adjacent to two integers, so having four distinct a, b, c, d is impossible. Therefore, N can't be 1 or -1. So m can't be 3.Similarly, moving on to m - 2 = 3. Then N * Q(k) = 3. So possible N could be ±1, ±3, and Q(k) would be ±3 or ±1. Let's see if N = 3. Then (k - a)...(k - d) = 3. But 3 is a prime. The prime factorization of 3 into four integers. The only ways are 1*1*1*3, or -1*-1*1*3, etc. But then, the product of four integers is 3. So three of them are 1 and one is 3, or other combinations with negatives. But if (k - a), (k - b), (k - c), (k - d) are 1, 1, 1, 3, then the numbers a, b, c, d would be k - 1, k - 1, k - 1, k - 3. But a, b, c, d need to be distinct. So that's a problem because you have repeated values. Similarly, if you have two -1s and two -3s, but the product would be positive 9, which is not 3. Wait, maybe another combination. For example, 1, -1, -1, 3: product is 1*(-1)*(-1)*3 = 3. Then the numbers a, b, c, d would be k -1, k +1, k +1, k -3. But again, duplicates here. So even if you try to get different factors, you end up with duplicates among a, b, c, d, which contradicts the fact that they are four distinct integers. Therefore, N can't be 3. Similarly, if N = -3, then Q(k) = -1. But then (k - a)...(k - d) = -3. The same problem occurs: the factors would have to include an odd number of negative factors, but trying to split -3 into four integers with distinct a, b, c, d is impossible without duplication.Alternatively, if N = 1 and Q(k) = 3, but as before, N =1 requires the product of four integers, each ±1, but that's impossible due to the distinctness. Similarly for N = -1 and Q(k) = -3. So all possibilities lead to contradictions, meaning m - 2 = 3 is impossible. Therefore, m cannot be 5.Similarly, for m - 2 = 5. Then N * Q(k) = 5. So possible N values: ±1, ±5. Let's check N =5. Then (k - a)...(k - d) =5. The prime 5 can be written as 1*1*1*5, but again, duplicates in a, b, c, d. If we try with negative numbers: (-1)*(-1)*1*5=5, but same duplication. So again, duplicates. If N = -5, then Q(k) = -1. So (k - a)...(k - d) = -5. Then possible factors: (-1)*1*1*5, but duplicates. Again, impossible to have four distinct a, b, c, d. So N can't be ±5. If N = ±1, same as before, impossible. Hence, m - 2 =5 is impossible. So m can't be 7.Similarly, for m -2=7. N * Q(k) =7. Then N could be ±1, ±7. If N=7, then (k - a)...(k - d)=7. Again, factors would need to be 1*1*1*7, leading to duplicates. If N=-7, similar issue. N=±1 again impossible. So m=9 is ruled out.Therefore, in all cases, the assumption that P(k) is 1,3,5,7,9 leads to a contradiction because it would require four distinct integers a,b,c,d such that (k - a)(k - b)(k - c)(k - d) is a divisor of a small prime or ±1, which is impossible due to the necessity of having distinct factors without repetition.Alternatively, maybe there's another approach using the fact that for integer x, P(x) ≡ P(y) mod (x - y) if P has integer coefficients. But I think the first approach is solid. Let me check if I missed something.Wait, another angle: consider that P(x) - 2 has four distinct integer roots, so it's divisible by (x - a)(x - b)(x - c)(x - d). Then for any integer k, P(k) - 2 is divisible by (k - a)(k - b)(k - c)(k - d). Therefore, if we suppose that P(k) = m, then m - 2 must be divisible by the product (k - a)(k - b)(k - c)(k - d). But since m - 2 is one of -1,1,3,5,7, the only divisors are ±1 and the prime itself. However, the product of four integers (k - a)...(k - d) must divide m - 2. But since m - 2 is small, the absolute value of the product must be 1 or the prime. But as we saw earlier, the product can't be ±1 because that would require four factors of ±1, leading to duplicate roots. And if the product is a prime, then it's impossible to factor it into four distinct integers without duplicates. Hence, such a product cannot exist, leading to a contradiction. Therefore, P(k) cannot be 1,3,5,7,9 for any integer k.Yes, this seems to cover all the cases. So the key idea is that the difference P(k) - 2 must be a product of four integers (differences) times another integer, which can't be achieved when the result is a small prime or ±1 due to the requirement of distinct factors.

✅:To prove that the polynomial ( P(x) ) with integer coefficients does not take the values 1, 3, 5, 7, or 9 for any integer ( x ), we proceed as follows:Step 1: Factorize ( P(x) - 2 )Since ( P(x) = 2 ) at four distinct integer values ( a, b, c, d ), we write:[P(x) - 2 = (x - a)(x - b)(x - c)(x - d)Q(x),]where ( Q(x) ) is a polynomial with integer coefficients. Thus,[P(x) = 2 + (x - a)(x - b)(x - c)(x - d)Q(x).]Step 2: Assume ( P(k) = m ) for ( m in {1, 3, 5, 7, 9} )For contradiction, suppose there exists an integer ( k ) such that ( P(k) = m ). Substituting into the polynomial:[m = 2 + (k - a)(k - b)(k - c)(k - d)Q(k).]Rearranging gives:[(k - a)(k - b)(k - c)(k - d)Q(k) = m - 2.]Here, ( m - 2 in {-1, 1, 3, 5, 7} ).Step 3: Analyze the Product of FactorsLet ( N = (k - a)(k - b)(k - c)(k - d) ), an integer product. Then:[N cdot Q(k) = m - 2.]Since ( m - 2 ) is small (a prime or ±1), ( N ) must divide ( m - 2 ). However:- If ( |m - 2| = 1 ): ( N = pm 1 ), requiring four distinct integers ( a, b, c, d ) such that each ( (k - a), (k - b), (k - c), (k - d) ) is ±1. This is impossible, as ( k ) can only differ by 1 from two distinct integers.- If ( |m - 2| ) is prime (3, 5, 7): ( N = pm |m - 2| ). Factoring a prime into four integers requires duplicates (e.g., ( 3 = 1 cdot 1 cdot 1 cdot 3 )), contradicting the distinctness of ( a, b, c, d ).Step 4: ConclusionAll cases lead to contradictions. Therefore, no integer ( k ) exists such that ( P(k) in {1, 3, 5, 7, 9} ).Final AnswerThe polynomial ( P(x) ) cannot take the values 1, 3, 5, 7, or 9 for any integer ( x ). (boxed{P(x) notin {1, 3, 5, 7, 9} text{ for any } x in mathbf{Z}})

🔑:1. Define and Consider Polynomial: Let ( P(x) ) be a polynomial with integer coefficients that takes the value 2 at four different integers ( a, b, c, text{ and } d ). Define a new polynomial ( Q(x) = P(x) - 2 ). 2. Polynomial Roots: Since ( P(a) = P(b) = P(c) = P(d) = 2 ), it follows that ( Q(a) = Q(b) = Q(c) = Q(d) = 0 ). Thus, ( a, b, c, d ) are roots of ( Q(x) ).3. Factoring using Roots: By the Factor Theorem (a specific case of the Remainder Theorem), we can factorize ( Q(x) ) as: [ Q(x) = (x-a)(x-b)(x-c)(x-d)R(x) ] where ( R(x) ) is a polynomial with integer coefficients resulting from the division. 4. Leading Coefficient: Since ( (x-a)(x-b)(x-c)(x-d) ) is a product of four first-degree polynomials and has leading coefficient 1, ( R(x) ) must likewise have integer coefficients to ensure ( Q(x) ) has integer coefficients.5. Properties of ( Q(x) ): Consider any integer ( x_0 ) different from ( a, b, c, text{ and } d ). Then: [ Q(x_0) = (x_0 - a)(x_0 - b)(x_0 - c)(x_0 - d)R(x_0) ] Since ( R(x_0) ) is an integer, ( Q(x_0) ) is the product of four distinct integers and hence is either zero or a composite number.6. Q(x) Cannot Be Specific Numbers: Since all products of four distinct non-zero integers result in composite numbers unless any factor is ( pm 1 ), ( Q(x) ) cannot be (-1, 1, 3, 5, 7). Therefore, absolute values of ( Q(x) ) cannot be (|1|, |3|, |5|, |7|).7. Conclusion: If ( Q(x) = P(x) - 2 ) cannot be (-1, 1, 3, 5, 7), then ( P(x) ) cannot equate to the values ( 1, 3, 5, 7, 9).[boxed{text{Hence, } P(x)text{ cannot equal } 1, 3, 5, 7, text{or } 9.}]

🧠:Okay, so I need to find the smallest positive integer n greater than 1 such that the mean of the squares of the first n natural numbers is an integer. Hmm, let me parse the question again. The mean here is defined as the square root of the average of the squares. Wait, the average of the squares is (1² + 2² + ... + n²)/n, and then we take the square root of that. And we need this square root to be an integer. So essentially, (1² + 2² + ... + n²)/n must be a perfect square.Right, so the formula for the sum of the squares of the first n natural numbers is n(n + 1)(2n + 1)/6. Therefore, the average is [n(n + 1)(2n + 1)/6]/n = (n + 1)(2n + 1)/6. So we have that (n + 1)(2n + 1)/6 must be a perfect square. Let me verify this step.Sum of squares formula: 1² + 2² + ... + n² = n(n + 1)(2n + 1)/6. Correct. Then dividing by n gives (n + 1)(2n + 1)/6. So indeed, that expression must be a perfect square. Let me denote this expression as K², so (n + 1)(2n + 1)/6 = K². Therefore, (n + 1)(2n + 1) = 6K². Our task is to find the smallest n > 1 for which this equation holds, where K is a positive integer.So we need to solve the equation (n + 1)(2n + 1) = 6K². Let me think about how to approach this. Since n and K are integers, we can look for integer solutions to this equation. Let's note that n + 1 and 2n + 1 are consecutive terms in a sense. Let me compute their greatest common divisor (gcd). Let d = gcd(n + 1, 2n + 1). Then d divides (2n + 1) - 2(n + 1) = 2n + 1 - 2n - 2 = -1. Therefore, gcd(n + 1, 2n + 1) = 1. So n + 1 and 2n + 1 are coprime.Since they are coprime and their product is 6K², each of them must be a square times a factor of 6. But since 6 factors into 2 * 3, and they are coprime, each of n + 1 and 2n + 1 must be of the form 2a² and 3b², or 3a² and 2b², or some multiple where each factor is accounted for. Wait, because their product is 6K², and they are coprime, so each must individually be a multiple of a square times the respective factors. Let me formalize this.Since n + 1 and 2n + 1 are coprime, they must each be proportional to squares times factors of 6. However, since their product is 6K², and they are coprime, one of them must be twice a square and the other three times a square. Because 6 is 2 * 3, and since they are coprime, one factor will take the 2 and the other the 3. So possible cases:Case 1: n + 1 = 2a² and 2n + 1 = 3b²Case 2: n + 1 = 3a² and 2n + 1 = 2b²Where a and b are positive integers. Let me check these cases.Starting with Case 1: n + 1 = 2a², so n = 2a² - 1. Then substituting into 2n + 1 = 3b²:2*(2a² - 1) + 1 = 4a² - 2 + 1 = 4a² - 1 = 3b²So 4a² - 1 = 3b². Let's rearrange this: 4a² - 3b² = 1. This is a Diophantine equation. Similarly, for Case 2:n + 1 = 3a², so n = 3a² - 1. Substitute into 2n + 1 = 2b²:2*(3a² - 1) + 1 = 6a² - 2 + 1 = 6a² - 1 = 2b²Hence, 6a² - 1 = 2b², which simplifies to 3a² - b² = 0.5. Wait, that can't be, since a and b are integers, the left-hand side must be an integer, but 0.5 is not an integer. Therefore, this case is impossible. Therefore, Case 2 leads to a contradiction, so only Case 1 is possible.So the only possible case is Case 1: 4a² - 3b² = 1. Now, we need to solve this equation for positive integers a and b, and then compute n = 2a² - 1.So solving 4a² - 3b² = 1. Let me see if I can find small integer solutions. Let's try small a and b.For a=1: 4*1 -3b² =1 → 4 -3b²=1 → 3b²=3 → b²=1 → b=1. So a=1, b=1 is a solution. Then n=2*(1)^2 -1=2-1=1. But n must be greater than 1. So next solution.We need to find the next solution. This equation is a form of Pell's equation. Let's see. 4a² -3b²=1. Pell-type equation. Let me write it as 4a² - 3b² =1. Let me rearrange:(2a)² - 3b² =1. Let x = 2a, then x² - 3b² =1. That's the Pell equation x² - 3b² =1. The minimal solution for this Pell equation is x=2, b=1, which gives a=1, which is the solution we already found. Then the subsequent solutions can be generated using the recurrence relations for Pell equations.Recall that for Pell equations x² - Dy² =1, solutions can be generated using the fundamental solution. For D=3, the minimal solution is (2,1). Then the solutions are generated by x + b√3 = (2 + √3)^k for k=1,2,3,...Therefore, the next solution is (2 + √3)^2 = (2 + √3)(2 + √3) = 4 +4√3 +3 =7 +4√3. Therefore, x=7, b=4. Then a=x/2=7/2. But a must be integer. Hmm, this is a problem. So perhaps the solutions for a and b aren't all generated by this method? Wait, perhaps we need to consider that x must be even, since x=2a. So the Pell equation solutions x must be even. Let's check.Starting with x=2, b=1 (k=1). Then next solution k=2: x=7, b=4. But x=7 is odd, which would make a=7/2, not integer. So this is invalid. Next solution: k=3: (2 + √3)^3. Let's compute that.(2 + √3)^3 = (2 + √3)*(7 +4√3) = 14 +8√3 +7√3 +4*3=14 +15√3 +12=26 +15√3. Therefore, x=26, b=15. Then a=26/2=13. So a=13, b=15. Check 4*(13)^2 -3*(15)^2=4*169 -3*225=676-675=1. Correct. So this is a valid solution. Therefore, next solution is a=13, b=15. Then n=2a² -1=2*(169)-1=338-1=337. So n=337. Let me check if this is the minimal n>1.Wait, but before that, is there any solution between a=1 and a=13? Let's check a=2: 4*4 -3b²=16 -3b²=1→3b²=15→b²=5, not integer.a=3: 4*9 -3b²=36 -3b²=1→3b²=35→b²=35/3, not integer.a=4:4*16=64 -3b²=1→3b²=63→b²=21→b=√21, not integer.a=5:4*25=100 -3b²=1→3b²=99→b²=33→not integer.a=6:4*36=144 -3b²=1→3b²=143→not integer.a=7:4*49=196 -3b²=1→3b²=195→b²=65→nope.a=8:4*64=256 -3b²=1→3b²=255→b²=85→nope.a=9:4*81=324 -3b²=1→3b²=323→323/3≈107.666, no.a=10:4*100=400 -3b²=1→3b²=399→b²=133→nope.a=11:4*121=484 -3b²=1→3b²=483→b²=161→nope.a=12:4*144=576 -3b²=1→3b²=575→575/3≈191.666, no.a=13:4*169=676 -3*225=676 -675=1. Yes, works.Therefore, the next solution after a=1 is a=13, giving n=337. Let me check if that works.Compute (n +1)(2n +1)/6. For n=337: (338)(675)/6. Let's compute that. 338 divided by 6 is 338/6=56.333… Wait, but let's do prime factorization:338=2*13² (since 13²=169, 2*169=338). 675=25*27=25*3³=5²*3³. So 338*675=2*13²*5²*3³. Divided by 6=2*3. So (2*13²*5²*3³)/(2*3)=13²*5²*3²=(13*5*3)²=(195)². Therefore, it is indeed a perfect square. So (n +1)(2n +1)/6 =195², so the square root is 195, which is an integer. Therefore, n=337 is a solution.Is there a smaller n>1? Since the next solution after a=1 is a=13, giving n=337, and all a in between 1 and 13 do not give integer b, then n=337 is the minimal solution. Therefore, the answer is 337.But wait, let me check if there are any other cases. Earlier, we dismissed Case 2 because it led to a non-integer equation, but let me double-check. Case 2: n +1=3a², 2n +1=2b². Then n=3a² -1. Substitute into 2n +1=2b²:2*(3a² -1)+1=6a² -2 +1=6a² -1=2b². So 6a² -1=2b². Rearranged as 6a² -2b²=1→2(3a² -b²)=1. But the left-hand side is even, and the right-hand side is 1, which is odd. Contradiction. Hence, no solution in this case. Therefore, only Case 1 is valid, and the minimal solution is indeed n=337.Therefore, after thorough analysis, the smallest positive integer n>1 is 337.Final AnswerThe smallest positive integer ( n ) is boxed{337}.

✅:To find the smallest positive integer ( n ) such that the mean of the squares of the first ( n ) natural numbers is an integer, we start by noting that the mean is given by the square root of the average of the squares. This requires that the expression (frac{(n + 1)(2n + 1)}{6}) be a perfect square. The sum of the squares of the first ( n ) natural numbers is (frac{n(n + 1)(2n + 1)}{6}). Dividing this by ( n ) gives the average (frac{(n + 1)(2n + 1)}{6}). We need this to be a perfect square, so we set up the equation:[frac{(n + 1)(2n + 1)}{6} = K^2]Multiplying both sides by 6, we get:[(n + 1)(2n + 1) = 6K^2]Since ( n + 1 ) and ( 2n + 1 ) are coprime, each must be a multiple of a square. We consider the cases where one factor is twice a square and the other is three times a square. We find that solving the equation ( 4a^2 - 3b^2 = 1 ) (a form of Pell's equation) leads us to the solutions. Using the Pell equation ( x^2 - 3y^2 = 1 ) where ( x = 2a ) and ( y = b ), the fundamental solution is ( (2, 1) ). Subsequent solutions are generated using powers of ( (2 + sqrt{3}) ). The next valid solution after ( a = 1 ) is ( a = 13 ), leading to ( n = 2a^2 - 1 = 2 times 13^2 - 1 = 337 ).Verification shows that (frac{(337 + 1)(2 times 337 + 1)}{6} = 195^2), confirming it is a perfect square. Thus, the smallest positive integer ( n ) is:[boxed{337}]

🔑:To find the smallest integer n > 1 such that the root mean square (RMS) of the first n natural numbers is an integer, we follow these steps:1. Define the RMS of the first n natural numbers: [ sqrt{frac{1^2 + 2^2 + cdots + n^2}{n}} = m quad text{where } m in mathbb{Z} ]2. Square both sides to eliminate the square root: [ frac{1^2 + 2^2 + cdots + n^2}{n} = m^2 ]3. Use the formula for the sum of squares of the first n natural numbers: [ sum_{k=1}^n k^2 = frac{n(n+1)(2n+1)}{6} ] Thus, the equation becomes: [ frac{n(n+1)(2n+1)}{6n} = m^2 ] This simplifies to: [ frac{(n+1)(2n+1)}{6} = m^2 ]4. Multiply both sides by 6 to clear the denominator: [ (n+1)(2n+1) = 6m^2 ]5. Notice that 6m^2 is an even number and 2n+1 is an odd number, so n+1 must be even. Hence, n is odd.6. Consider n in terms of multiples of 6: ( n = 6p + 3 ), ( n = 6p - 1 ), and ( n = 6p + 1 ). Case 1: ( n = 6p + 3 )1. Substitute ( n = 6p + 3 ): [ (6p + 4)(12p + 7) = 6m^2 ]2. Simplify: [ 72p^2 + 90p + 28 = 6m^2 ] There is no suitable p that satisfies this equation for m in mathbb{Z}, so there is no solution in this case. Case 2: ( n = 6p - 1 )1. Substitute ( n = 6p - 1 ): [ 6p(12p - 1) = 6m^2 ]2. Simplify: [ m^2 = p(12p - 1) ]3. Since p and 12p - 1 are coprime, both must be perfect squares. Suppose: [ p = s^2 quad text{and} quad 12p - 1 = t^2 ]4. Substitute ( p = s^2 ): [ t^2 = 12s^2 - 1 ] This simplifies to: [ t^2 = 4(3s^2 - 1) + 3 ] This equation cannot hold since t^2 equiv 0 text{ or } 1 pmod{4}, while the right-hand side is equiv 3 pmod{4}. Thus, there is no solution in this case. Case 3: ( n = 6p + 1 )1. Substitute ( n = 6p + 1 ): [ (6p + 2)(12p + 3) = 6m^2 ]2. Simplify: [ m^2 = (3p + 1)(4p + 1) ]3. Since 3p + 1 and 4p + 1 are coprime, both must be perfect squares. Suppose: [ 3p + 1 = u^2 quad text{and} quad 4p + 1 = v^2 ]4. Substitute ( 3p + 1 = u^2 ): [ p = frac{u^2 - 1}{3} ]5. Substitute ( 4p + 1 = v^2 ): [ 4left(frac{u^2 - 1}{3}right) + 1 = v^2 implies frac{4u^2 - 4 + 3}{3} = v^2 implies 4u^2 - 4 = 3v^2 ]6. Re-arrange: [ 4u^2 - 3v^2 = 1 ]7. It is evident that the smallest solution to this equation occurs when ( u=v=1 ): [ p = 0 quad text{or} quad n = 1 quad text{not valid since } n > 1 ] Thus, find the smallest solution for ( u geq 2 ):8. Assume ( v = 2q + 1 ): [ u^2 = 3(q+1)(q + 1) + 1 ]9. Due to coprimality considerations and constraints the minimum solution found is: [ u = 13, v = 15 implies n = 6 cdot 56 + 1 = 337 ] Conclusion: [ boxed{337} ]

🧠:Okay, so I need to find all natural numbers n less than 50 where the expressions 4n + 5 and 7n + 6 have a common divisor greater than 1. Hmm, let's break this down step by step. First, I remember that if two numbers have a common divisor greater than 1, then their greatest common divisor (GCD) must be greater than 1. So, the problem is asking for all n < 50 such that GCD(4n + 5, 7n + 6) > 1. How do I compute the GCD of these two expressions? Well, the Euclidean algorithm is a reliable method for finding the GCD of two numbers. Maybe I can apply that here. Let me recall how the Euclidean algorithm works. If I have two numbers, say a and b, then GCD(a, b) = GCD(b, a mod b). I can keep applying this until the remainder is zero, and the last non-zero remainder is the GCD.So let me set a = 7n + 6 and b = 4n + 5. Then, I need to compute GCD(7n + 6, 4n + 5). Let's perform the Euclidean algorithm steps.First step: Compute (7n + 6) divided by (4n + 5). The quotient here would be 1, since 4n + 5 fits once into 7n + 6. The remainder is (7n + 6) - 1*(4n + 5) = 7n + 6 - 4n -5 = 3n + 1. So, GCD(7n + 6, 4n + 5) = GCD(4n + 5, 3n + 1).Second step: Now compute GCD(4n + 5, 3n + 1). Divide 4n + 5 by 3n + 1. The quotient is 1 again, and the remainder is (4n + 5) - 1*(3n + 1) = 4n +5 -3n -1 = n + 4. So now, GCD(4n +5, 3n +1) = GCD(3n +1, n +4).Third step: Compute GCD(3n +1, n +4). Divide 3n +1 by n +4. The quotient is 3, since n +4 goes into 3n +12 three times, but we have 3n +1. So 3*(n +4) = 3n +12. The remainder is (3n +1) - 3*(n +4) = 3n +1 -3n -12 = -11. So the remainder here is -11, but since we take absolute value in GCD, it's 11. Therefore, GCD(3n +1, n +4) = GCD(n +4, 11).Fourth step: GCD(n +4, 11). This is straightforward now. The GCD of (n +4) and 11 is either 1 or 11, depending on whether 11 divides (n +4). So putting it all together, the original GCD(7n +6, 4n +5) simplifies down to GCD(n +4, 11). Therefore, the GCD is 11 if 11 divides (n +4), otherwise it's 1. Since we want the GCD to be greater than 1, we need GCD(n +4, 11) = 11. That happens exactly when 11 divides (n +4). So, n +4 ≡ 0 mod 11. Which implies n ≡ -4 mod 11. But since n is a natural number, we can write this as n ≡ 7 mod 11 (because -4 mod 11 is equivalent to 7 mod 11). Therefore, n must be congruent to 7 modulo 11. Now, since n is a natural number less than 50, we need to find all such n in that range. Let me list them out. Starting with n = 7, then adding 11 each time:First term: 7Second term: 7 + 11 = 18Third term: 18 + 11 = 29Fourth term: 29 + 11 = 40Fifth term: 40 + 11 = 51But 51 is greater than 50, so we stop at 40. Therefore, the values of n are 7, 18, 29, 40. Wait, let me verify that. If n is 7, then 4n +5 = 28 +5 = 33, and 7n +6 = 49 +6 = 55. GCD(33,55) is 11, which is correct. n =18: 4*18 +5 =72 +5=77; 7*18 +6=126 +6=132. GCD(77,132). Let's compute it. 132 divided by 77 is 1 with remainder 55. Then 77 divided by 55 is 1 with remainder 22. Then 55 divided by 22 is 2 with remainder 11. Then 22 divided by 11 is 2 with remainder 0. So GCD is 11. Correct.n=29: 4*29 +5=116 +5=121; 7*29 +6=203 +6=209. GCD(121,209). 209 divided by 121 is 1 with remainder 88. 121 divided by 88 is 1 with remainder 33. 88 divided by 33 is 2 with remainder 22. 33 divided by 22 is 1 with remainder 11. 22 divided by 11 is 2 with remainder 0. GCD is 11. Good.n=40: 4*40 +5=160 +5=165; 7*40 +6=280 +6=286. GCD(165,286). 286 divided by 165 is 1 with remainder 121. 165 divided by 121 is 1 with remainder 44. 121 divided by 44 is 2 with remainder 33. 44 divided by 33 is 1 with remainder 11. 33 divided by 11 is 3 with remainder 0. GCD is 11. Correct.So these four numbers are indeed solutions. Are there any others?Wait, let me check if there's a possibility that GCD(n +4,11) could be a multiple of 11. But since 11 is prime, the GCD is either 1 or 11. So no, there are no other divisors. Therefore, the only possible n's are those congruent to 7 mod 11, less than 50.But just to be thorough, let's check n=7 +11k where k=0,1,2,3,4:k=0: 7, which is less than 50.k=1: 18, less than 50.k=2:29, less than 50.k=3:40, less than 50.k=4:51, which is over 50. So only 7,18,29,40. That's four numbers. Wait, but maybe there's a mistake in my initial approach. Let me think again. Is there any other way that the GCD could be greater than 1? For example, maybe I missed a step in the Euclidean algorithm?Let me retrace the steps. Starting with GCD(7n +6, 4n +5). Then subtract 1*(4n +5) from 7n +6 to get 3n +1. Then GCD(4n +5, 3n +1). Subtract 1*(3n +1) from 4n +5 to get n +4. Then GCD(3n +1, n +4). Then subtract 3*(n +4) from 3n +1 to get -11. So GCD(n +4, 11). That seems correct.Therefore, unless my calculation in the Euclidean algorithm is wrong, the GCD reduces down to 11 or 1. Therefore, only multiples where n +4 is a multiple of 11. So n ≡7 mod11. Therefore, n=7,18,29,40. But just to check, let me try a different approach. Suppose d is a common divisor of 4n +5 and 7n +6. Then d divides (7n +6) - 7*(4n +5)/4? Wait, maybe not. Alternatively, d divides any linear combination. Let's consider that d divides both 4n +5 and 7n +6. Then d divides (7n +6) - 1*(4n +5) =3n +1. Then d divides (4n +5) -1*(3n +1)=n +4. Then d divides (3n +1)-3*(n +4)= -11. Therefore, d divides 11. Since 11 is prime, d must be 11. So the common divisor can only be 11. Therefore, the only possible common divisor greater than 1 is 11, which occurs when 11 divides both 4n +5 and 7n +6. Which, as above, occurs when n ≡7 mod11. Therefore, the same conclusion. So the answer should be n=7,18,29,40.Is there a possibility that n=0? Wait, n is a natural number. Depending on the definition, natural numbers start from 1 or 0. The problem says "natural number less than 50". If natural numbers start at 1, then n=7,18,29,40. If 0 is included, check n=0: 4*0 +5=5, 7*0 +6=6. GCD(5,6)=1. So even if n=0 is allowed, it's not a solution. So the answer remains the same.Therefore, the values of n are 7, 18, 29, and 40. Let me just confirm once more that these are all less than 50. Yes: 7,18,29,40—all less than 50. The next one would be 51, which is over 50, so excluded. So, the final answer should be these four numbers. I think that's all. I don't see any mistakes in the reasoning. The key insight was using the Euclidean algorithm to reduce the GCD step by step until we found that the common divisor must be 11, leading to the congruence condition on n.

✅:To find all natural numbers ( n < 50 ) such that ( 4n + 5 ) and ( 7n + 6 ) share a common divisor greater than 1, we compute their greatest common divisor (GCD) using the Euclidean algorithm:1. Initial GCD: ( gcd(7n + 6, 4n + 5) ).2. First Step: Subtract ( 1 times (4n + 5) ) from ( 7n + 6 ): ( gcd(4n + 5, 3n + 1) ).3. Second Step: Subtract ( 1 times (3n + 1) ) from ( 4n + 5 ): ( gcd(3n + 1, n + 4) ).4. Third Step: Subtract ( 3 times (n + 4) ) from ( 3n + 1 ): ( gcd(n + 4, 11) ).The GCD reduces to 11 if ( 11 mid (n + 4) ), implying ( n equiv -4 equiv 7 pmod{11} ). Thus, ( n ) must satisfy ( n = 11k + 7 ) for integers ( k geq 0 ).Valid ( n ) values less than 50: - ( k = 0 ): ( n = 7 )- ( k = 1 ): ( n = 18 )- ( k = 2 ): ( n = 29 )- ( k = 3 ): ( n = 40 )Verification: For each ( n ), ( 4n + 5 ) and ( 7n + 6 ) are multiples of 11, confirming their GCD is 11.Final Answer: ( boxed{7, 18, 29, 40} )

🔑:Given the problem involves finding the natural numbers n less than 50 such that the expressions 4n + 5 and 7n + 6 share a common divisor greater than 1.1. Let (d) be a common divisor such that ((4n + 5, 7n + 6) = d > 1). Then, it follows that:[d mid (4n + 5) quad text{and} quad d mid (7n + 6)]2. Because (d) divides both (4n + 5) and (7n + 6), it must also divide any integer linear combination of these two expressions. Therefore:[d mid (7n + 6) - (4n + 5) = 3n + 1][d mid (4n + 5) - (3n + 1) = n + 4][d mid (3n + 1) - 2(n + 4) = n - 7]3. Since (d) also divides (n + 4) and (n - 7), it must also divide:[d mid (n + 4) - (n - 7) = 11]4. Since 11 is a prime number, the only possible value for (d) that is greater than 1 is (d = 11).5. Given (d = 11), we need to find the values of (n) such that (n - 7) is a multiple of 11:[n - 7 = 11k quad text{for some integer } k][n = 11k + 7]6. To ensure (n) is a natural number less than 50:[0 < 11k + 7 < 50]Solving for (k) within this range,[0 < 11k + 7 < 50][-7 < 11k < 43][-frac{7}{11} < k < frac{43}{11}][-0.636 < k < 3.909]Since (k) must be an integer, possible values of (k) are (0, 1, 2, 3).7. Plugging these values back into the equation (n = 11k + 7), we get:[k = 0 Rightarrow n = 11(0) + 7 = 7][k = 1 Rightarrow n = 11(1) + 7 = 18][k = 2 Rightarrow n = 11(2) + 7 = 29][k = 3 Rightarrow n = 11(3) + 7 = 40] Conclusion:The values of (n) that satisfy the given conditions are:[boxed{7, 18, 29, 40}]

🧠:Okay, so I have this problem about quadratic sequences. Let me try to understand what it's asking. A quadratic sequence is a finite sequence of integers where each term differs from the previous one by exactly i², where i is the position in the sequence. So for the first term, i=1, the difference is 1²=1, then for the next term, i=2, the difference is 2²=4, and so on. But wait, the problem says the absolute value of the difference is i². So each step, you can either add or subtract i². That's important. So starting from a₀, each subsequent term is a₀ ±1², then that term ±2², etc. Part (a) asks to prove that for any two integers b and c, there exists a natural number n and a quadratic sequence starting at b and ending at c. So no matter what b and c are, we can find such a sequence. Hmm. How do I approach this? Maybe induction? Or maybe show that we can reach any integer by choosing the signs appropriately? Let's think.First, let's note that the sum of the first n squares is S = 1² + 2² + ... + n² = n(n+1)(2n+1)/6. The maximum possible difference between b and c would be S if all the steps are in the same direction. But since we can choose the sign of each step, the possible reachable values from b would be b plus any combination of ±1², ±2², ..., ±n². So the problem is similar to showing that by choosing signs appropriately, we can reach any integer c from b. But how?Wait, but integers can be any distance apart. So even if we can choose signs, the total sum is fixed once n is fixed. So for example, if n is 1, the maximum distance is 1. If n is 2, maximum is 1+4=5. But since we can subtract, the minimum is -1-4=-5. So the possible reachable points are from -5 to +5 in steps of 2? Wait, no. Because each term is added or subtracted. So the total displacement is the sum of ±k². So the total displacement can be any integer of the form S = ±1 ±4 ±9 ±...±n². So we need to show that for some n, this sum can be equal to c - b. Because starting at b, adding S will get to c. So problem reduces to: Given any integer d = c - b, there exists n such that d can be expressed as a sum of ±1² ±2² ... ±n².So part (a) is equivalent to proving that the set of all such sums is the entire set of integers, given a sufficiently large n. That is, for any integer d, there exists n where d can be written as a combination of ±1 ±4 ±9 ... ±n². Is that true? How can we prove that?Maybe using the fact that the squares modulo some number generate all residues, or using some kind of density argument. Alternatively, perhaps using induction. Let's think of parity first. Let's see:Each square is either 0 or 1 mod 2. Because even²=0 mod 2, odd²=1 mod 2. So the sum of squares will have some parity. If we can flip signs, then the total sum can have the same parity as the sum of the squares, because flipping the sign of a term changes the sum by 2k², which is even if k is even (since k² is even), and 2*odd = even if k is odd. So changing the sign of any term changes the total sum by an even number. Therefore, the parity of the sum is fixed. The parity of the sum of 1² + 2² + ... + n² is equal to the sum modulo 2 of each term. Since 1 mod 2 is 1, 4 mod 2 is 0, 9 mod 2 is 1, 16 mod 2 is 0, etc. So the sum modulo 2 is equal to the number of odd squares in 1² to n². The odd squares come from odd k. So for each odd k from 1 to n, we add 1 modulo 2. So the total sum modulo 2 is equal to the number of odd numbers from 1 to n, modulo 2. The number of odd numbers from 1 to n is ceil(n/2). So if n is even, it's n/2; if n is odd, (n+1)/2. Wait, actually, for n, the number of odds is floor((n+1)/2). Let's take n=1: 1 odd. n=2: 1 odd. n=3: 2 odds. n=4: 2 odds. So the number of odds is ⌈n/2⌉. Therefore, the sum modulo 2 is ⌈n/2⌉ mod 2. So depending on n, the parity of the total sum is fixed. But when we flip signs, as we saw, the parity doesn't change. Because flipping the sign of a term with k even: 2k² is divisible by 2, so parity remains. Flipping the sign of a term with k odd: the term was 1 mod 2, and flipping it subtracts 2*(1 mod 2) which is 0 mod 2. So overall, the parity of the total sum is fixed regardless of sign choices. Therefore, the possible displacements d = c - b must have the same parity as the total sum of 1² + 2² + ... +n². Therefore, to reach any integer d, we need to have that d has the same parity as the sum of the first n squares. But the problem says "for any two integers b and c", so d can be any integer. Therefore, we need to choose n such that d has the same parity as the sum of the first n squares. So maybe, for a given d, we can choose n such that sum_{k=1}^n k² ≡ d mod 2. Since d can be even or odd. So if d is even, we need sum of first n squares even. If d is odd, sum of first n squares odd. But sum of first n squares is even or odd depending on n. Let's check:sum_{k=1}^n k² mod 2 is equal to the number of odd squares in 1² to n² mod 2. As we saw earlier, the number of odd k from 1 to n is ⌈n/2⌉. So sum is ⌈n/2⌉ mod 2. So for example:If n ≡ 0 or 1 mod 4? Let's see:Wait, maybe let's compute sum_{k=1}^n k² mod 2 for some small n:n=1: 1²=1 mod 2 → 1. ⌈1/2⌉=1 mod 2=1.n=2: 1²+2²=1+0=1 mod 2. ⌈2/2⌉=1 mod 2=1.n=3: 1+0+1=2≡0 mod 2. ⌈3/2⌉=2 mod 2=0.n=4: 1+0+1+0=2≡0 mod 2. ⌈4/2⌉=2 mod 2=0.n=5: 1+0+1+0+1=3≡1 mod 2. ⌈5/2⌉=3 mod 2=1.n=6: sum up to 6: 1+0+1+0+1+0=3≡1 mod 2. ⌈6/2⌉=3 mod 2=1.n=7: sum up to 7: 1+0+1+0+1+0+1=4≡0 mod 2. ⌈7/2⌉=4 mod 2=0.Hmm, so the parity of the sum is equal to ⌈n/2⌉ mod 2. Let me check for n=3: ⌈3/2⌉=2, which is 0 mod 2. Correct. For n=5: ⌈5/2⌉=3, 1 mod 2. Correct. So in general, if n is even, then ⌈n/2⌉ = n/2. If n is odd, it's (n+1)/2. So parity alternates every two numbers:n=1: 1 → oddn=2: 1 → oddn=3: 2 → evenn=4: 2 → evenn=5: 3 → oddn=6: 3 → oddn=7: 4 → evenSo the parity of the sum is odd when n ≡ 1 or 2 mod 4? Wait, n=1:1, n=2:1, n=5:3, n=6:3. Wait, n=1: 1 mod 4, parity 1. n=2: 2 mod 4, parity 1. n=3: 3 mod 4, parity 0. n=4: 0 mod 4, parity 0. Then n=5:1 mod 4, parity 1. So it seems that the parity is 1 when n ≡1 or 2 mod 4, and 0 when n ≡3 or 0 mod 4. Let me verify:n=1:1 mod4 → parity 1n=2:2 mod4 → parity1n=3:3 mod4 → parity0n=4:0 mod4 → parity0n=5:1 mod4 → parity1Yes, that seems to hold. So the parity of the sum is 1 if n ≡1 or 2 mod4, 0 if n≡3 or 0 mod4.Therefore, if we need the displacement d to have a certain parity, we can choose n such that n ≡1 or 2 mod4 for parity 1, or n≡3 or 0 mod4 for parity0.Therefore, for any d, we can choose n to be congruent to 1,2,3, or 0 mod4 depending on the parity of d. So that's possible. So if d is even, pick n ≡0 or3 mod4. If d is odd, pick n≡1 or2 mod4.But even if we can match the parity, how do we know that the sum can reach any integer? For example, with coins problem, if we have coins of 1,4,9,...,n², with each coin being able to be added or subtracted, can we reach any integer? This is similar to the concept of the set of numbers generated by these coins with ± signs. In the case of 1 and 4, can we reach any integer?Wait, let's take a simple case. Suppose n=2: possible steps ±1, ±4. The possible sums are ±1±4. Let's see: 1+4=5, 1-4=-3, -1+4=3, -1-4=-5. So possible sums: ±3, ±5. So you can't reach numbers like 1, 2, 4, etc. So with n=2, you can't reach all integers. Similarly, for higher n, maybe the reachable numbers become more dense?Wait, but when n increases, the steps get larger. But with more steps, you can adjust more finely? Wait, for n=3: steps ±1, ±4, ±9. Let's see possible sums. Maybe by combining these, you can reach more numbers. For example:Suppose we have 1,4,9. Let's see if we can reach 2. Let's try 1 -4 +9 =6. Not 2. Hmm. 1 +4 -9= -4. 9 -4 -1=4. Hmm, not sure. Maybe 1 +4 +9=14, -1 -4 -9=-14. It's not obvious. Maybe 9 -4 +1=6. 9 -4 -1=4. Maybe it's still not enough. So even with n=3, maybe you can't reach 2.Wait, but perhaps with larger n? For example, if n=4, adding ±16. Then maybe with more terms, the number of combinations increases. But how do we know that eventually, for some n, we can reach any integer?Alternatively, maybe use induction. Suppose that for some n, the set of reachable numbers is a certain range, and by adding another term ±(n+1)², we can extend that range. Let's try.Base case: n=1. Reachable numbers: ±1. So from -1 to 1, step 2. Not much. Then n=2: reachable numbers ±1±4. As before, -5,-3,3,5. So ranges from -5 to 5, but only even steps? Wait, no: -5, -3, 3,5. So gaps in between. Then n=3: adding ±9. So possible sums are previous sums ±9. So previous sums were -5,-3,3,5. Adding 9: 4,6,12,14. Subtracting 9: -14,-12,-6,-4. So combined with previous, now we have -14,-12,-6,-5,-4,-3,3,4,5,6,12,14. Still gaps. Hmm. Then n=4: adding ±16. Take each previous sum and add/subtract 16. So ranges from -14-16=-30 to 14+16=30, but still with gaps. It's not obvious that this will eventually cover all integers. Alternatively, maybe the problem is similar to the postage stamp problem, where with sufficient denominations, you can cover all amounts beyond a certain point. But here, since we can subtract as well, it's different. Maybe using the fact that consecutive squares grow quadratically, so their differences can cover linear gaps?Alternatively, consider modular arithmetic. If we can reach every residue modulo some number, then with the large steps, we can adjust the higher digits. For example, using the Chinese Remainder Theorem. If we can cover all residues modulo m for some m, then using multiples of m to reach any number. Let's try m=2 first. Since the parity is fixed, as we saw. So if d has the same parity as the sum, we can reach it. But if we can choose n such that the sum has the required parity, then maybe we can adjust the higher bits. But how?Alternatively, think of the problem as a graph where each node is an integer, and edges connect a node to node ±k². Then the question is whether this graph is connected, given sufficiently many steps. But since the steps can be of any size up to n², and n can be as large as needed, perhaps by choosing n large enough, we can reach any node. But how to formalize this?Alternatively, consider that the steps 1², 2², ..., n² are generators of the integer additive group under addition and subtraction. Since the integers are a cyclic group under addition, and if the generators have gcd 1, then they can generate all integers. Wait, but the squares may not have gcd 1. For example, 1²=1, 2²=4, 3²=9, etc. The gcd of 1 and 4 is 1, so their combination can generate any integer. Wait, actually, the set {1,4} can generate any integer because 1 is there. For example, to get 2: 1 +1, but we can't use 1 twice. Wait, but in our case, each square can be used only once, with a sign. So it's a signed sum of distinct squares. So the problem is, given a target d, can we find n and signs ε_i ∈ {±1} such that Σ_{i=1}^n ε_i i² = d.This is similar to the subset sum problem with signed sums. In standard subset sum, you want a subset of numbers that add up to a target. Here, it's a signed sum, so equivalent to choosing a subset to add and a subset to subtract. The difference between the two subsets is the target.It is known that if the numbers are 1, 2, 4, 8, ..., you can reach any integer up to the sum. But here the numbers are squares, which are more sparse. However, since 1 is among them, and 1 can be used to adjust by 1. So if we have 1 in our sum, we can change the parity and make small adjustments. Wait, but once we fix n, the maximum adjustment is 1. But if we can choose n, then perhaps we can use 1 as a fine adjustment. For example, if we need to reach a number d, first approximate it with larger squares, then adjust with smaller ones. Let's think about this strategy.Suppose we want to reach d. Let's take the largest square less than or equal to |d|, say k². Then subtract or add it to get closer. Wait, but this is a greedy algorithm. However, in our case, we can choose the signs of each square, so it's more flexible. But since each square can only be used once, we can't repeat squares. But n can be as large as needed. So if we can use squares up to some k² where k is sufficiently large, then perhaps we can reach d by including or excluding squares as needed.But how do we formalize that any integer can be expressed as such a sum? Maybe use induction. Suppose that for any integer up to S(n) = 1² + 2² + ... +n², we can reach it. Then adding (n+1)², which is larger than S(n) for n ≥3, since S(n) ~ n³/3 and (n+1)² ~n². Wait, actually S(n) is O(n³), so for large n, S(n) is much larger than (n+1)². Wait, maybe not. Wait, S(n) = n(n+1)(2n+1)/6. So for n=3, S(3)=14, (4)^2=16. So S(3)=14 < 16. For n=4, S(4)=30, 5²=25. S(4)=30>25. For n=5, S(5)=55, 6²=36. So S(n) grows faster than n². So for n≥4, S(n) > (n+1)^2. Therefore, once we have S(n) greater than some value, we can cover numbers beyond that by adding the next square. But how to use this?Alternatively, think in terms of the following: Since 1 is in the sequence, we can adjust the sum by ±1. So if we can reach a number that's congruent to d modulo 2, then by adding or subtracting 1, we can reach d. Wait, but 1 is only available once. So maybe use 1 to adjust the parity, and other squares to build up the number.Alternatively, consider that once you have 1 and 4, you can reach numbers that are 1 mod 4, etc. But I'm not sure. Let's try to see for small d.Suppose d=2. Can we find a combination of squares with signs that sum to 2? Let's see:Try n=3: 1,4,9. Let's see: 1 +4 -9 = -4. Not helpful. 9 -4 -1=4. 9 -4 +1=6. 4 -1=3. Not 2.n=4: 1,4,9,16. Let's try combinations:16 -9 -4 -1=2. Yes! So 16 -9 -4 -1=2. So using n=4, signs: +16 -9 -4 -1. Therefore, sequence would be:Start at b, then add 1, add 4, subtract 9, add 16. Wait, no. Wait, the sequence is a₀, a₁,...,a₄. Each term differs by i². So to get the sum 16 -9 -4 -1, we need to arrange the signs. Let's see:The total displacement is 16 -9 -4 -1 =2. So the sequence would be: a₀ = b, a₁ = b ±1, a₂ = a₁ ±4, a₃ = a₂ ±9, a₄ = a₃ ±16. So to get the total displacement of +2, we need to have: Let's see:Start at b. Then first step: +1 (i=1). Second step: +4 (i=2). Third step: -9 (i=3). Fourth step: -16 (i=4). Wait, but the displacement would be 1 +4 -9 -16 = -20. Not 2. Wait, maybe I need to arrange the signs in a different order. Wait, no—each step corresponds to i=1 to n, so the displacement is the sum of ±i² from i=1 to n. So the order in which you choose the signs doesn't matter; the total displacement is the sum. So if you have displacement 16 -9 -4 -1=2, then regardless of the order, the total displacement is 2. But the sequence is a path where each step is determined by the previous term. So the actual sequence of terms would be b, b+1, b+1+4, b+1+4-9, b+1+4-9+16= b +1+4-9+16= b +12. Hmm, that's not displacement 2. Wait, maybe I messed up the signs.Wait, the displacement is the sum of the signed steps. To get a total displacement of 2, the sum of ±1 ±4 ±9 ±16 must be 2. For example: +1 -4 -9 +16= 1 -4= -3, -3 -9= -12, -12 +16=4. No. Or -1 +4 +9 -16= -1 +4=3, 3 +9=12, 12 -16= -4. Hmm. Wait, maybe another combination:+1 +4 +9 -16=1+4=5, 5+9=14, 14-16=-2. No. +1 -4 +9 -16=1-4=-3, -3+9=6, 6-16=-10. No. -1 +4 +9 -16= same as above. Wait, how do we get +2? Let's think:We need x + y + z + w =2, where x=±1, y=±4, z=±9, w=±16. Let's solve this equation:Possible values:1. Let's try w=+16. Then x + y + z =2 -16= -14.Need x + y + z= -14. With x=±1, y=±4, z=±9.Possible combinations:If z=+9, then x + y= -14 -9= -23. But x and y can only be ±1 and ±4. Max negative sum: -1 -4= -5. Not enough.If z=-9, then x + y= -14 +9= -5. So x + y= -5.Possible: x=-1, y=-4: -1 -4= -5. Yes. So z=-9. So solution: x=-1, y=-4, z=-9, w=+16. Sum: -1 -4 -9 +16=2. Yes. So the sequence would be:Start at b.Step 1: a₁ = b -1.Step 2: a₂ = a₁ -4 = b -5.Step 3: a₃ = a₂ -9 = b -14.Step 4: a₄ = a₃ +16 = b +2.So yes, with n=4, we can reach b +2. Therefore, d=2 is reachable with n=4.Similarly, if we need d=3, can we find a combination? Let's see:Solve x + y + z + w=3. Let's try similar approach.Take w=+16. Then x + y + z=3 -16= -13.If z=+9: x + y= -13 -9= -22 (impossible).If z=-9: x + y= -13 +9= -4.Possible x=-1, y=-3: but y can only be ±4. So x + y= -1 -4= -5. Close but not -4. If x=+1, y=-5: but y can't be -5. Hmm. Maybe different approach.Try w= -16. Then x + y + z=3 +16=19.With x=±1, y=±4, z=±9. Max sum is 1 +4 +9=14. Not enough. So no.Alternatively, take z=+9, w=+16: x + y=3 -16 -9= -22. Impossible.Alternatively, z=+9, w=-16: x + y=3 +16 -9=10. x=+1, y=+9: but y can only be ±4. So x + y=1 +4=5. Not 10. Doesn't work.Wait, maybe n=5. Let's try adding 25.With n=5, we have ±1, ±4, ±9, ±16, ±25. Let's see if we can get 3.Suppose we take +25 -16 -4 -1 -9=25 -16=9, 9 -4=5, 5 -1=4, 4 -9=-5. Not 3.Wait, let's set up the equation: ±1 ±4 ±9 ±16 ±25=3.Trying combinations:Start with +25. Then we need ±1 ±4 ±9 ±16=3 -25= -22.But sum of ±1 ±4 ±9 ±16. Let's see: the max negative sum is -1 -4 -9 -16= -30. So -30 to 30. To get -22, need to sum to -22. Let's see:-1 -4 -9 -16= -30. If we flip some signs:-1 -4 -9 +16= -1 -4= -5, -5 -9= -14, -14 +16=2.-1 -4 +9 -16= -1 -4= -5, -5 +9=4, 4 -16= -12.-1 +4 -9 -16= -1 +4=3, 3 -9= -6, -6 -16= -22. Yes! So x=-1, y=+4, z=-9, w=-16. Then total sum: -1 +4 -9 -16= -22. Adding to +25 gives 3. Therefore, the combination is +25 -1 +4 -9 -16=3. So the sequence would be:Start at b.Step 1: a₁ = b -1.Step 2: a₂ = a₁ +4= b +3.Step 3: a₃= a₂ -9= b -6.Step 4: a₄= a₃ -16= b -22.Step 5: a₅= a₄ +25= b +3.So yes, with n=5, we can reach d=3. Therefore, it seems that by increasing n, we can reach any integer by appropriately choosing signs. The key is that with larger squares, we can make up the difference after approximating with smaller ones. So perhaps by induction, we can show that for any d, there exists an n such that d can be expressed as a sum of ±1² ±2² ... ±n².Alternatively, we can use the fact that the squares generate a subgroup of the integers under addition. Since the integers are a cyclic group and the subgroup must be generated by the gcd of the squares. The gcd of all squares is 1, since 1 is a square. Therefore, the subgroup is the entire integers. Therefore, any integer can be expressed as a linear combination of squares with integer coefficients. But in our case, the coefficients are restricted to ±1. However, since we can use each square once with either +1 or -1, it's a finite combination, but we can choose n as large as needed. So since the infinite set of squares generates the integers, for any integer d, there exists a finite n such that d is a combination of ±1² ... ±n². Therefore, part (a) is true.But how to formalize this into a proof?Maybe by induction on |d|. Suppose for all integers with absolute value less than |d|, we can find such an n. Then for d, we can take the largest square less than or equal to |d|, say k². Then consider d - k² and d + k². If |d -k²| < |d|, then by induction hypothesis, we can express d -k² as a combination of squares, so d = (d -k²) +k². Similarly, if |d +k²| < |d|, then d = (d +k²) -k². But since k² <= |d| < (k+1)², then |d -k²| <= |d| -k² < (k+1)² -k² = 2k +1. Not sure if that helps. Alternatively, use the fact that for d >=0, we can use a greedy algorithm: subtract the largest square less than or equal to d, and repeat. But since we can also add squares, it's more flexible.Alternatively, since 1 is among the squares, and 1 can adjust the sum by 1, once we can get within 1 of the target, we can reach it. But how to get close?Alternatively, use the fact that the sum of squares grows cubically, so for sufficiently large n, the sum S(n) exceeds |d|, and then by flipping signs of the larger squares, we can subtract them to reach the target. But I need a better approach.Wait, here's a theorem: If you have a set of numbers that includes 1 and is closed under addition, then you can reach every integer beyond a certain point. But here, we can subtract as well. But since we can use each number once, it's different. However, if we can choose n large enough so that the sum S(n) is greater than |d|, then by flipping the signs of some squares, we can adjust the total sum to reach d. The key idea is that the difference between the maximum sum S(n) and the target d must be even, which can be achieved by adding 1 or another square to adjust parity. Wait, but we already established that parity can be matched by choosing appropriate n. So here's a possible plan:For any integer d, choose n such that:1. S(n) has the same parity as d.2. S(n) >= |d|.Then, by the subset sum problem with signed sums, since S(n) - |d| is even (since S(n) and d have the same parity), we can split the difference into 2k, which can be achieved by flipping signs of some squares. Each sign flip changes the sum by 2k². So we need to express k as a sum of some squares. However, I'm not sure if this works directly, but there's a theorem called the Greedy Algorithm for the Signed Sum Problem.Alternatively, use the fact that the squares form a complete set of residues modulo some number. For example, modulo 4: squares are 0 or 1. So if d ≡0 or1 mod4, but since we can have any parity by choosing n, maybe not helpful.Alternatively, consider that the problem is similar to representing numbers in binary, but with base squares. Not sure.Alternatively, use mathematical induction. Let's try.Base case: d=0. Trivial, n=0. But n must be natural number. So n=1: sum ±1. Can't get 0. Wait, no. If n=1, the sequence is a₀ and a₁=a₀±1. So to get from b to b, we need n=0. But the problem states n is a natural number, so n≥1? Wait, the problem says "natural number n" and "quadratic sequence a₀,...,a_n". So n must be at least 1. Wait, but for d=0, can we have n=1? Then a₁=a₀±1. So a₁ -a₀=±1. So unless b=c, but the problem allows n to be any natural number. If b=c, then d=0. How do we get d=0? If n=1, we have a₁ =b±1≠b. So we need n=2? Let's see: a₂ = a₁ ±4. Starting from a₀=b, a₁=b±1, a₂= a₁ ±4. To get back to b, we need ±1 ±4=0. That would require ±1=∓4. But 1≠4, so impossible. For n=3: ±1 ±4 ±9=0. Is that possible? Let's see: 1 +4 -9= -4. 9 -4 -1=4. -1 -4 +9=4. 9 -1 -4=4. Doesn't work. For n=4: ±1 ±4 ±9 ±16=0. Let's see:-1 -4 -9 +16=2. No. 1 +4 +9 -16= -2. 16 -9 -4 -1=2. 16 -9 -4 +1=4. 16 -9 +4 -1=10. Not zero. Hmm. n=5: ±1 ±4 ±9 ±16 ±25=0.For example: 25 -16 -9 -4 +1= -3. No. 25 -16 -9 -4 -1= -5. 25 -16 -9 +4 +1=5. 25 -16 +9 -4 -1=13. Not zero. Maybe n=12? Not sure. Wait, but the problem says "for any two integers b and c". If b=c, then d=0. How to get d=0? Maybe n=0? But n must be a natural number. The problem says "natural number n", which usually starts at 1. If n=0 is allowed, then it's trivial, but I think n=0 is excluded. So there might be a special case for d=0, but the problem states "any two integers b and c", which includes b=c. Therefore, perhaps part (a) requires n=0? But the problem says "natural number n", which usually is 1,2,... So maybe there's a mistake in the problem statement, or perhaps the sequence can have n=0, which is just a single element a₀. If n=0 is allowed, then yes, for b=c, n=0 works. Otherwise, if n must be at least 1, then it's impossible to have a sequence from b to b with n≥1. So perhaps the problem allows n=0. Let me check the original problem statement."A finite sequence of integers a₀, a₁,...,aₙ is called quadratic if for each i ∈ {1,...,n} we have |a_i -a_{i-1}|=i². (a) Prove that for any two integers b and c, there exists a natural number n and a quadratic sequence with a₀=b and aₙ=c."Natural number n. So n ∈ ℕ, which typically is {1,2,3,...}, but sometimes ℕ includes 0. Depending on the convention. If ℕ includes 0, then n=0 is allowed. If not, then not. However, the sequence is a₀,...,aₙ. If n=0, the sequence has only a₀. So to have a₀=b and aₙ=c, we need b=c. Therefore, if n=0 is allowed, then for b=c, n=0 works. For b≠c, we need n≥1. If the problem allows n=0, then it's okay. But since the problem says "natural number n", and in some definitions, natural numbers start at 1, then for b=c, we would need to have a sequence of length 1, but that would require a₁ =a₀ ±1, which can't be equal. Therefore, there's an inconsistency. Therefore, likely the problem allows n=0, and the answer for part (a) would be: for b≠c, use a suitable n≥1, and for b=c, use n=0. But since the problem says "natural number n", which may or may not include 0, but in the context of sequences, n=0 would correspond to a single element, which is allowed. Therefore, the answer is that for any b and c, there exists n∈ℕ and a quadratic sequence from b to c. If b=c, take n=0. If b≠c, construct a sequence with n≥1.But since part (b) asks for the minimal n for a specific case, probably part (a) allows n to be any natural number including 0. So moving on.Given that, for part (a), to show that for any d =c -b, there exists n∈ℕ and signs ε_i ∈{±1} such that Σ ε_i i² =d. To prove this, we can use the following approach:1. Since 1 is among the squares, we can adjust the sum by ±1.2. For any d, choose n sufficiently large such that S(n) = Σ_{i=1}^n i² ≥ |d|.3. The difference S(n) - |d| is even (by choosing n such that S(n) ≡d mod2).4. Since the difference is even, it can be written as 2k. We need to flip signs of some squares to subtract 2k from S(n). Flipping the sign of square i² changes the sum by -2i². So we need to find a subset of squares whose sum is k. If such a subset exists, then flipping their signs will reduce the total sum by 2k, achieving the desired d.But how do we know that such a subset exists? This is similar to the subset sum problem, which is NP-hard, but in our case, the numbers are squares. However, with 1 available, and the numbers growing quadratically, we can use a greedy approach: subtract the largest possible square less than or equal to k, then iterate.However, to guarantee that such a subset exists, we might need to use the fact that every integer can be expressed as a sum of four squares (Lagrange's theorem), but here we need a subset of the squares up to n², which is different. But since we can use any subset (due to the flexibility of choosing n), and for sufficiently large k, such representations are possible, perhaps this approach works.Alternatively, we can use induction. Suppose that for any integer m, there exists a subset of squares whose sum is m. But this is not necessarily true. However, with the addition of 1, we can adjust by 1 each time.But this is getting too vague. Let's instead think constructively. Given any d, we can proceed as follows:- Choose n such that S(n) ≥ |d| and S(n) ≡d mod2.- Let k = (S(n) -d)/2. Then we need to find a subset of the squares up to n² that sum to k. If we can find such a subset, then flipping their signs will change the total sum from S(n) to S(n) -2k =d.But how to ensure that k can be expressed as a sum of some subset of the squares up to n². Since k can be any non-negative integer up to S(n)/2, and S(n) is large, by Lagrange's four-square theorem, every natural number can be expressed as the sum of four squares. Therefore, if k can be expressed as a sum of four squares, and those squares are present in our sequence up to n², then we can represent k. However, since our sequence includes all squares up to n², if n is sufficiently large to include the four squares needed for k.But Lagrange's theorem allows for up to four squares, not necessarily distinct or non-consecutive. However, since our squares are distinct and consecutive, but we can choose any subset. Therefore, for sufficiently large n, which includes all squares up to a size greater than k, we can certainly find a subset that sums to k by breaking it into four squares or fewer. Therefore, this approach would work.Therefore, the steps are:1. Given d, choose n such that S(n) ≥ |d| and S(n) ≡d mod2.2. Compute k=(S(n)-d)/2.3. Express k as a sum of at most four squares (possible by Lagrange's theorem).4. If those squares are present in {1², 2², ..., n²}, flip their signs to subtract 2k from S(n), resulting in d.5. If the squares needed for k are not present, increase n until they are included.Since we can make n as large as needed, step 5 ensures that eventually, all required squares are included. Therefore, this method constructs the required quadratic sequence.Therefore, part (a) is proved.For part (b), we need to find the minimal natural number n such that there exists a quadratic sequence starting at 0 and ending at 1996. So we need the minimal n where 1996 can be expressed as a signed sum of squares from 1² to n².To find the minimal n, we need to find the smallest n such that 1996 can be written as Σ_{i=1}^n ε_i i², where ε_i=±1.Approach: The minimal n is the smallest n where the sum of the first n squares is at least 1996, and the difference between the sum and 1996 is even, and that difference can be expressed as twice a sum of some subset of the squares (to flip their signs).First, compute S(n) =n(n+1)(2n+1)/6. Find the smallest n such that S(n) ≥1996.Calculate S(n):Start with n=20: S(20)=20*21*41/6=20*21=420, 420*41=17220, 17220/6=2870. So S(20)=2870.2870 is greater than 1996. Let's check smaller n:n=18: S(18)=18*19*37/6=18/6=3, 3*19=57, 57*37=2109.n=17: S(17)=17*18*35/6=17*3*35=17*105=1785.1785 <1996 <2109. So n=18 gives S(18)=2109. 2109 -1996=113. So k=(2109 -1996)/2=113/2=56.5. Not integer. So invalid. Therefore, parity mismatch. Since d=1996, check S(18) parity.S(18)=2109. 2109 mod2=1 (since 2109 is odd). d=1996 is even. So parity is different. Therefore, need to choose n such that S(n) is even. Let's check n=19: S(19)=19*20*39/6=19*20=380, 380*39=14820, 14820/6=2470. S(19)=2470, which is even. 2470 -1996=474. k=474/2=237. So need to express 237 as a sum of some subset of squares from 1² to19². Then flip those signs.Is 237 representable as a sum of squares up to19²=361? Let's try.But 237 is less than 361, so we can use squares up to 15²=225.Use a greedy approach:237. The largest square less than 237 is 225 (15²). 237-225=12.12. Largest square less than 12 is 9 (3²). 12-9=3.3. Largest square less than 3 is 1 (1²). 3-1=2.2. Largest square less than 2 is 1. 2-1=1.1. 1-1=0.So 237=15²+3²+1²+1². But we can only use each square once. Therefore, duplicates are not allowed. So this approach fails.Alternative decomposition:237. Let's try 14²=196. 237-196=41.41: 36 (6²)=36. 41-36=5.5:4 (2²)+1 (1²).So 237=14²+6²+2²+1². All distinct squares:14,6,2,1. These are all ≤14²=196. Wait, 14² is 196,6²=36, etc. So total sum:196+36+4+1=237. Yes. So 237=14²+6²+2²+1². Therefore, we can flip the signs of these squares in the total sum S(19)=2470 to get 2470 -2*237=2470-474=1996. Therefore, n=19 works.But wait, we need to check if these squares (1,2,6,14) are all ≤19. Yes, since n=19, the squares up to19² are allowed. Therefore, by flipping the signs of 1², 2²,6²,14² in the sum S(19)=2470, we get 2470 -2*(1+4+36+196)=2470 -2*237=2470 -474=1996. Therefore, n=19 is possible.But is there a smaller n? Let's check n=18. S(18)=2109, which is odd. 2109-1996=113, which is odd, so k=56.5 invalid. So no.n=17: S(17)=1785, which is odd. 1785 <1996. So insufficient.n=19 is the first n where S(n) ≥1996 and has the same parity as 1996 (even). Therefore, n=19 is the minimal n.But wait, let's check if n=18 can work with some other approach. Even though S(18)=2109 is odd and 1996 is even, the difference is 113, which is odd. So k=56.5, which is not integer. So impossible. Therefore, n=19 is the minimal.But to confirm, we need to ensure that with n=19, we can actually construct the sequence. Since we found that 237 can be expressed as sum of 14²+6²+2²+1², then flipping those signs in the sequence would work. Therefore, the minimal n is 19.Wait, but let's verify if there's a smaller n. Suppose n=20, which gives S(20)=2870. Then k=(2870 -1996)/2=437. Need to express 437 as sum of squares up to20². But since n=20 is larger than 19, but we're looking for minimal n. Since n=19 works, n=20 is not needed. Therefore, the answer is n=19.But let's check if n=16 could work. S(16)=16*17*33/6=16*17=272, 272*33=8976, 8976/6=1496. S(16)=1496<1996. So insufficient.n=17:1785<1996.n=18:2109≥1996, but parity mismatch.n=19:2470≥1996, parity match, and achievable. So yes, n=19 is minimal.Therefore, the answer to part (b) is 19.

✅:(a) To prove that for any integers ( b ) and ( c ), there exists a natural number ( n ) and a quadratic sequence from ( b ) to ( c ), we observe the following:1. Parity Consideration: The sum ( S(n) = sum_{i=1}^n i^2 ) modulo 2 determines the parity of the displacement. By choosing ( n ) such that ( S(n) equiv c - b pmod{2} ), we align the parities.2. Sufficiently Large ( n ): Select ( n ) such that ( S(n) geq |c - b| ). The difference ( k = frac{S(n) - (c - b)}{2} ) must be an integer. By Lagrange's four-square theorem, ( k ) can be expressed as a sum of at most four squares. These squares are part of ( {1^2, 2^2, ldots, n^2} ) for sufficiently large ( n ). Flipping the signs of these squares in the sequence adjusts the total displacement to ( c - b ).Thus, such an ( n ) exists, proving part (a).(b) To find the minimal ( n ) for ( a_0 = 0 ) and ( a_n = 1996 ):1. Compute ( S(n) ): The smallest ( n ) where ( S(n) geq 1996 ) and ( S(n) equiv 1996 pmod{2} ). Calculations show ( n = 19 ) with ( S(19) = 2470 ).2. Subset Sum: Verify ( k = frac{2470 - 1996}{2} = 237 ). Express ( 237 ) as a sum of distinct squares: ( 14^2 + 6^2 + 2^2 + 1^2 ). Flipping these signs in ( S(19) ) achieves the displacement ( 1996 ).No smaller ( n ) satisfies the conditions. Thus, the minimal ( n ) is:Answer(a) For any integers ( b ) and ( c ), there exists a natural number ( n ) and a quadratic sequence connecting them. (b) The smallest natural number ( n ) is (boxed{19}).

🔑:# Part (a)1. Objective: Prove that, given any integers ( b ) and ( c ), there exists a natural number ( n ) and a quadratic sequence starting from ( a_0 = b ) and ending at ( a_n = c ).2. Initialization: Without loss of generality, it is sufficient to show that for any integer ( c ), one can construct a quadratic sequence starting from ( a_0 = 0 ) and ending at ( a_m = c ).3. Expression of Sequence Differences: [ a_i - a_{i-1} = pm i^2 ] This implies: [ a_i = a_{i-1} pm i^2 ]4. Key Observation: Show that it’s possible to express any integer ( c ) as: [ c = pm 1^2 pm 2^2 pm cdots pm n^2 ]5. Inductive Step: - Note the relationship: [ (n+1)^2 - (n+2)^2 - (n+3)^2 + (n+4)^2 = 4 ] - This implies that if a number ( c ) can be represented using sums of squares up to ( n ), then we can also represent ( c pm 4 ) by adjusting the signs and including the squares up to ( n+4 ).6. Base Cases: Verify for ( c = 0, 1, 2, 3 ): - ( c = 1: quad 1 = 1^2 ) - ( c = 2: quad 2 = (-1^2) + (-2^2) + (-3^2) + 4^2 ) - ( c = 3: quad 3 = (-1^2) + 2^2 ) - ( c = 0 ): This is trivially ( 0 ).7. Conclusion: By proving the inductive step and verifying the base cases, we conclude that any integer ( c ) can indeed be written as a sum of squares ( pm 1^2 pm 2^2 pm cdots pm n^2 ).blacksquare# Part (b)1. Objective: Find the smallest natural number ( n ) such that there exists a quadratic sequence with ( a_0 = 0 ) and ( a_n = 1996 ).2. Sum of Squares Bound: [ 1^2 + 2^2 + cdots + n^2 = frac{n(n+1)(2n+1)}{6} ]3. Estimate ( n ): - Find ( n ) such that: [ frac{n(n+1)(2n+1)}{6} geq 1996 ] - Try ( n = 18 ): [ frac{18 cdot 19 cdot 37}{6} = 2109 geq 1996 ] However, check ( a_{18} ) explicitly, and note it must match the parity of 1996: [ 1^2 + 2^2 + cdots + 18^2 = 2109 quad (text{odd, 1996 is even}) ]4. Verify ( n = 19 ): - Parity of the sum ( 1^2 + 2^2 + cdots + 19^2 = 2470 ): [ text{Sum of Squares up to } 19^2 = 2470 - 1996 = 474 quad (text{even}) ]5. Construct Required Sequence: - Since ( 2470 = 1996 + 237 ) and ( 237 = 14^2 + 5^2 + 4^2 ): [ 1996 = (1^2 + 2^2 + 3^2 + cdots + 13^2) - 14^2 + 15^2 + cdots + 19^2 ]6. Conclusion: Thus, the smallest ( n ) is: [ boxed{19} ]

🧠:Okay, let's see. I need to find the value of sin(2x - y) where x and y are solutions to the given equations. The equations are x + sinx cosx - 1 = 0 and 2cosy - 2y + π + 4 = 0. Hmm, both equations look a bit tricky. Let me start by understanding each equation and figuring out how to solve them.First, the equation for x: x + sinx cosx - 1 = 0. I know that sinx cosx can be written as (1/2) sin2x, so maybe rewriting it would help. Let's try that. So substituting, the equation becomes x + (1/2) sin2x - 1 = 0. That simplifies to x = 1 - (1/2) sin2x. Hmm, this is a transcendental equation, meaning it can't be solved algebraically. I might need to use numerical methods or look for obvious solutions.Let me check if there are any simple values of x that satisfy this equation. Let's try x = 0. Plugging in, 0 + sin0 cos0 -1 = 0 + 0 -1 = -1 ≠ 0. Not good. How about x = π/4? Then sin(π/4)cos(π/4) = (√2/2)(√2/2) = 1/2. So x + 1/2 -1 = π/4 - 1/2. π is approximately 3.14, so π/4 is about 0.785. 0.785 - 0.5 = 0.285 ≈ 0.285 ≠ 0. So that's not a solution. Maybe x = 1? Let's check. x=1: 1 + sin1 cos1 -1 = sin1 cos1. Sin1 is about 0.8415, cos1 is about 0.5403. Multiplying gives ≈0.4546. So 0.4546 ≈ 0.45, which isn't zero. So x=1 isn't a solution either. Maybe x=0.5? Let's compute sin(0.5)cos(0.5). Sin0.5 ≈ 0.4794, cos0.5≈0.8776. Product≈0.420. Then 0.5 +0.420 -1 ≈ -0.08. Still not zero. So maybe a value between 0.5 and 1? Let's see. Let's try x=0.6. Then sin0.6≈0.5646, cos0.6≈0.8253. Product≈0.5646*0.8253≈0.466. Then x + product -1 =0.6 +0.466 -1≈0.066. So that's positive. So at x=0.5, we had -0.08, at x=0.6, +0.066. So crossing zero between 0.5 and 0.6. Maybe use linear approximation? Let me set up f(x) = x + sinx cosx -1. We have f(0.5)≈-0.08, f(0.6)≈0.066. Let's approximate the root. The difference in x is 0.1, difference in f is 0.146. To go from -0.08 to 0, need 0.08/0.146≈0.547 of the interval. So approximate root at 0.5 + 0.547*0.1≈0.5 +0.0547≈0.5547. Let's check x=0.5547. Sinx≈sin0.5547≈0.529, cosx≈cos0.5547≈0.848. Product≈0.529*0.848≈0.449. Then x + product -1 ≈0.5547 +0.449 -1≈0.0037. Close to zero. Maybe x≈0.554. Let's try x=0.554. Sin0.554≈approx. Let's compute 0.554 radians. Let's use calculator: sin(0.554)≈sin(31.7 degrees)≈0.527, cos(0.554)≈0.850. Product≈0.527*0.850≈0.447. So x + product -1≈0.554 +0.447 -1≈0.001. Very close. So x≈0.554. Maybe even closer. So approximate solution x≈0.554 radians.Alternatively, maybe there's an exact solution. Let me think. Suppose x=0.5? Wait, not likely. Since pi is transcendental, but maybe not. Alternatively, maybe x=pi/6? Let's check. Pi/6≈0.5236. Then sin(pi/6)=0.5, cos(pi/6)=√3/2≈0.866. Product≈0.5*0.866≈0.433. Then x + product -1≈0.5236 +0.433 -1≈-0.043. Not zero. Hmm. So maybe the solution is indeed approximately 0.554 radians. Let's note that down as x≈0.554.Now, moving to the second equation: 2cosy -2y + pi +4 =0. Let's rearrange terms: 2cosy = 2y - pi -4. Divide both sides by 2: cosy = y - (pi +4)/2. Let me compute (pi +4)/2. Pi≈3.1416, so pi +4≈7.1416, divided by 2≈3.5708. So the equation becomes cosy = y -3.5708.Again, transcendental equation. Let's see if we can find y such that cosy = y -3.5708. Let's try some values. First, note that y must be such that y -3.5708 is between -1 and 1 because cosine can only take values in that interval. So y must be in [2.5708, 4.5708]. Let's check y=3. cos3≈-0.989, and 3 -3.5708≈-0.5708. So -0.989 vs -0.5708. Not equal. y=3.5: cos3.5≈-0.936, 3.5 -3.5708≈-0.0708. Not close. y=4: cos4≈-0.6536, 4 -3.5708≈0.4292. Not equal. y=2.5708: cos(2.5708)≈-0.841, 2.5708 -3.5708≈-1. So equation would be cosy = -1. cosy=-1 when y=pi≈3.1416. Wait, but 3.1416 is approximately pi. Let me check y=pi. Then cos(pi)= -1. On the right-hand side: y -3.5708≈3.1416 -3.5708≈-0.4292. So -1 vs -0.4292. Not equal. So maybe another approach.Alternatively, maybe there's a value of y where y -3.5708 is equal to cosy. Let me set up f(y)=cosy - y +3.5708. We need f(y)=0. Let's check y=3: f(3)=cos3 -3 +3.5708≈-0.989 +0.5708≈-0.418. Negative. y=3.5: f(3.5)=cos3.5 -3.5 +3.5708≈-0.936 +0.0708≈-0.865. Still negative. y=4: f(4)=cos4 -4 +3.5708≈-0.6536 -0.4292≈-1.0828. Hmm, getting more negative. What about y=2.5708: f(2.5708)=cos(2.5708) -2.5708 +3.5708≈-0.841 +1≈0.159. Positive. So between y=2.5708 and y=3, the function goes from +0.159 to -0.418. So there's a root between 2.5708 and 3. Let's try y=2.8. f(2.8)=cos2.8≈-0.9422 -2.8 +3.5708≈-0.9422 +0.7708≈-0.1714. Still negative. So between 2.5708 and 2.8. y=2.7: cos2.7≈-0.9041 -2.7 +3.5708≈-0.9041 +0.8708≈-0.0333. Close to zero. y=2.68: cos2.68≈cos(153.6 degrees)≈-0.892 -2.68 +3.5708≈-0.892 +0.8908≈-0.0012. Almost zero. y=2.675: cos(2.675)≈cos(153.3 degrees)≈-0.8912 -2.675 +3.5708≈-0.8912 +0.8958≈0.0046. So between 2.675 and 2.68. Let's use linear approximation. At y=2.675, f(y)=0.0046; y=2.68, f(y)=-0.0012. The difference in y is 0.005, difference in f is -0.0058. To find where f(y)=0: 0.0046 - (0.0058)(delta)/0.005 =0. Solving for delta: delta=0.0046*(0.005)/0.0058≈0.0046*0.862≈0.00396. So approximate root at y=2.675 +0.00396≈2.679. So y≈2.679 radians. Let me verify: y=2.679. cos(2.679)≈cos(153.5 degrees)≈-0.890. Then f(y)= -0.890 -2.679 +3.5708≈-0.890 +0.8918≈0.0018. Close enough. So approximate solution y≈2.679 radians.Alternatively, maybe there's an exact solution here. Let me check if y=pi/2 + something, but since the result is approximately 2.679 which is about 153 degrees, which isn't a standard angle. So likely, it's an approximate solution.So now, we have x≈0.554 radians and y≈2.679 radians. Now, need to compute sin(2x - y). Let's compute 2x - y first. 2x≈2*0.554≈1.108. Then 1.108 - y≈1.108 -2.679≈-1.571 radians. What is -1.571 radians? That's approximately -pi/2 (since pi≈3.1416, so pi/2≈1.5708). So -1.571≈-pi/2. So sin(-pi/2)= -1. But let's verify with the approximate values.Compute sin(2x - y)=sin(-1.571). Since sin(-theta)=-sin(theta), sin(-1.571)= -sin(1.571). 1.571≈pi/2, so sin(pi/2)=1. Therefore, sin(-1.571)= -1. But is 2x - y exactly -pi/2? Let's check.If 2x - y = -pi/2, then y=2x + pi/2. Let's check if this holds with our approximate x and y. If x≈0.554, then 2x + pi/2≈1.108 +1.5708≈2.6788, which is very close to our approximate y≈2.679. That's interesting. So it's possible that the exact solution is y=2x + pi/2. Let me check if this is indeed the case.Suppose y=2x + pi/2. Let's substitute this into the second equation: 2cosy -2y + pi +4=0. Substitute y=2x + pi/2:2cos(2x + pi/2) -2(2x + pi/2) + pi +4=0.Simplify cos(2x + pi/2). Using the identity cos(A + pi/2)= -sinA. So this becomes:2*(-sin2x) -4x - pi + pi +4=0.Simplify terms:-2 sin2x -4x +4=0.Divide both sides by -2:sin2x +2x -2=0.But wait, from the first equation, we have x + sinx cosx -1=0. Which we rewrote as x + (1/2) sin2x -1=0, or 2x + sin2x -2=0. Which is exactly the equation sin2x +2x -2=0. So if we substitute y=2x + pi/2 into the second equation, we end up with the first equation. That means that if x satisfies the first equation, then y=2x + pi/2 satisfies the second equation. Therefore, the exact relation is y=2x + pi/2. Therefore, 2x - y=2x - (2x + pi/2)= -pi/2. Therefore, sin(2x - y)=sin(-pi/2)= -1.Wow, that's clever! So instead of approximating, there's an exact relationship between x and y. By substituting y=2x + pi/2 into the second equation, we reduce it to the first equation, which shows that the solutions are connected in such a way that 2x - y= -pi/2. Therefore, the sine of that is -1. So the exact value is -1.So even though numerically we approximated x≈0.554 and y≈2.679, which gives 2x - y≈-1.571≈-pi/2, the exact value is -pi/2, so sin(-pi/2)= -1. Therefore, the answer is -1.

✅:To solve for ( sin(2x - y) ) given the equations ( x + sin x cos x - 1 = 0 ) and ( 2 cos y - 2y + pi + 4 = 0 ):1. Analyze the First Equation: [ x + sin x cos x - 1 = 0 implies 2x + sin 2x - 2 = 0 quad text{(using ( sin 2x = 2 sin x cos x ))}. ]2. Analyze the Second Equation: [ 2 cos y - 2y + pi + 4 = 0 implies cos y = y - frac{pi + 4}{2}. ]3. Identify Relationship Between ( x ) and ( y ): Assume ( y = 2x + frac{pi}{2} ). Substitute into the second equation: [ 2 cosleft(2x + frac{pi}{2}right) - 2left(2x + frac{pi}{2}right) + pi + 4 = 0. ] Simplify using ( cosleft(2x + frac{pi}{2}right) = -sin 2x ): [ -2 sin 2x - 4x - pi + pi + 4 = 0 implies -2 sin 2x - 4x + 4 = 0 implies sin 2x + 2x - 2 = 0. ] This matches the first equation, confirming ( y = 2x + frac{pi}{2} ).4. Compute ( 2x - y ): [ 2x - y = 2x - left(2x + frac{pi}{2}right) = -frac{pi}{2}. ]5. Evaluate ( sin(2x - y) ): [ sinleft(-frac{pi}{2}right) = -1. ]Final Answer:[boxed{-1}]

🔑:Given the equations:1. ( x + sin(x) cdot cos(x) - 1 = 0 )2. ( 2 cos(y) - 2y + pi + 4 = 0 )We need to determine the value of ( sin(2x - y) ).Step 1: Simplify the first equationThe given equation is:[ x + sin(x) cdot cos(x) - 1 = 0 ]Using the double-angle identity (sin(2x) = 2 sin(x) cos(x)), we can rewrite:[ sin(x) cos(x) = frac{1}{2} sin(2x) ]Thus, the equation becomes:[ x + frac{1}{2} sin(2x) - 1 = 0 ]Multiply the entire equation by 2:[ 2x + sin(2x) - 2 = 0 ]Or,[ 2x + sin(2x) = 2 ]Step 2: Simplify the second equationThe second equation is:[ 2 cos(y) - 2y + pi + 4 = 0 ]Rearrange the terms:[ 2 cos(y) = 2y - pi - 4 ]Divide by 2:[ cos(y) = y - frac{pi}{2} - 2 ]Step 3: Introducing function ( f(x) )Define the function:[ f(x) = x + sin(x) ]We need the properties of ( f(x) ):[ f'(x) = frac{d}{dx}(x + sin(x)) = 1 + cos(x) ]Since ( 1 - cos(x) ) is non-negative for all ( x ) (because (cos(x) leq 1)):[ f'(x) geq 0 ]Therefore, ( f(x) ) is non-decreasing on (mathbb{R}).Step 4: Properties of function (f(x))The function ( f ) is an odd function since:[ f(-x) = -x - sin(-x) = -x + sin(x) = -(x + sin(x)) = -f(x) ]Step 5: Using equations (1) and (2)From the first equation:[ f(2x) = 2 ]From the second equation:[ fleft(frac{pi}{2} - yright) = -2 ]By the odd function nature of (f(x)):[ fleft(frac{pi}{2} - yright) = -fleft(y - frac{pi}{2}right) ]Hence:[ fleft(y - frac{pi}{2} right) = 2 ]Thus:[ 2x + frac{pi}{2} - y = 0 ]or,[ 2x - y = -frac{pi}{2} ]Step 6: Calculating (sin(2x - y))Since:[ 2x - y = -frac{pi}{2} ]We have:[ sin(2x - y) = sinleft(-frac{pi}{2}right) ](sin) is an odd function, thus:[ sinleft(-frac{pi}{2}right) = -sinleft(frac{pi}{2}right) = -1 ]Conclusion:[boxed{-1}]

🧠:Okay, so I need to prove that point E is the midpoint of segment AI, given all these equilateral triangles. Let me try to visualize the problem first. We have points A, B, C on a plane. Then there are points D, E, F, G, H, I such that triangles ABD, BAE, CAF, DFG, ECH, and GHI are all positively oriented equilateral triangles. Positive orientation probably means they are rotated counterclockwise from their respective bases. First, I should sketch this out mentally. Let me start with triangle ABD. Since it's a positively oriented equilateral triangle, if I consider segment AB, then point D should be such that when you rotate AB 60 degrees counterclockwise, you reach D. Similarly, triangle BAE: starting with BA, rotating 60 degrees counterclockwise gives point E. Wait, but triangle BAE is also an equilateral triangle. So point E is constructed by rotating BA 60 degrees counterclockwise. Similarly, triangle CAF: rotating CA 60 degrees counterclockwise gives point F.Then DFG is another equilateral triangle. So starting with DF, rotating 60 degrees counterclockwise gives G. Then ECH: rotating EC 60 degrees gives H, and finally GHI: rotating GH 60 degrees gives I.The goal is to show that E is the midpoint of AI. So if I can find coordinates for all these points and compute the coordinates of E and AI's midpoint, then check if they coincide, that might work. Alternatively, maybe there's a geometric approach using properties of equilateral triangles, midpoints, and rotations.Let me consider setting up a coordinate system. Let me assign coordinates to points A, B, and C to simplify calculations. Let's set point A at the origin (0,0). Let me let point B be at (1,0). Then, point C can be arbitrary, but maybe to make it easier, let me place C somewhere else, say (c_x, c_y). Wait, but maybe choosing coordinates for A, B, and C in such a way that calculations are manageable. Alternatively, maybe place A at (0,0), B at (1,0), and C at (0,1). But maybe that would complicate things. Alternatively, since all constructions are based on equilateral triangles, perhaps using complex numbers would be helpful, as rotations can be easily represented with complex multiplications.Yes, complex plane might be a good approach. Let me model the plane as the complex plane. Let me denote points as complex numbers. Let’s let A be the complex number a, B be b, C be c, and so on.Given that, constructing equilateral triangles with positive orientation (counterclockwise) can be done by multiplying by e^(iπ/3), which is a rotation of 60 degrees. Similarly, the other direction would be multiplying by e^(-iπ/3). Let me confirm: to rotate a vector by 60 degrees counterclockwise, multiply by e^(iπ/3) = cos(π/3) + i sin(π/3) = 0.5 + i (√3)/2.So for example, point D in triangle ABD is such that vector AD is AB rotated 60 degrees counterclockwise. Wait, but triangle ABD is equilateral and positively oriented. So starting from A to B to D to A. So to get from AB to BD, we need a 60-degree rotation. Wait, perhaps it's better to express D in terms of A and B.Alternatively, the standard formula for the third vertex of an equilateral triangle given two vertices. If we have two points, say A and B, then the third point D can be obtained by rotating vector AB by 60 degrees around point B or A. Wait, actually, for triangle ABD to be equilateral and positively oriented, the rotation should be around point A or B? Let me clarify.If triangle ABD is positively oriented, the order is A, B, D. So the rotation from AB to AD should be 60 degrees. Wait, no. The orientation is determined by the order of the points. So in triangle ABD, the vertices are ordered A, B, D, so the rotation at each vertex is such that when you traverse A to B to D, the interior is on the left. So to construct D, we can take vector AB and rotate it 60 degrees counterclockwise around point B, or around point A? Hmm, perhaps need a better approach.Alternatively, in complex numbers, if we have two points a and b, then the third point d such that triangle abd is equilateral and positively oriented can be found by d = a + (b - a) * e^(iπ/3). Is that correct? Let's see. If we take vector AB, which is b - a, rotate it 60 degrees counterclockwise, then add it to point A (which is a), giving the new point D. So yes, that formula would give point D. Similarly, if we have triangle BAE, which is equilateral and positively oriented. So points B, A, E. So starting from BA, which is vector a - b, rotating 60 degrees counterclockwise, and adding to point B to get E. Wait, no. Let's see. The triangle BAE is ordered B, A, E. So the rotation is from BA to BE, but I need to make sure.Wait, maybe another approach. Let me consider each equilateral triangle and express the third vertex in terms of the other two using rotation.For triangle ABD: starting at A, going to B, then to D. So the vector from A to B is (b - a). To get from B to D, we need to rotate the vector AB by -60 degrees (since the triangle is positively oriented). Wait, no. The direction of rotation depends on the order of the vertices. Hmm.Alternatively, the third vertex can be found by rotating one vertex around another by 60 degrees. For example, in triangle ABD, if we rotate point B around point A by 60 degrees counterclockwise, we get point D. Similarly, in triangle BAE, rotating point A around point B by 60 degrees counterclockwise gives point E. Let me confirm.Yes, if you have two points, say A and B, and you want to construct an equilateral triangle ABD with positive orientation, you can rotate point B around point A by 60 degrees counterclockwise to get D. Similarly, for triangle BAE, rotating point A around point B by 60 degrees counterclockwise gives E. Similarly, for triangle CAF, rotating point A around point C by 60 degrees counterclockwise gives F. Is that correct?Wait, maybe. Let me verify with an example. Suppose A is at (0,0), B is at (1,0). Rotating B around A by 60 degrees counterclockwise would give D at (0.5, (√3)/2). Then triangle ABD would indeed be equilateral with positive orientation. Similarly, rotating A around B by 60 degrees counterclockwise: A is (0,0), so translating B to the origin, A becomes (-1,0), rotating 60 degrees counterclockwise gives (-1*cos60 - 0*sin60, -1*sin60 + 0*cos60) = (-0.5, -√3/2). Then translating back by adding B's coordinates (1,0) gives (0.5, -√3/2). So point E would be at (0.5, -√3/2). Then triangle BAE has vertices at B (1,0), A (0,0), and E (0.5, -√3/2). Let me check the distances. BA is 1, BE is distance from (1,0) to (0.5, -√3/2): sqrt((0.5)^2 + (√3/2)^2) = sqrt(0.25 + 0.75) = sqrt(1) = 1. AE is from (0,0) to (0.5, -√3/2): same distance. So yes, that works.Therefore, generalizing, if we have two points X and Y, then the third point Z of the positively oriented equilateral triangle XYZ can be obtained by rotating Y around X by 60 degrees counterclockwise. So Z = X + (Y - X) * e^(iπ/3). In complex numbers, that would be z = x + (y - x) * (cos60 + i sin60). Which is the same as z = y rotated 60 degrees around x.So, applying this:- Point D is obtained by rotating B around A by 60 degrees: D = A + (B - A) * e^(iπ/3)- Point E is obtained by rotating A around B by 60 degrees: E = B + (A - B) * e^(iπ/3)Wait, no. Wait, the formula for rotating a point Y around point X by θ degrees is: Z = X + (Y - X) * e^(iθ). So if we want to rotate Y around X by 60 degrees counterclockwise, θ = 60°, which is π/3 radians. So yes, Z = X + (Y - X) * e^(iπ/3).Therefore, for triangle ABD: rotating B around A by 60 degrees gives D. So D = A + (B - A) * e^(iπ/3).Similarly, triangle BAE: rotating A around B by 60 degrees gives E. So E = B + (A - B) * e^(iπ/3).Similarly, triangle CAF: rotating A around C by 60 degrees gives F. So F = C + (A - C) * e^(iπ/3).Then moving on to triangle DFG: which is an equilateral triangle with positive orientation. So starting from D to F to G. Wait, the triangle is DFG, so the order is D, F, G. So to construct G, we need to rotate F around D by 60 degrees? Or is it rotate D around F? Hmm.Wait, similar logic: to construct the third vertex G given DFG with positive orientation, we can rotate F around D by 60 degrees counterclockwise. So G = D + (F - D) * e^(iπ/3).But let's confirm. If triangle DFG is ordered D, F, G, then the rotation should be from DF to FG. So to go from DF to FG, we rotate DF by 60 degrees counterclockwise around point F. Wait, maybe the third point is obtained by rotating DF around F by 60 degrees? Hmm, this is a bit confusing.Alternatively, if triangle DFG is positively oriented, then the rotation from DF to DG should be 60 degrees. Wait, maybe I need to think in terms of the order of the points. If the triangle is DFG, then the edges are D to F, F to G, G to D. So to maintain positive orientation, when moving from D to F to G, the turn should be to the left. Therefore, the point G is obtained by rotating vector FD around point F by -60 degrees? Or maybe rotating vector DF around point D by 60 degrees? Hmm, perhaps I need a better method.Alternatively, using the same formula as before: given two points X and Y, to find the third point Z such that triangle XYZ is positively oriented equilateral, Z = X + (Y - X) * e^(iπ/3). Wait, but in the case of DFG, the two points are D and F, so G would be D + (F - D)*e^(iπ/3). Let me check with the previous example.Suppose D is (0.5, √3/2), F is some point. If I compute G as D + (F - D)*e^(iπ/3), would that give a positively oriented triangle DFG? Let me take specific coordinates. Suppose D is (0,0), F is (1,0). Then G would be (0,0) + (1,0)*(0.5 + i√3/2) = (0.5, √3/2). So triangle DFG would be (0,0), (1,0), (0.5, √3/2), which is indeed a positively oriented equilateral triangle. So yes, that formula works. Therefore, for triangle DFG, G = D + (F - D)*e^(iπ/3).Similarly, triangle ECH: points E, C, H. So H is obtained by rotating C around E by 60 degrees? Wait, using the same formula: H = E + (C - E)*e^(iπ/3).Wait, no. Wait, triangle ECH is ordered E, C, H. So starting at E, going to C, then to H. So to find H, given E and C, H is E + (C - E)*e^(iπ/3). Let me verify. If E is (0.5, -√3/2), C is, say, (0,1). Then H would be (0.5, -√3/2) + ( (0,1) - (0.5, -√3/2) ) * e^(iπ/3). Let's compute this vector:(0 - 0.5, 1 - (-√3/2)) = (-0.5, 1 + √3/2). Then multiplying by e^(iπ/3) which is (0.5, √3/2). So in complex numbers, multiplying (-0.5 + i(1 + √3/2)) by (0.5 + i√3/2):First, let's compute the real and imaginary parts:Real part: (-0.5)(0.5) - (1 + √3/2)(√3/2) = -0.25 - [ (√3/2 + (√3)^2/(2*2) ) ] = -0.25 - [ (√3/2 + 3/4 ) ] = -0.25 - √3/2 - 0.75 = -1 - √3/2Imaginary part: (-0.5)(√3/2) + (1 + √3/2)(0.5) = (-√3/4) + (0.5 + √3/4) = 0.5So the result is (-1 - √3/2) + i 0.5. Then adding E's coordinates (0.5, -√3/2):Real part: -1 - √3/2 + 0.5 = -0.5 - √3/2Imaginary part: 0.5 + (-√3/2) = 0.5 - √3/2So H would be at (-0.5 - √3/2, 0.5 - √3/2). Hmm, complicated. But the key point is that with complex numbers, we can express each point in terms of previous ones.Similarly, triangle GHI: points G, H, I. So I is obtained by rotating H around G by 60 degrees: I = G + (H - G)*e^(iπ/3).So now, if I express all points in terms of A, B, C, then perhaps I can find an expression for I and then check if E is the midpoint of A and I.Let me proceed step by step. Let me denote e^(iπ/3) as ω for simplicity. So ω = e^(iπ/3) = 0.5 + i√3/2. Note that ω^2 = e^(i2π/3) = -0.5 + i√3/2, and ω^3 = -1, etc.Given that, let's define all points:1. D = A + (B - A)ω2. E = B + (A - B)ω3. F = C + (A - C)ω4. G = D + (F - D)ω5. H = E + (C - E)ω6. I = G + (H - G)ωOur goal is to compute AI's midpoint and check if it equals E.First, let's express all these points in terms of A, B, C.Starting with D:D = A + (B - A)ω = A(1 - ω) + BωSimilarly, E:E = B + (A - B)ω = B(1 - ω) + AωF:F = C + (A - C)ω = C(1 - ω) + AωNow, G is D + (F - D)ω. Let's compute F - D:F - D = [C(1 - ω) + Aω] - [A(1 - ω) + Bω] = C(1 - ω) + Aω - A(1 - ω) - Bω = C(1 - ω) - A(1 - ω - ω) + ... Wait, let's compute term by term.F - D = (C(1 - ω) + Aω) - (A(1 - ω) + Bω) = C(1 - ω) - A(1 - ω) + Aω - Bω = (C - A)(1 - ω) + (A - B)ωTherefore, G = D + (F - D)ω = [A(1 - ω) + Bω] + [ (C - A)(1 - ω) + (A - B)ω ] ωLet me expand that:G = A(1 - ω) + Bω + (C - A)(1 - ω)ω + (A - B)ω^2Similarly, let's compute each term:First term: A(1 - ω)Second term: BωThird term: (C - A)(1 - ω)ω = (C - A)(ω - ω^2)Fourth term: (A - B)ω^2So combining all terms:G = A(1 - ω) + Bω + (C - A)(ω - ω^2) + (A - B)ω^2Let me expand this:= A(1 - ω) + Bω + C(ω - ω^2) - A(ω - ω^2) + Aω^2 - Bω^2Now, let's collect like terms:A terms:A(1 - ω) - A(ω - ω^2) + Aω^2= A[1 - ω - ω + ω^2 + ω^2]= A[1 - 2ω + 2ω^2]B terms:Bω - Bω^2C terms:C(ω - ω^2)So G = A(1 - 2ω + 2ω^2) + B(ω - ω^2) + C(ω - ω^2)Hmm, this seems complicated. Let me recall that ω^2 = ω - 1, since ω^2 + ω + 1 = 0 (since ω is a primitive third root of unity). Wait, actually, ω^3 = 1, and ω ≠ 1, so ω^2 + ω + 1 = 0. Therefore, ω^2 = -ω -1.Let me use that to simplify expressions. So replace ω^2 with -ω -1 wherever it appears.So let's rewrite G:G = A(1 - 2ω + 2ω^2) + B(ω - ω^2) + C(ω - ω^2)Substituting ω^2 = -ω -1:First term: 1 - 2ω + 2ω^2 = 1 - 2ω + 2(-ω -1) = 1 - 2ω -2ω -2 = -1 -4ωSecond term: ω - ω^2 = ω - (-ω -1) = ω + ω +1 = 2ω +1Third term: same as second term, 2ω +1Therefore:G = A(-1 -4ω) + B(2ω +1) + C(2ω +1)Hmm, interesting. Let me write that as:G = (-1 -4ω)A + (1 + 2ω)B + (1 + 2ω)CNow, let's compute H. H = E + (C - E)ω.E is already defined as E = B(1 - ω) + Aω.Therefore, C - E = C - B(1 - ω) - AωThen H = E + (C - E)ω = [B(1 - ω) + Aω] + [C - B(1 - ω) - Aω]ωLet me expand this:= B(1 - ω) + Aω + Cω - B(1 - ω)ω - Aω^2Again, let's substitute ω^2 = -ω -1 where necessary.First term: B(1 - ω)Second term: AωThird term: CωFourth term: -B(1 - ω)ω = -B(ω - ω^2) = -B(ω - (-ω -1)) = -B(2ω +1)Fifth term: -Aω^2 = -A(-ω -1) = Aω + ASo combining all terms:= B(1 - ω) + Aω + Cω - B(2ω +1) + Aω + ANow, let's expand:= B - Bω + Aω + Cω -2Bω - B + Aω + ACombine like terms:B terms: B - B -2Bω - Bω = -3BωA terms: Aω + Aω + A = 2Aω + AC terms: CωSo H = -3Bω + 2Aω + A + CωFactor ω terms:= A + (2A -3B + C)ωHmm, that seems manageable. Now, moving on to I = G + (H - G)ω.Let's compute H - G first:H - G = [A + (2A -3B + C)ω] - [ (-1 -4ω)A + (1 + 2ω)B + (1 + 2ω)C ]Wait, let me rewrite G and H:G = (-1 -4ω)A + (1 + 2ω)B + (1 + 2ω)CH = A + (2A -3B + C)ωTherefore, H - G = [A + (2A -3B + C)ω] - [ (-1 -4ω)A + (1 + 2ω)B + (1 + 2ω)C ]Let me distribute the subtraction:= A + (2A -3B + C)ω - (-1 -4ω)A - (1 + 2ω)B - (1 + 2ω)CExpand each term:= A + (2A -3B + C)ω + (1 + 4ω)A - (1 + 2ω)B - (1 + 2ω)CNow, distribute the coefficients:= A + (1 + 4ω)A + (2A -3B + C)ω - (1 + 2ω)B - (1 + 2ω)CCombine like terms:For A:A + (1 + 4ω)A = A[1 +1 +4ω] = A[2 +4ω]For B:(2A -3B + C)ω - (1 + 2ω)B = 2Aω -3Bω + Cω - B -2ωBFor C:- (1 + 2ω)CSo H - G = A(2 +4ω) + B(-3ω -1 -2ω) + C(ω -1 -2ω) + 2AωWait, wait. Let me recompute:Breaking down term by term:1. A terms:Original A term: AFrom (1 +4ω)A: +A +4ωAFrom (2A -3B + C)ω: +2AωTotal A terms: A + A +4ωA +2Aω = 2A +6ωA2. B terms:From (2A -3B + C)ω: -3BωFrom -(1 +2ω)B: -B -2ωBTotal B terms: -3Bω -B -2ωB = -B -5Bω3. C terms:From (2A -3B + C)ω: +CωFrom -(1 +2ω)C: -C -2ωCTotal C terms: Cω -C -2ωC = -C -CωTherefore, H - G = 2A +6ωA - B -5Bω -C -CωNow, factor terms:= A(2 +6ω) + B(-1 -5ω) + C(-1 -ω)Now, I = G + (H - G)ωWe have G and H - G already expressed. Let me compute I:I = G + (H - G)ω= [ (-1 -4ω)A + (1 + 2ω)B + (1 + 2ω)C ] + [2A +6ωA - B -5Bω -C -Cω] ωFirst, let's compute the second term: [2A +6ωA - B -5Bω -C -Cω] ωMultiply each term by ω:= 2Aω +6ω^2A - Bω -5ω^2B -Cω -ω^2CAgain, substitute ω^2 = -ω -1:= 2Aω +6(-ω -1)A - Bω -5(-ω -1)B -Cω -(-ω -1)C= 2Aω -6ωA -6A - Bω +5ωB +5B -Cω +ωC + CNow, expand:= (2Aω -6Aω) + (-6A) + (-Bω) + (5ωB) +5B + (-Cω + ωC) + CSimplify each group:A terms:2Aω -6Aω = -4Aω-6A remainsB terms:-Bω +5ωB = 4ωB5B remainsC terms:-Cω + ωC = 0 (they cancel)C remainsSo overall:= -4Aω -6A +4ωB +5B + CTherefore, the second term is -4Aω -6A +4ωB +5B + CNow, add this to G:G = (-1 -4ω)A + (1 + 2ω)B + (1 + 2ω)CSo I = G + (H - G)ω = [ (-1 -4ω)A + (1 + 2ω)B + (1 + 2ω)C ] + [ -4Aω -6A +4ωB +5B + C ]Let me combine term by term:A terms:(-1 -4ω)A + (-4ωA -6A) = -A -4ωA -4ωA -6A = -7A -8ωAB terms:(1 + 2ω)B +4ωB +5B = (1 +5)B + (2ω +4ω)B = 6B +6ωBC terms:(1 + 2ω)C + C = (1 +1)C +2ωC = 2C +2ωCThus, I = -7A -8ωA +6B +6ωB +2C +2ωCFactor terms:= A(-7 -8ω) + B(6 +6ω) + C(2 +2ω)Now, our goal is to compute the midpoint of AI. Since A is the origin in complex plane (we didn't set A to 0, but in general, A is a complex number a). Wait, no, in our setup, we didn't fix A at the origin. Wait, actually, in complex numbers, all points are variables unless fixed. Since we assigned A, B, C as arbitrary points, unless we set coordinates. Wait, but in the problem statement, points A, B, C are given on a plane, no specific coordinates. So perhaps the proof should hold for any positions of A, B, C.But in our complex number approach, we can express everything in terms of A, B, C. So the midpoint of AI is (A + I)/2. We need to check if this equals E.Given that E = B(1 - ω) + AωSo let's compute (A + I)/2 and see if it equals E.First, express I:I = -7A -8ωA +6B +6ωB +2C +2ωCSo A + I = A + (-7A -8ωA +6B +6ωB +2C +2ωC) = (-6A -8ωA) +6B +6ωB +2C +2ωCTherefore, (A + I)/2 = [ -6A -8ωA +6B +6ωB +2C +2ωC ] /2 = (-3A -4ωA +3B +3ωB +C +ωC)Now, let's compare this with E:E = B(1 - ω) + Aω = B - Bω + AωSo, E = Aω + B(1 - ω)Therefore, we need to check if:(-3A -4ωA +3B +3ωB +C +ωC)/2 = Aω + B - BωMultiply both sides by 2:-3A -4ωA +3B +3ωB +C +ωC = 2Aω +2B -2BωBring all terms to left side:-3A -4ωA +3B +3ωB +C +ωC -2Aω -2B +2Bω = 0Combine like terms:A terms: -3A -4ωA -2Aω = -3A -6ωAB terms: 3B -2B +3ωB +2Bω = B +5ωBC terms: C +ωCSo total:(-3 -6ω)A + (1 +5ω)B + (1 +ω)C = 0For this equation to hold for all A, B, C, the coefficients must be zero.But unless the coefficients themselves are zero, which they are not:Coefficient of A: -3 -6ωCoefficient of B: 1 +5ωCoefficient of C: 1 +ωThese are not zero unless ω takes specific values, which it doesn't. ω is e^(iπ/3), approximately 0.5 + i0.866.Therefore, this suggests that our initial approach might have an error, or perhaps miscalculations during expansion. Given the complexity of the expressions, this is plausible.Alternatively, perhaps there's a symmetry or a property we're missing that can simplify this.Wait, perhaps instead of using complex numbers with A, B, C as variables, we can assign coordinates to simplify. Let me try setting A at the origin, B at (1,0), and C at some point (p, q). Then compute all points in coordinates.Let me attempt this coordinate approach.Set A at (0,0), B at (1,0), C at (p, q).Compute D: rotating B around A by 60 degrees counterclockwise. So D is obtained by rotating (1,0) 60 degrees around (0,0). The rotation matrix is [cos60 -sin60; sin60 cos60] = [0.5 -√3/2; √3/2 0.5]. So D = (0.5*1 - √3/2*0, √3/2*1 + 0.5*0) = (0.5, √3/2)Compute E: rotating A around B by 60 degrees counterclockwise. To rotate A(0,0) around B(1,0) by 60 degrees, first translate A by (-1,0), rotate, then translate back.Original point: (0 -1, 0 -0) = (-1, 0)Rotation: [0.5 -√3/2; √3/2 0.5] * (-1, 0) = (0.5*(-1) - √3/2*0, √3/2*(-1) + 0.5*0) = (-0.5, -√3/2)Translate back by adding (1,0): (-0.5 +1, -√3/2 +0) = (0.5, -√3/2). So E is at (0.5, -√3/2)Compute F: rotating A around C by 60 degrees counterclockwise. Let C be (p, q). To rotate A(0,0) around C(p,q) by 60 degrees:Translate A by (-p, -q): (-p, -q)Apply rotation: [0.5 -√3/2; √3/2 0.5] * (-p, -q) = (0.5*(-p) - √3/2*(-q), √3/2*(-p) + 0.5*(-q)) = (-0.5p + (√3/2)q, -√3/2 p -0.5 q)Translate back by adding (p, q): x-coordinate: (-0.5p + (√3/2)q) + p = 0.5p + (√3/2)qy-coordinate: (-√3/2 p -0.5 q) + q = -√3/2 p + 0.5 qThus, F = (0.5p + (√3/2)q, -√3/2 p + 0.5 q)Now compute G: triangle DFG, which is an equilateral triangle with positive orientation. So G is obtained by rotating F around D by 60 degrees.First, D is (0.5, √3/2), F is (0.5p + (√3/2)q, -√3/2 p + 0.5 q)To rotate F around D by 60 degrees:Translate F by subtracting D: (0.5p + (√3/2)q -0.5, -√3/2 p + 0.5 q - √3/2)Apply rotation matrix:x’ = 0.5*(x) - (√3/2)*yy’ = (√3/2)*x + 0.5*yThen translate back by adding D.Let’s compute x and y:x = 0.5p + (√3/2)q -0.5y = -√3/2 p + 0.5 q - √3/2So,x’ = 0.5*(0.5p + (√3/2)q -0.5) - (√3/2)*(-√3/2 p + 0.5 q - √3/2)Compute term by term:First term: 0.25p + (√3/4) q -0.25Second term: - (√3/2)*( -√3/2 p + 0.5 q - √3/2 )= (√3/2)*(√3/2 p -0.5 q + √3/2 )= ( (√3 * √3)/4 p ) - ( (√3)/4 q ) + ( (√3 * √3)/4 )= (3/4 p) - (√3/4 q) + (3/4)So x’ total:0.25p + (√3/4) q -0.25 +3/4 p - (√3/4 q) +3/4Combine like terms:p terms: 0.25p +0.75p = pq terms: (√3/4 q - √3/4 q) = 0Constants: -0.25 +0.75 = 0.5Thus, x’ = p +0.5Similarly, compute y’:y’ = (√3/2)*x +0.5*y= (√3/2)*(0.5p + (√3/2)q -0.5) +0.5*(-√3/2 p +0.5 q - √3/2)Compute term by term:First term:√3/2 *0.5p = (√3/4)p√3/2*(√3/2 q) = (3/4)q√3/2*(-0.5) = -√3/4Second term:0.5*(-√3/2 p) = -√3/4 p0.5*(0.5 q) = 0.25 q0.5*(-√3/2) = -√3/4Combine terms:p terms: √3/4 p - √3/4 p = 0q terms: 3/4 q +0.25 q = (3/4 +1/4) q = qConstants: -√3/4 -√3/4 = -√3/2Thus, y’ = q -√3/2Now, translate back by adding D's coordinates (0.5, √3/2):x’ +0.5 = p +0.5 +0.5 = p +1y’ +√3/2 = q -√3/2 +√3/2 = qTherefore, G is at (p +1, q)Wait, that's interesting. G is at (p +1, q). Which is C + (1,0). Since C is (p, q), so G is (p+1, q). That seems like a significant simplification. How did that happen?Similarly, let's compute H: triangle ECH, which is rotating C around E by 60 degrees. E is (0.5, -√3/2). Let's rotate C(p, q) around E by 60 degrees.Translate C by subtracting E: (p -0.5, q +√3/2)Apply rotation matrix:x’ = 0.5*(p -0.5) - (√3/2)*(q +√3/2)y’ = (√3/2)*(p -0.5) +0.5*(q +√3/2)Translate back by adding E.Compute x’:=0.5p -0.25 - (√3/2)q - (√3/2)*(√3/2)=0.5p -0.25 - (√3/2)q - 3/4=0.5p - (√3/2)q -1Translate back x’ +0.5 =0.5p - (√3/2)q -1 +0.5=0.5p - (√3/2)q -0.5Compute y’:=(√3/2)p - (√3/2)*0.5 +0.5q + (√3/4)= (√3/2)p - √3/4 +0.5q +√3/4= (√3/2)p +0.5qTranslate back y’ + (-√3/2) = (√3/2)p +0.5q -√3/2Thus, H is at (0.5p - (√3/2)q -0.5, (√3/2)p +0.5q -√3/2)Now compute I: triangle GHI, which is rotating H around G by 60 degrees. G is at (p +1, q). Let's rotate H around G.Translate H by subtracting G: (0.5p - (√3/2)q -0.5 - (p +1), (√3/2)p +0.5q -√3/2 - q)Simplify coordinates:x =0.5p - (√3/2)q -0.5 -p -1 = -0.5p - (√3/2)q -1.5y = (√3/2)p +0.5q -√3/2 - q = (√3/2)p -0.5q -√3/2Apply rotation matrix:x’ =0.5*(-0.5p - (√3/2)q -1.5) - (√3/2)*( (√3/2)p -0.5q -√3/2 )Compute x’:= -0.25p - (√3/4)q -0.75 - (√3/2)*(√3/2 p) + (√3/2)*(0.5 q) + (√3/2)*(√3/2)= -0.25p - (√3/4)q -0.75 - (3/4)p + (√3/4)q + (3/4)Simplify:p terms: -0.25p -0.75p = -1pq terms: -√3/4 q + √3/4 q =0constants: -0.75 +0.75=0So x’ = -pSimilarly, compute y’:y’ = (√3/2)*(-0.5p - (√3/2)q -1.5) +0.5*( (√3/2)p -0.5q -√3/2 )Compute term by term:First part:√3/2*(-0.5p) = - (√3/4)p√3/2*(-√3/2 q) = - (3/4) q√3/2*(-1.5) = - (3√3)/4Second part:0.5*(√3/2 p) = (√3/4)p0.5*(-0.5q) = -0.25 q0.5*(-√3/2) = -√3/4Combine terms:p terms: -√3/4 p + √3/4 p =0q terms: -3/4 q -0.25 q = -qconstants: - (3√3)/4 -√3/4 = -√3Thus, y’ = -q -√3Translate back by adding G's coordinates (p +1, q):x’ + p +1 = -p +p +1 =1y’ + q = -q -√3 + q = -√3Therefore, point I is at (1, -√3)Wow, that's a constant point! Regardless of where C is (since C was (p, q)), I is always at (1, -√3). Wait, but in our coordinate system, point B is at (1,0), and point E is at (0.5, -√3/2). Let me check what the midpoint of AI is. Point A is at (0,0), point I is at (1, -√3). The midpoint would be ((0 +1)/2, (0 -√3)/2) = (0.5, -√3/2), which is exactly point E. Therefore, E is indeed the midpoint of AI.So regardless of the position of point C, point I always ends up at (1, -√3), making the midpoint of AI at (0.5, -√3/2), which is E. Therefore, the proof holds.But wait, in this coordinate system, we fixed A at (0,0) and B at (1,0), and C was arbitrary (p, q). Through the construction of all the equilateral triangles, point I ended up at (1, -√3), which is a fixed point relative to A and B. Therefore, AI is the segment from (0,0) to (1, -√3), whose midpoint is (0.5, -√3/2), which is E. Hence, proved.But why does this happen? It seems that despite the complex construction involving multiple equilateral triangles and the position of C, the final point I is always fixed at (1, -√3), making E the midpoint. This suggests some underlying geometric invariance or property that isn't immediately obvious but emerges through the iterative construction of equilateral triangles.Therefore, by choosing specific coordinates and computing each step meticulously, we've demonstrated that E is indeed the midpoint of AI. This coordinate approach leveraged the power of assigning coordinates to simplify the problem, allowing us to compute each point's position explicitly and observe the result.Final Answerboxed{E text{ is the midpoint of segment } AI}

✅:Given points (A), (B), and (C) on a plane, we need to prove that point (E) is the midpoint of segment (AI) given the construction of several positively oriented equilateral triangles. We start by assigning coordinates to simplify the calculations. Let (A) be at ((0,0)), (B) at ((1,0)), and (C) at an arbitrary point ((p, q)). 1. Constructing Points (D), (E), and (F): - (D) is obtained by rotating (B) around (A) by 60 degrees counterclockwise. This gives (D = left(0.5, frac{sqrt{3}}{2}right)). - (E) is obtained by rotating (A) around (B) by 60 degrees counterclockwise. This gives (E = left(0.5, -frac{sqrt{3}}{2}right)). - (F) is obtained by rotating (A) around (C) by 60 degrees counterclockwise. This gives (F = left(0.5p + frac{sqrt{3}}{2}q, -frac{sqrt{3}}{2}p + 0.5qright)).2. Constructing Points (G), (H), and (I): - (G) is obtained by rotating (F) around (D) by 60 degrees counterclockwise. This results in (G = (p + 1, q)). - (H) is obtained by rotating (C) around (E) by 60 degrees counterclockwise. This results in (H = left(0.5p - frac{sqrt{3}}{2}q - 0.5, frac{sqrt{3}}{2}p + 0.5q - frac{sqrt{3}}{2}right)). - (I) is obtained by rotating (H) around (G) by 60 degrees counterclockwise. This results in (I = (1, -sqrt{3})).3. Midpoint Calculation: - The midpoint of segment (AI) is calculated as (left(frac{0 + 1}{2}, frac{0 - sqrt{3}}{2}right) = left(0.5, -frac{sqrt{3}}{2}right)), which is exactly the coordinates of point (E).Thus, we have shown through coordinate geometry and rotational transformations that point (E) is indeed the midpoint of segment (AI).[boxed{E text{ is the midpoint of segment } AI}]

🔑:1. Visualize the Plane and Initial Setup: Points ( A ), ( B ), and ( C ) lie on a plane. Points ( D ), ( E ), ( F ), ( G ), ( H ), and ( I ) are also on this plane, forming equilateral triangles ( triangle ABD ), ( triangle BAE ), ( triangle CAF ), ( triangle DFG ), ( triangle ECH ), and ( triangle GHI ).2. Transformation via Rotations: Consider the triangles ( triangle ADF ) and ( triangle CGF ). Since ( triangle DFG ) and ( triangle CAF ) are equilateral: [ AF = CF quad text{and} quad DF = GF ] Also, [ angle DFG = angle AFC = 60^circ ] This means that ( triangle ADF ) can be mapped to ( triangle CGF ) by rotating around point ( F ) counterclockwise by ( 60^circ ).3. Second Transformation: Similarly, consider triangles ( triangle HCG ) and ( triangle HEI ). Since ( triangle GHI ) and ( triangle ECH ) are both equilateral: [ HC = EC quad text{and} quad HG = IG ] and also [ angle HCG = angle HEI = 60^circ ] Thus, ( triangle HCG ) maps to ( triangle HEI ) by rotating around point ( H ) counterclockwise by ( 60^circ ).4. Rotation and Midpoint Positions: If we rotate line segment ( AD ) by ( 120^circ ) around point ( A ), line segment ( AD ) becomes line segment ( AE ). Due to the properties of equilateral triangles, we have: [ AE = AD = CG = EI ] 5. Equidistance and Angle Consistency: Therefore, point ( E ) lies on line ( AI ). Since ( AE ) and ( EI ) are of equal length and the angle formed (considering the equilateral triangle properties) is consistent, conclusion: Since ( E ) is equidistant to ( A ) and ( I ), it is thus the midpoint of ( AI ). Conclusion:Hence, point ( E ) is the midpoint of segment ( AI ).[boxed{E}]