🧠:Okay, so I have these two inequalities to prove involving sine and cosine functions. Let me start with the first one. The problem states that when α₁, α₂, α₃ are in the interval [0, π], the sum of their sines is less than or equal to three times the sine of their average. Hmm, that reminds me of Jensen's inequality, which applies to convex or concave functions. Let me recall: if a function is concave on an interval, then the average of the function at some points is less than or equal to the function of the average of the points. Conversely, for convex functions, the inequality flips.Since we're dealing with sine here, I should check if sine is concave or convex on [0, π]. The second derivative of sin(x) is -sin(x), right? So on [0, π], sin(x) is non-negative because the sine of any angle between 0 and π is between 0 and 1. Therefore, the second derivative is non-positive, which means the sine function is concave on this interval. If that's the case, then by Jensen's inequality for a concave function, we have:( f(α₁) + f(α₂) + f(α₃) ) / 3 ≤ f( (α₁ + α₂ + α₃)/3 )Multiplying both sides by 3 gives exactly the inequality we need to prove for part (1). So that seems straightforward once I recall the concavity of sine on [0, π]. But wait, let me double-check if all conditions for Jensen's are satisfied. The variables α₁, α₂, α₃ are in [0, π], and the function is concave there. So yes, Jensen's applies directly. Therefore, part (1) is proven using Jensen's inequality.Moving on to part (2). Now we have α₁, α₂, α₃ in [-π/2, π/2], and we need to show that the sum of their cosines is less than or equal to three times the cosine of their average. Again, this looks like another application of Jensen's inequality. Let's check the concavity or convexity of cosine on this interval.First derivative of cos(x) is -sin(x), second derivative is -cos(x). So the second derivative is -cos(x). Now, in the interval [-π/2, π/2], cos(x) is non-negative because cosine is positive in the first and fourth quadrants (i.e., between -π/2 and π/2). Therefore, the second derivative of cosine is -cos(x) which is non-positive in this interval. That means cosine is concave on [-π/2, π/2].Again, applying Jensen's inequality for a concave function:( f(α₁) + f(α₂) + f(α₃) ) / 3 ≤ f( (α₁ + α₂ + α₃)/3 )Multiplying by 3 gives the required inequality. So part (2) also follows directly from Jensen's inequality once we establish the concavity of cosine on the given interval.Wait, but let me make sure there's no catch here. For Jensen's inequality, the weights should be non-negative and sum to 1. In the standard Jensen's inequality, yes, but here we are taking an unweighted average, so the weights are all 1/3, which are positive and sum to 1. So that's fine.Is there any possibility that the angles could be outside the interval when averaged? For part (1), since each α_i is in [0, π], their average (α₁ + α₂ + α₃)/3 would also be in [0, π]. Similarly, for part (2), if each α_i is in [-π/2, π/2], then the average is also in [-π/2, π/2]. So the argument of the sine or cosine in the right-hand side is still within the domain where we established concavity. Hence, there's no issue here.Therefore, both inequalities are indeed direct applications of Jensen's inequality, leveraging the concavity of sine on [0, π] and cosine on [-π/2, π/2]. Alternatively, if I wanted to approach this without invoking Jensen's directly, maybe I could use the method of Lagrange multipliers or consider using the AM-GM inequality, but those might be more complicated. Let me think if there's another way, just to verify.For part (1), suppose I consider the function f(α) = sin α. Since f''(α) = -sin α ≤ 0 on [0, π], as before. So the function is concave. Then, by the definition of concavity, for any three points in the interval and weights summing to 1, the function of the weighted average is at least the weighted average of the functions. Here, weights are equal (1/3 each), so indeed, the inequality holds.Alternatively, if I expand it out, maybe using trigonometric identities. Let's say set β = (α₁ + α₂ + α₃)/3. Then, perhaps express each α_i as β + δ_i, where δ₁ + δ₂ + δ₃ = 0. Then, expand sin(β + δ_i) and sum them up. But this approach might get messy. Let's see:sin(β + δ_i) = sin β cos δ_i + cos β sin δ_i.Sum over i:Sum sin α_i = sin β (cos δ₁ + cos δ₂ + cos δ₃) + cos β (sin δ₁ + sin δ₂ + sin δ₃).But since δ₁ + δ₂ + δ₃ = 0, can we use that to simplify the sum of sines and cosines of δ_i?Hmm, but this seems complicated. For example, sin δ₁ + sin δ₂ + sin δ₃. Since δ₁ + δ₂ + δ₃ = 0, maybe we can relate this to some trigonometric identity. Alternatively, note that if δ₃ = - (δ₁ + δ₂), then sin δ₃ = -sin(δ₁ + δ₂). Then the sum sin δ₁ + sin δ₂ - sin(δ₁ + δ₂). But I don't know if that helps.Alternatively, maybe using the identity for sum of sines. But this seems more involved. Maybe it's better to stick with Jensen's.Similarly for the cosine inequality. So perhaps the simplest and most straightforward way is to use Jensen's inequality once we confirm concavity.Therefore, both proofs rely on recognizing the concavity of the respective trigonometric functions on the given intervals and applying Jensen's inequality appropriately.Just to recap:1. For the sine inequality, since sin is concave on [0, π], Jensen gives the result.2. For the cosine inequality, cos is concave on [-π/2, π/2], so Jensen applies again.Therefore, both inequalities are proven using Jensen's inequality after verifying concavity.I should also consider if there are cases where equality holds. For Jensen's inequality, equality occurs when all the variables are equal, i.e., α₁ = α₂ = α₃. So in both cases, equality holds when all angles are the same, which makes sense. For example, if α₁ = α₂ = α₃ = β, then the left side is 3 sin β and the right side is 3 sin β, so equality. Same for cosine.Are there other cases where equality might hold? In Jensen's inequality, for strictly concave functions, equality holds only when all variables are equal. Since sin is strictly concave on [0, π] (since its second derivative is strictly negative except at 0 and π where it's zero), so if all α_i are equal or if two of them are at the endpoints? Wait, but if two angles are at 0 and one is at π, but their average would be (0 + 0 + π)/3 = π/3. Then sin(0) + sin(0) + sin(π) = 0 + 0 + 0 = 0, while 3 sin(π/3) = 3*(√3/2) ≈ 2.598, so 0 < 2.598. So in this case, equality doesn't hold. Hence, only when all angles are equal does equality occur. Similarly for cosine.Therefore, the inequalities are strict unless all α_i are equal.Hence, the proofs are complete using Jensen's inequality. I don't see any gaps in this reasoning. The key was recognizing the concavity of the functions on the given intervals, which allows the application of Jensen's. The intervals are crucial because outside of them, the concavity might change. For example, cosine is convex on [π/2, 3π/2], but here we're only concerned with [-π/2, π/2], where it's concave. Similarly, sine is concave on [0, π] but convex on [π, 2π]. So the intervals given in the problem are essential for the concavity needed for Jensen's.Therefore, the solutions are correct.

✅:Both inequalities can be proven using Jensen's inequality by leveraging the concavity of the respective trigonometric functions over the given intervals.Proof of (1):1. Concavity of Sine: For ( alpha in [0, pi] ), the second derivative of ( sin alpha ) is ( -sin alpha leq 0 ). Thus, ( sin alpha ) is concave on this interval.2. Apply Jensen's Inequality: For concave functions, Jensen's inequality states: [ frac{sin alpha_1 + sin alpha_2 + sin alpha_3}{3} leq sinleft(frac{alpha_1 + alpha_2 + alpha_3}{3}right). ] Multiplying both sides by 3 yields: [ sin alpha_1 + sin alpha_2 + sin alpha_3 leq 3 sin frac{alpha_1 + alpha_2 + alpha_3}{3}. ]3. Equality Condition: Equality holds if and only if ( alpha_1 = alpha_2 = alpha_3 ).Proof of (2):1. Concavity of Cosine: For ( alpha in left[-frac{pi}{2}, frac{pi}{2}right] ), the second derivative of ( cos alpha ) is ( -cos alpha leq 0 ). Hence, ( cos alpha ) is concave here.2. Apply Jensen's Inequality: By Jensen's inequality for concave functions: [ frac{cos alpha_1 + cos alpha_2 + cos alpha_3}{3} leq cosleft(frac{alpha_1 + alpha_2 + alpha_3}{3}right). ] Multiplying by 3 gives: [ cos alpha_1 + cos alpha_2 + cos alpha_3 leq 3 cos frac{alpha_1 + alpha_2 + alpha_3}{3}. ]3. Equality Condition: Equality occurs precisely when ( alpha_1 = alpha_2 = alpha_3 ).Final Answers:(1) (boxed{sin alpha_{1}+sin alpha_{2}+sin alpha_{3} leqslant 3 sin frac{alpha_{1}+alpha_{2}+alpha_{3}}{3}})(2) (boxed{cos alpha_{1}+cos alpha_{2}+cos alpha_{3} leqslant 3 cos frac{alpha_{1}+alpha_{2}+alpha_{3}}{3}})

🔑:Part 1: When alpha_{1}, alpha_{2}, alpha_{3} in [0, pi], to show that:[ sin alpha_{1} + sin alpha_{2} + sin alpha_{3} leqslant 3 sin frac{alpha_{1} + alpha_{2} + alpha_{3}}{3} ]1. Expand the sum sin alpha_{1} + sin alpha_{2} + sin alpha_{3}.[A = sin alpha_{1} + sin alpha_{2}][B = alpha_{1}+ alpha_{2}]Hence, [ sin alpha_{1} + sin alpha_{2} = 2 sin frac{B}{2} cos frac{alpha_{1} - alpha_{2}}{2}]Where cos frac{alpha_{1} - alpha_{2}}{2} leq 1.2. Add other trigonometric expressions for simplification and apply the identity for sum-to-product:[sin alpha_{1} + sin alpha_{2} + sin alpha_{3} = 2 sin frac{B}{2} cos frac{alpha_{1} - alpha_{2}}{2} + 2 sin left( frac{alpha_{3} + frac{alpha_{1} + alpha_{2} + alpha_{3}}{3}}{2} right) cos left( frac{alpha_{3} - frac{alpha_{1} + alpha_{2} + alpha_{3}}{3}}{2} right)]3. Due to the range of functions (0 to π), always maximizes cos terms:[cos frac{alpha_{1} - alpha_{2}}{2} leq 1]Hence,[ sin frac{B}{2} cos frac{alpha_{1} - alpha_{2}}{2} leq sin frac{B}{2}]Similarly, we constrict:[ leq 2 left( sin frac{alpha_{1} + alpha_{2}}{2} + sin frac{alpha_{1} + alpha_{2} + 4alpha_{3}}{6} right)]4. Due to trigonometric bounds and behavior:[leq 4 sin frac{frac{alpha_{1} + alpha_{2}}{2} + frac{alpha_{1} + alpha_{2} + 4alpha_{3}}{6}}{2} cos frac{frac{alpha_{1} + alpha_{2}}{2} - frac{alpha_{1} + alpha_{2} + 4alpha_{3}}{6}}{2}][= 4 sin frac{alpha_{1} + alpha_{2} + alpha_{3}}{3}]Hence,[sin alpha_{1} + sin alpha_{2} + sin alpha_{3} leq 3 sin frac{alpha_{1} + alpha_{2} + alpha_{3}}{3}] Conclusion:[boxed{sin alpha_{1} + sin alpha_{2} + sin alpha_{3} leq 3 sin frac{alpha_{1} + alpha_{2} + alpha_{3}}{3}}]Part 2: When alpha_{1}, alpha_{2}, alpha_{3} in left[-frac{pi}{2}, frac{pi}{2} right], to show that:[ cos alpha_{1} + cos alpha_{2} + cos alpha_{3} leqslant 3 cos frac{alpha_{1} + alpha_{2} + alpha_{3}}{3}]This can be shown similarly by mimicking the proof steps for Part 1, noting the cosine function stays positive in the prescribed range:1. Given alpha_{1}, alpha_{2}, alpha_{3} in left[-frac{pi}{2}, frac{pi}{2} right] every cos value within the range is positive and decreasing. 2. The inequality follows from Jensen's Inequality for concave functions.[ boxed{cos alpha_{1} + cos alpha_{2} + cos alpha_{3} leq 3 cos frac{alpha_{1} + alpha_{2} + alpha_{3}}{3}}]

🧠:Okay, let's see. There are 100 prisoners, right? They get one night to discuss a strategy, and after that, they're in separate cells with no communication. The warden will occasionally take one of them to a room with a lamp that's initially off. When they leave, they can choose to leave the lamp on or off. If someone correctly declares that all have been to the room, they go free; otherwise, it's crocodile food. Also, the warden says that if they stay silent, eventually everyone will visit the room, and no one's last visit will be the final one. So we need a strategy that guarantees someone can eventually make that correct declaration.Hmm, classic problem? I remember something about a counter and others being scouts. Let me think. So the main challenge is that prisoners can't communicate once they're in cells, except through the state of the lamp. The lamp is the only means of passing information. Since the lamp starts off, and each time a prisoner is taken in, they can toggle it or leave it as is.The key is that one prisoner will be designated as the counter. The counter's job is to count the number of times they've seen the lamp turned on by others. Once the counter has counted 99 (since they don't count themselves), they can declare that all have been in the room. But how do the others signal to the counter that they've been there?So, the other 99 prisoners are supposed to act as "scouts." Each of them can turn the lamp on only once in their lifetime. The first time a scout enters the room and finds the lamp off, they turn it on. Subsequent times, if they find the lamp on, they leave it off (but wait, no—if they've already turned it on once, they shouldn't do it again). Wait, maybe each non-counter prisoner is allowed to turn the lamp on only once. So the first time they see the lamp off, they turn it on. After that, they never turn it on again. The counter, on the other hand, whenever they see the lamp on, turns it off and increments their count. Once the count reaches 99, they know all others have been in the room at least once.But there's a problem here. What if the same non-counter prisoner is brought multiple times before the counter has a chance to count their signal? For example, if a non-counter turns the lamp on, then the counter hasn't seen it yet. Then another non-counter comes in and also wants to turn it on. Wait, but each non-counter is only allowed to turn it on once. So if the lamp is already on, even if another non-counter hasn't turned it on before, they can't do anything. Wait, no. The idea is that each non-counter can only turn the lamp on once. So when they first find the lamp off, they turn it on. After that, regardless of the lamp's state, they don't turn it on again. So if the lamp is on when they come in again, they just leave it on. But the counter's job is to turn it off each time they see it on, incrementing the count each time.So the process would be:1. Choose one prisoner as the counter. The rest are scouts.2. Scouts: When a scout enters the room, if the lamp is off and they have never turned it on before, they turn it on. Otherwise, they do nothing (leave the lamp as is).3. Counter: Every time the counter enters the room and finds the lamp on, they turn it off and increment their count. Once their count reaches 99, they declare that all have been to the room.This way, each scout can only signal once by turning the lamp on when it's off. The counter counts each time they turn the lamp off (each signal from a scout). Since there are 99 other prisoners, once the counter has counted 99 signals, they can be sure all scouts have been in the room at least once. Therefore, all prisoners have been in the room.But wait, what if a scout is taken multiple times before the counter can count their signal? For example, a scout turns the lamp on, then before the counter comes in, the same scout is taken again. Since they've already turned the lamp on once, they can't do it again. If the lamp is on when they come in again, they leave it on. So the lamp remains on until the counter comes in and turns it off. So even if the same scout is taken multiple times, they only contribute one signal. So that's okay.Another possible issue: The counter might be taken multiple times and have the lamp off each time, so their count doesn't increase. But as long as every other prisoner has turned the lamp on once, eventually the counter will see it on 99 times. Since the warden ensures that everyone will visit the room infinitely often, the counter will get to see all 99 signals eventually.But also, the problem statement says "if you remain silent, then everyone will visit the room, and for no one will a visit to the room be the last." So the warden is guaranteeing that the process is fair: every prisoner will be taken infinitely often. Therefore, the counter will eventually be taken enough times to count all 99 signals.But let me check again. Suppose a scout is taken first, turns the lamp on. Then another scout is taken, sees the lamp on, but since they haven't turned it on before, can they turn it on? Wait, no. Each scout is allowed to turn the lamp on only once, and only when it's off. So if a scout enters and the lamp is on, regardless of their previous actions, they can't turn it on again. So only the first time they see it off, they turn it on. All subsequent times, they leave it as is.Therefore, the total number of times the lamp can be turned on by scouts is exactly 99 (once per scout). The counter turns it off each time they see it on, counting each such occurrence. Once the counter has done this 99 times, they know all scouts have been in the room.Therefore, this strategy should work. The main points are:- One counter, others are scouts.- Scouts only turn the lamp on once (their first time seeing it off).- Counter counts each time they turn the lamp off.- Since every prisoner is taken infinitely often, all scouts will eventually have their chance to turn the lamp on once, and the counter will count all 99.Therefore, the counter can declare once their count reaches 99.But wait, the counter also has to have been in the room at least once, right? Because the problem says "all of you have already been in the room." So if the counter has counted 99, but hasn't been in the room themselves, then technically, they haven't been in the room. Wait, no. The counter is a prisoner, so they must have been in the room when they were taken. Because the counter is part of the 100 prisoners. So the counter's own presence in the room is accounted for because they are a prisoner. So when the counter has counted 99, that means the other 99 have been in the room, and the counter themselves has been in the room (since they're the one making the declaration). Therefore, all 100 have been in the room.Wait, but the counter may have been in the room multiple times. Each time they go in, they check if the lamp is on. If it's on, they turn it off and count. If it's off, they do nothing. So the counter's own visits don't contribute to the count. So the count is only incremented when another prisoner has signaled. Therefore, when the counter has counted 99, that means 99 other prisoners have signaled, each at least once, so they have all been in the room. The counter themselves have been in the room multiple times, so all 100 prisoners have indeed been in the room.Yes, that makes sense. So the key is that the counter's own visits don't need to be counted because they are the one keeping track. As long as all others have been counted, and the counter is part of the group, then all have been in the room.Therefore, the strategy is solid. Let me try to think of edge cases. For example, what if the counter is the first one to enter the room? The lamp is off. Since they are the counter, they do nothing (they only act if the lamp is on). So they leave it off. Then a scout comes in, sees the lamp off, turns it on. Then maybe the counter comes in again, sees it on, turns it off, increments to 1. Then another scout comes in, turns it on, etc. So that works.Another edge case: all scouts are taken multiple times before the counter can count. But since the warden guarantees that everyone is taken infinitely often, the counter will eventually get to see all 99 signals. The only delay is if the counter isn't taken often enough, but the problem states that if they remain silent, everyone will visit the room infinitely often. Therefore, the counter will have infinitely many chances to count, so eventually, all 99 signals will be received.Yes, this seems to cover it. So the strategy is:1. Elect one prisoner as the counter. The rest are informers.2. Informers: When entering the room, if the lamp is off and you have never turned it on before, turn it on. Otherwise, leave it as is.3. Counter: Each time you enter the room, if the lamp is on, turn it off and increment your count. Once your count reaches 99, declare that all have been in the room.This strategy ensures that each informer signals exactly once, and the counter counts each signal. Once all 99 signals are received, the counter knows all informers (and themselves) have been in the room, so they can safely declare.

✅:Solution:To guarantee the prisoners' release, they can adopt the following strategy:1. Designate a Counter: - Choose one prisoner (e.g., Prisoner 1) to act as the Counter. Their role is to track how many unique prisoners have visited the room by counting signals from others.2. Role of Non-Counters (Scouts): - The remaining 99 prisoners act as Scouts. Each Scout may turn the lamp on only once in their lifetime. Specifically: - If a Scout enters the room and finds the lamp off, and they have never turned it on before, they turn it on. - In all other cases (lamp already on or they have previously turned it on), they leave the lamp unchanged.3. Counter’s Protocol: - Every time the Counter enters the room: - If the lamp is on, they turn it off and increment their internal count by 1. - If the lamp is off, they do nothing. - Once the Counter’s count reaches 99, they confidently declare that all prisoners have visited the room.Why This Works:- Signaling Mechanism: Each Scout signals their first visit by turning the lamp on exactly once. This ensures the Counter receives 99 unique signals (one per Scout).- Fairness Guarantee: The warden ensures every prisoner visits the room infinitely often. Thus, all Scouts will eventually trigger their one-time signal, and the Counter will encounter all 99 "on" states over time.- Completeness: When the Counter reaches 99, they know all Scouts (99 prisoners) have visited. Since the Counter themself is also a prisoner, all 100 have necessarily been in the room.This strategy guarantees a correct declaration with no risk of error, ensuring the prisoners’ eventual release.

🔑:1. Assign Roles: - Select one prisoner to be the “counter.” - All other prisoners will be “regular” prisoners.2. Initial Setup: - The light in the room starts in the OFF position.3. Regular Prisoners' Duties: - Each regular prisoner will only turn on the light (if it is off) the first time they enter the room and will never turn it on again in any subsequent visits. - If the light is already on, they don’t change it.4. Counter's Duties: - The counter will only turn off the light each time they see it on. - The counter will keep a count of how many times they have turned off the light. - Initially, the count is set to zero.5. Counting Criterion: - Each time the counter turns off the light, the counter increments their count by one.6. Completion of Counting: - Once the counter’s count reaches 99, it indicates all 99 prisoners have turned on the light at least once. - The counter can then safely declare that all prisoners have been to the room.7. Detailed Steps: - Step 1: Regular Prisoner’s Visit: - If light is OFF and the regular prisoner has never turned on the light before, turn the light ON. - If light is ON, do nothing and leave. - Step 2: Counter’s Visit: - If light is ON, turn it OFF and increment the count by one. - If light is OFF, do nothing and leave.8. Ensuring the Strategy Works: - Each regular prisoner is responsible for turning on the light exactly once. There are 99 regular prisoners, hence the light will be turned on exactly 99 times. - The counter's role is to count these 99 activations by turning off the light each time it is found on.9. Mathematical and Logical Proof: - Let's assume the counter is initially zero. Whenever a regular prisoner turns on the light, the counter eventually turns it off and increments their count. - Since each of the 99 regular prisoners turns on the light exactly once, the counter will turn off the light exactly 99 times.10. Handling the Starting Condition: - If at the beginning, the state of the light is unknown, the counter can start by turning it OFF once to ensure the known initial state. This does not change the logic of the counting strategy described. Conclusion:With this strategy, the counter will accurately count to 99. Once the count reaches 99, the counter will declare that all prisoners have visited the room. Thus, the release condition is met, ensuring the prisoners' freedom.[boxed{text{Strategy Successfully Guarantees Freedom}}]

🧠:Okay, so I need to find all continuous functions f(x) that satisfy this equation where they're multiplied by coefficients 0, 1, 2, ..., n and each subsequent function is composed with x squared iterated more times. Let me parse the equation again.The equation is:0*f(x) + 1*f(x²) + 2*f(x⁴) + ... + (n-1)*f(x^{2^{n-1}}) + n*f(x^{2^n}) = 0.Wait, hold on. The coefficients are 0, 1, 2, ..., n? So starting from 0 up to n. That means there are n+1 terms in the sum. Let me check the indices. The first term is 0*f(x), then 1*f(x²), up to n*f(x^{2^n}).So, the general term is k*f(x^{2^k}) where k ranges from 0 to n. But when k=0, the coefficient is 0, so the first term is zero. So effectively, the equation is:Sum_{k=1 to n} (k) * f(x^{2^k}) = 0.Wait, but the original problem says from k=0 to n, but with coefficients (0), (1), ..., (n). So the first term is 0*f(x), then 1*f(x²), etc. So the sum is from k=0 to n, with coefficient k multiplied by f(x^{2^k}). Therefore, the equation is:Sum_{k=0}^n k * f(x^{2^k}) = 0.But when k=0, the term is 0*f(x), so it's 0. Therefore, the equation simplifies to:Sum_{k=1}^n k * f(x^{2^k}) = 0.So, we need to find all continuous functions f such that for all x in the domain, this sum equals zero.First thought: Maybe the only solution is the zero function? Let's test simple cases.Let's consider n=1. Then the equation would be 0*f(x) + 1*f(x²) = 0. So f(x²) = 0 for all x. Since x² can take any non-negative real number (assuming domain is real numbers), and if f is continuous, then f must be zero on [0, ∞). But if the domain is all real numbers, then x² covers [0, ∞), but f(x²) = 0 implies f is zero on [0, ∞). However, for x negative, we might have f(x) being something else? Wait, but the original equation for n=1 is f(x²) = 0 for all x. So for x negative, x² is positive, so f(x²) = 0. But if the function is continuous, then as x approaches 0 from the left, f(x) must approach f(0) = 0. Similarly, if the function is defined on all real numbers, then for x negative, f(x) is not determined directly by the equation, unless the function's continuity forces it to be zero. Because suppose f is zero on [0, ∞), but for x < 0, maybe f(x) could be non-zero? But wait, if the function is continuous at x=0, then as x approaches 0 from the left, f(x) approaches f(0) = 0. However, the equation doesn't impose any condition on f(x) for x < 0, except that through the functional equation, perhaps?Wait, no. For n=1, the equation is 1*f(x²) = 0 for all x. So for any real x, x² is non-negative, so f must be zero on [0, ∞). But for x negative, x is not in [0, ∞), so the equation doesn't impose any condition on f(x) for x < 0. However, the problem states "find all continuous functions f(x)", so if f is continuous on ℝ, then even though the equation only forces f to be zero on [0, ∞), but for x < 0, f could be any continuous function that approaches zero as x approaches 0 from the left. Wait, but if f is continuous on ℝ, and f(x) is zero on [0, ∞), then for x < 0, the limit as x approaches 0 from the left is 0, so to maintain continuity, f(x) must approach 0. However, f(x) for x < 0 could be non-zero but approaching 0. Wait, but then the function could be non-zero on (-∞, 0), but still continuous. However, does the functional equation apply to all x, including x negative?Wait, let's take x negative. For example, x = -1. Then x² = 1, so f(1) = 0. But x² is always non-negative, so regardless of whether x is positive or negative, x^{2^k} is non-negative. Therefore, f is forced to be zero on all x^{2^k}, which are non-negative. Thus, for the equation to hold for all real x (including negative x), f must be zero on [0, ∞). But for x < 0, since x^{2^k} is positive, the equation doesn't impose any condition on f(x) for x < 0. However, the problem says "find all continuous functions f(x)". If the domain is all real numbers, then f must be zero on [0, ∞) and can be arbitrary on (-∞, 0), but to be continuous, f must approach zero as x approaches 0 from the left. However, if the problem's domain is only [0, ∞), then f(x) ≡ 0 is the only solution. But the problem statement doesn't specify the domain. Hmm.Wait, the problem says "natural number n" and "continuous functions f(x)". It doesn't specify the domain. In math problems, unless specified, sometimes the default is real numbers, but sometimes it's assumed to be the domain where the expressions make sense. For x^{2^k}, if x is negative, then x^{2^k} is defined (since exponent is an integer power of 2, which is even, so x^{2^k} is positive). So if the domain is all real numbers, then x can be any real number, but the functional equation only constrains f on [0, ∞). However, since the problem says "continuous functions", if the domain is all real numbers, then f could be non-zero on the negative side, but continuous. However, let's check the problem statement again. The original equation must hold for all x in the domain. If the domain is all real numbers, then even though x^{2^k} is non-negative, the equation must hold for all x. Therefore, f must be zero on [0, ∞), but for x < 0, f(x) is unrestricted except for continuity. However, if the problem requires the equation to hold for all x, including x < 0, but the left-hand side only involves f evaluated at non-negative arguments (since x^{2^k} is non-negative for any real x). Therefore, f needs to be zero on [0, ∞) to satisfy the equation for all real x. Then, for x < 0, even though the equation doesn't involve f(x) directly, but the function's definition on negatives is unrestricted except for continuity.But wait, if the problem is over the entire real line, and f is continuous, then even though the equation only forces f to be zero on [0, ∞), the function could be non-zero on (-∞, 0). But the problem asks to "find all continuous functions f(x)" such that the equation holds for all x. However, if the equation only involves f evaluated at non-negative points (since x^{2^k} is non-negative), then even if x is negative, the arguments of f are non-negative. Therefore, the equation reduces to f being zero on [0, ∞). So regardless of x being positive or negative, the arguments inside f are non-negative, so f must be zero on [0, ∞). Then, on (-∞, 0), f can be any continuous function, but since f is continuous on the entire real line, and f is zero on [0, ∞), the limit as x approaches 0 from the left must be zero. Therefore, f(x) must satisfy f(x) → 0 as x approaches 0 from the left. But for x < 0, f(x) could be any continuous function that approaches zero at x=0. However, the problem statement says "Find all continuous functions f(x)", so if the domain is all real numbers, then the solutions are functions that are zero on [0, ∞) and any continuous function on (-∞, 0) that approaches zero at x=0. However, if the domain is [0, ∞), then the only solution is f(x) ≡ 0.But the problem didn't specify the domain. Hmm. Wait, in the original equation, x is raised to the power of 2^k. If we consider x to be a real number, then x must be non-negative to begin with, because otherwise, even though x^{2^k} is defined for even exponents, the function f might not be defined for negative x. Wait, but the problem statement is about f(x), so if f is defined on all real numbers, then even if x is negative, x^{2^k} is positive. However, if the domain of f is only [0, ∞), then x must be restricted to [0, ∞) to have x^{2^k} in the domain. Therefore, maybe the problem assumes the domain of f is [0, ∞). Otherwise, if the domain is all real numbers, the equation would still make sense because x^{2^k} is non-negative, so f is evaluated at non-negative points. However, in that case, the equation still forces f to be zero on [0, ∞), but f can be arbitrary on (-∞, 0), provided it's continuous. But since the problem didn't specify the domain, maybe the intended answer is f ≡ 0, assuming domain [0, ∞).Wait, but in the case of n=1, if the domain is all real numbers, then non-zero solutions exist. For example, f(x) = 0 for x ≥ 0, and f(x) = some continuous function for x < 0 that approaches 0 as x approaches 0. But the problem says "Find all continuous functions f(x)", so depending on the domain, the answer varies. However, in most mathematical contexts, unless specified otherwise, the domain is assumed to be where the expressions make sense. For x^{2^k}, if f is defined on [0, ∞), then x can be in [0, ∞). If f is defined on ℝ, then x can be in ℝ, but x^{2^k} is in [0, ∞). Therefore, unless restricted, the problem might be intended for functions defined on [0, ∞), because otherwise, if defined on ℝ, the solution isn't unique. But given that the problem is likely from an exam or textbook, the answer is probably f ≡ 0. Let me check with n=1.For n=1, the equation is f(x²) = 0 for all x. If x is in [0, ∞), then x² covers [0, ∞), so f must be zero on [0, ∞). If the domain is [0, ∞), then f ≡ 0. If the domain is ℝ, then f must be zero on [0, ∞), but can be non-zero on (-∞, 0). However, if we require the equation to hold for all real x, then even for x negative, the equation still holds because f(x²) = 0. Therefore, in that case, f is zero on [0, ∞), arbitrary on (-∞, 0) but continuous. So, in the real line case, the solutions are functions that are zero on [0, ∞) and continuous on (-∞, 0) with f(x) approaching 0 as x approaches 0 from the left. But maybe the problem expects the domain to be [0, ∞), so the only continuous solution is f ≡ 0.Given that the problem mentions "natural number n" and doesn't specify the domain, it's safer to assume the domain is [0, ∞), so that x^{2^k} is within the domain for all x in the domain. Therefore, in that case, f must satisfy the equation for all x in [0, ∞). Let me proceed under that assumption.So, assuming domain is [0, ∞), then for any x in [0, ∞), the equation Sum_{k=1}^n k * f(x^{2^k}) = 0 holds. Now, we need to find all continuous functions f: [0, ∞) → ℝ such that this holds for all x ≥ 0.Let me consider the case when n=1: Sum_{k=1}^1 k * f(x^{2^k}) = 1*f(x²) = 0. Therefore, f(x²) = 0 for all x ≥ 0. Since x² covers [0, ∞) as x ranges over [0, ∞), this implies f(y) = 0 for all y ≥ 0. Therefore, f ≡ 0 on [0, ∞).Similarly, for n=2: 1*f(x²) + 2*f(x⁴) = 0 for all x ≥ 0. Let me set y = x², then for y ≥ 0, the equation becomes 1*f(y) + 2*f(y²) = 0. So f(y) + 2f(y²) = 0 for all y ≥ 0.This is a functional equation. Let me see if f ≡ 0 is the only solution. Suppose f is not identically zero. Let me assume f is continuous. Let's analyze the equation f(y) = -2f(y²). Suppose y is in [0,1). Then y² is in [0, y). Let's iterate the equation. For y in [0,1), f(y) = -2f(y²) = -2*(-2)f(y⁴) = 4f(y⁴) = 4*(-2)f(y^8) = -8f(y^8), etc. Each time, we can express f(y) in terms of f(y^{2^{m}}). As m increases, y^{2^{m}} approaches 0 (for y in [0,1)). So taking the limit as m → ∞, since f is continuous at 0, f(y^{2^m}) → f(0). Therefore, f(y) = (-2)^m f(y^{2^m}) for any m. Taking m → ∞, if f(0) is not zero, then (-2)^m f(0) would oscillate or blow up unless f(0)=0. But (-2)^m goes to infinity in absolute value if m increases, so unless f(0)=0, the limit doesn't exist. But f is continuous, so f(0) must be the limit as y approaches 0 from the right. However, from the equation, if we take y approaching 0, then f(y) = -2f(y²). Taking y → 0, both sides approach f(0) = -2f(0), which implies f(0) = -2f(0) => 3f(0)=0 => f(0)=0. Therefore, f(0)=0.But then, for y in [0,1), f(y) = -2f(y²). Let's iterate this: f(y) = -2f(y²) = (-2)^2 f(y^{2^2}) = ... = (-2)^m f(y^{2^m}) → (-2)^m f(0) = 0 as m → ∞. But since f(0)=0, then f(y) = 0 for all y in [0,1). Similarly, for y ≥ 1, let's see. Take y ≥1. Then y² ≥ y ≥1. From the equation f(y) = -2f(y²). If y ≥1, then y² ≥1, so we can write f(y²) = -2f(y⁴). Therefore, f(y) = -2*(-2)f(y⁴) = 4f(y⁴). Similarly, f(y⁴) = -2f(y^8), so f(y) = 4*(-2)f(y^8) = -8f(y^8), etc. For y >1, y^{2^m} → ∞ as m → ∞. However, we don't have information about the behavior at infinity. But f is continuous on [0, ∞). Suppose f is zero on [0,1]. Then for y ≥1, f(y) = -2f(y²). But y² ≥1, so f(y²) = -2f(y⁴), so f(y) = -2*(-2)f(y⁴) =4f(y⁴). If y⁴ ≥1, then same applies. If y is such that y^{2^m} eventually exceeds 1, which it does for y >1. But if we assume f is zero on [0,1], then for y ≥1, f(y) = -2*0 = 0. Wait, no. If f is zero on [0,1], then for y ≥1, f(y²) is f evaluated at y² ≥1, which we don't know yet. Wait, let's assume f is zero on [0,1]. Then, for y ≥1, f(y) = -2f(y²). But y² ≥1, so f(y²) = -2f(y⁴). Thus, f(y) = -2*(-2)f(y⁴) =4f(y⁴). If y⁴ ≥1, same thing. If we keep iterating, then f(y) = (-2)^m f(y^{2^m}). But as m increases, y^{2^m} → ∞. However, if f is zero at infinity, but we don't know. However, since f is continuous, unless f tends to zero at infinity, but even so, unless f is zero everywhere, this recursion might not hold. Wait, suppose f is zero on [0,1], then for y ≥1, f(y) = -2f(y²). But if y ≥1, then y² ≥1, so f(y²) = -2f(y⁴). Thus, f(y) = (-2)^m f(y^{2^m}) for any m. If we take m such that y^{2^m} is very large, but unless f tends to zero at infinity, this could be problematic. However, even if f is non-zero at some point y ≥1, this recursive relation would require f(y^{2^m}) to be f(y)/ (-2)^m. As m increases, unless f(y^{2^m}) decays exponentially, which for continuous functions is not possible unless f(y^{2^m}) =0. But if f is continuous and non-zero at some point y0 ≥1, then near y0, f is non-zero, but the recursion would require f(y0^{2^m}) = f(y0)/ (-2)^m. However, as m increases, y0^{2^m} becomes a sequence going to infinity, and f evaluated at those points would need to approach zero. If f is continuous, but we don't have any constraints on behavior at infinity except continuity. Wait, but the problem only requires continuity, not necessarily boundedness or tending to zero. However, the relation f(y) = (-2)^m f(y^{2^m}) must hold for all m. But as m increases, unless f(y^{2^m}) =0, the right-hand side either blows up or oscillates if f(y^{2^m}) is non-zero. However, since f is continuous, suppose there exists some y ≥1 where f(y) ≠0. Then, f(y^{2^m}) = f(y)/ (-2)^m. As m increases, y^{2^m} → ∞, and f(y^{2^m}) must approach f(∞) if it exists. But unless f(∞) =0, this would not hold. But continuity on [0, ∞) doesn't imply anything about the limit at infinity. However, even if f(∞) doesn't exist, the relation f(y^{2^m}) = f(y)/ (-2)^m must hold. For this to be possible, as m increases, the left-hand side f(y^{2^m}) is forced to be f(y)/ (-2)^m, which tends to zero. However, for continuous functions, this would require that for any sequence z_m = y^{2^m} → ∞, f(z_m) →0. Therefore, if f is continuous and f(z) tends to zero as z→infty, then perhaps non-zero solutions exist? But even so, let's suppose f(z) tends to zero as z→infty. Then, for each y ≥1, f(y) = -2f(y²). Let's assume f(y) = C/y for some constant C. Then, f(y²) = C/y², so -2f(y²) = -2C/y². But f(y) = C/y, so equating: C/y = -2C/y² → 1 = -2/y → y= -2. But y ≥1, so this is impossible. Therefore, f(y) cannot be C/y. Alternatively, suppose f(y) = C(-1/2)^{log_2 log_y}. Not sure. Maybe another approach.Alternatively, suppose we define a function recursively. For y ≥1, define f(y) = -2f(y²). If we pick a function on [1, 2), then extend it to [2, 4) via f(y) = -2f(y²), and so on. However, to ensure continuity at each boundary, for example, at y=1, since y approaches 1 from below, f(1) must be 0 (from the left side, since f is zero on [0,1]). But from the right side, f(1) = -2f(1²) = -2f(1). Therefore, f(1) = -2f(1) => 3f(1)=0 => f(1)=0. So at y=1, f is 0. Similarly, at y=2, f(2) = -2f(4), and f(4) = -2f(16), etc. But if we try to define f on [1,2) arbitrarily, then f on [2,4) is determined by f(y) = -2f(√y). Wait, no. Wait, if y is in [2,4), then y² is in [4,16). Wait, but we need to express f(y) in terms of f(y²). For y in [1,2), y² is in [1,4). Then, for y in [2,4), y² is in [4,16), etc. So to define f on [1, ∞), we can define it piecewise. Let's say on [1,2), define f(y) = some continuous function g(y), then on [2,4), f(y) = -2g(√y), and on [4,16), f(y) = (-2)^2 g(√√y), etc. However, to ensure continuity at the boundaries, for example, at y=2, approaching from the left (y→2^-), f(y) approaches g(2), and approaching from the right (y→2^+), f(y) approaches -2g(√2). Therefore, continuity requires that g(2) = -2g(√2). Similarly, at y=4, approaching from the left, f(4^-) = -2g(2), and approaching from the right, f(4^+) = (-2)^2 g(√4) = 4g(2). Therefore, continuity at y=4 requires -2g(2) =4g(2) => -2g(2) -4g(2)=0 => -6g(2)=0 => g(2)=0. Similarly, at y=2, since g(2)=0, then continuity requires g(2) = -2g(√2) => 0 = -2g(√2) => g(√2)=0. Similarly, proceeding recursively, we find that all g(y) must be zero at points y=2^{2^{-m}} for integer m. But these points are dense in [1,2). If g is continuous on [1,2) and zero on a dense subset, then g must be identically zero on [1,2). Therefore, f is zero on [1,2), hence on [2,4), etc., so f is zero on [1, ∞). Therefore, the only continuous solution is f ≡0 on [0, ∞).Therefore, even for n=2, the only continuous solution is f ≡0. Similarly, for higher n, perhaps the same reasoning applies.Alternatively, let's consider general n. The equation is Sum_{k=1}^n k f(x^{2^k}) =0 for all x ≥0. Let's try to use induction on n. For n=1, we saw f ≡0. Suppose for some n ≥1, the only solution is f≡0. Then for n+1, the equation is Sum_{k=1}^{n+1} k f(x^{2^k}) =0. But not sure if induction applies here.Alternatively, set x=0. Then the equation becomes Sum_{k=1}^n k f(0^{2^k}) = Sum_{k=1}^n k f(0) =0. Since 0^{2^k}=0 for all k. Therefore, Sum_{k=1}^n k * f(0) = [n(n+1)/2] f(0) =0. Therefore, unless n(n+1)/2=0, which it isn't for natural n, we have f(0)=0.Therefore, f(0)=0.Next, let's consider x approaching 0. For example, set x=0, we have f(0)=0. For x approaching 0 from the right, x^{2^k} approaches 0 for all k. By continuity, f(x^{2^k}) approaches f(0)=0. But the equation is Sum_{k=1}^n k f(x^{2^k}) =0, which is automatically satisfied as x→0. Not helpful.Alternatively, consider substituting x such that x^{2^m} = y. For example, let’s set y = x^{2^m} for some m. Then x = y^{1/(2^m)}. Substituting into the equation, we get Sum_{k=1}^n k f(y^{2^{k}/2^m}) =0. If we set m=1, then x = y^{1/2}, so the equation becomes Sum_{k=1}^n k f(y^{2^{k-1}}) =0. But not sure if that helps.Alternatively, let's consider the function f as defined on [0, ∞) and consider the behavior under iterations of the squaring function. For x in [0,1), repeated squaring goes to 0. For x=1, it stays 1. For x >1, repeated squaring goes to infinity.Suppose we let x be in [0,1). Then x^{2^k} approaches 0 as k increases. The equation is Sum_{k=1}^n k f(x^{2^k}) =0. For x in [0,1), let's define y_k = x^{2^k}. Then y_{k} = y_{k-1}², starting from y_0 =x. So the sequence y_k is decreasing (if x <1) and approaches 0. The equation becomes Sum_{k=1}^n k f(y_{k}) =0. If we can express this recursively, maybe.Alternatively, consider the case when n=2: f(y) + 2f(y²) =0. As we saw earlier, this forces f≡0. Similarly, for n=3: f(y) + 2f(y²) + 3f(y⁴)=0. Maybe similar reasoning applies. Let's suppose f is non-zero somewhere. Let’s assume there exists a minimal y where f(y) ≠0. Let y0 be the infimum of y ≥0 where f(y)≠0. By continuity, f(y0)=0. If y0=0, then approaching 0 from the right, there are points where f≠0, but f(0)=0. However, from the equation, for x approaching 0, the sum becomes Sum_{k=1}^n k f(x^{2^k}) =0. But all terms f(x^{2^k}) approach f(0)=0, so the equation holds. If y0 >0, then near y0, f is zero on [0, y0). But then, for y just above y0, we can use the functional equation to express f(y) in terms of f(y²), etc. But if y0² < y0 (since y0 >1 would make y0² > y0, but if y0 <1, y0² < y0). Wait, if y0 is in (0,1), then y0² < y0, which contradicts y0 being the infimum where f is non-zero. If y0 is in (0,1), then y0² < y0, so y0² is in [0, y0), where f is zero. Then from the equation, f(y0) = -2f(y0²) - ... but since y0² < y0, f(y0²)=0, hence f(y0)=0, contradicting y0 being the infimum. Therefore, the infimum must be y0 ≥1. But for y0 ≥1, y0² ≥ y0, so y0² is in [y0, ∞). If y0 is the infimum where f is non-zero, then for y < y0, f(y)=0. But then, for y ≥ y0, y² ≥ y0² ≥ y0. From the equation, Sum_{k=1}^n k f(y^{2^k}) =0. For y just above y0, y^{2^k} ≥ y0^{2^k} ≥ y0. If y0=1, then y0²=1, etc. This seems complicated.Alternatively, suppose f is analytic and has a power series expansion around 0. Then, since f(0)=0, and all derivatives at 0 might be zero. But the problem only requires continuity, not analyticity.Another approach: consider the linear operator T acting on f defined by T(f)(x) = Sum_{k=1}^n k f(x^{2^k}). The equation is T(f) =0. We need to find the kernel of T. If T is injective, then the only solution is f=0. To check injectivity, suppose T(f)=0 implies f=0. For that, we might need to analyze the properties of T.Alternatively, consider applying T iteratively. For example, if T(f)=0, then T(T(f))=T(0)=0. But not sure.Alternatively, suppose we define a sequence of functions f_m(x) = f(x^{2^m}). Then the equation becomes Sum_{k=1}^n k f_{k}(x) =0. If we can express this as a recurrence relation, maybe. For example, for the highest term, n f_n(x) = - Sum_{k=1}^{n-1} k f_k(x). Then, f_n(x) = - (1/n) Sum_{k=1}^{n-1} k f_k(x). But f_n(x) = f(x^{2^n}). So f(x^{2^n}) = - (1/n) Sum_{k=1}^{n-1} k f(x^{2^k}). This relates the value of f at x^{2^n} to the values at x^{2^k} for k <n.But how can we use this? Let's consider substituting x with x^{2^m} for various m. For example, substitute x^{2} into the original equation:Sum_{k=1}^n k f( (x²)^{2^k} ) = Sum_{k=1}^n k f(x^{2^{k+1}} ) = Sum_{k=2}^{n+1} (k-1) f(x^{2^k}) =0.But the original equation is Sum_{k=1}^n k f(x^{2^k}) =0. Let's denote the original equation as E(x), and the substituted one as E(x²). Then E(x²): Sum_{k=2}^{n+1} (k-1) f(x^{2^k}) =0. If we subtract E(x²) multiplied by some factor from E(x), maybe we can eliminate terms.Let me write down E(x): Sum_{k=1}^n k f(x^{2^k}) =0.E(x²): Sum_{k=2}^{n+1} (k-1) f(x^{2^k}) =0.Notice that E(x²) has terms from k=2 to n+1, but our original sum goes up to k=n. So if we consider E(x) - E(x²) = [Sum_{k=1}^n k f(x^{2^k})] - [Sum_{k=2}^{n+1} (k-1) f(x^{2^k})].This equals f(x²) + Sum_{k=2}^n [k - (k-1)] f(x^{2^k}) - (n) f(x^{2^{n+1}}).Simplifying, f(x²) + Sum_{k=2}^n 1*f(x^{2^k}) -n f(x^{2^{n+1}}).But this equals f(x²) + Sum_{k=2}^n f(x^{2^k}) -n f(x^{2^{n+1}}).But since in E(x), we have Sum_{k=1}^n k f(x^{2^k}) =0, and we are trying to relate E(x) and E(x²). Not sure if this helps.Alternatively, consider a system of equations. For each m ≥0, substitute x^{2^m} into the original equation. Then we have for each m:Sum_{k=1}^n k f(x^{2^{k + m}} ) =0.This gives an infinite system of equations. But this might not help unless we can find a pattern.Another idea: consider the function f restricted to the interval [0,1]. For x in [0,1], x^{2^k} approaches 0 as k increases. Similarly, for x in [1, ∞), x^{2^k} approaches infinity. Maybe split the analysis into these two intervals.For x in [0,1], let's define y = x^{2^{n}}. Then, as k increases from 1 to n, x^{2^k} = y^{2^{n -k}}}. Wait, maybe not. Let me see:If y = x^{2^n}, then x = y^{1/(2^n)}. Then x^{2^k} = y^{2^k / 2^n} = y^{2^{k -n}}.But for k from 1 to n, k -n ranges from 1 -n to 0. So exponents are 2^{-m} where m from 0 to n-1.But not sure.Alternatively, let's make a substitution. Let’s set t = -ln(ln(1/x)) or something to turn the exponents into linear terms, but this might complicate.Alternatively, suppose we look for a function f that is zero everywhere except at zero. But since f is continuous, the only such function is f≡0.Alternatively, consider the behavior under iteration. Suppose we define a sequence of points x, x², x⁴, ..., x^{2^n}, ... For x in [0,1), this sequence approaches 0. For x=1, it stays 1. For x>1, it approaches infinity.If f is continuous and zero at 0 and at infinity, perhaps f must be zero everywhere. But continuity doesn't necessarily require f to tend to zero at infinity.Wait, but in our previous analysis for n=2, even without assuming behavior at infinity, the recursive relations forced f to be zero. Similarly, perhaps for general n, the recursive application of the functional equation forces f to be zero everywhere.Let me attempt this for general n. Suppose f is continuous and satisfies Sum_{k=1}^n k f(x^{2^k}) =0 for all x ≥0.First, f(0)=0 as shown before.For x in [0,1], x^{2^k} approaches 0 as k increases. For x in [0,1), let's consider the function on intervals [0, ε], ε small. For x in [0, ε], the equation becomes Sum_{k=1}^n k f(x^{2^k}) =0. Since x^{2^k} ≤ x^{2} ≤ ε² for k ≥1. Therefore, all terms in the sum involve f evaluated at points ≤ ε². If ε is small, by continuity, f is near zero. But this might not help.Alternatively, define a function g(t) = f(e^{-t}) for t ≥0. Then, the equation becomes Sum_{k=1}^n k g(t + ln(2^k)) =0. Because x^{2^k} = e^{-t * 2^k} if x= e^{-t}. Wait, no:If x = e^{-t}, then x^{2^k} = e^{-t * 2^k}, so ln(x^{2^k}) = -t * 2^k. Therefore, g(t) = f(e^{-t}), so f(x^{2^k}) = g(t * 2^k). Wait, maybe not the best substitution.Alternatively, use logarithmic substitution. Let x = e^{-2^{-s}} for s ∈ ℝ. Then x^{2^k} = e^{-2^{-s +k}}}. Not sure.Alternatively, take logarithms. Let’s set y = -ln x (for x in (0,1)). Then x^{2^k} = e^{-2^k y}. But not sure.Alternatively, consider the Mellin transform. The Mellin transform of f is M(s) = ∫₀^∞ f(x) x^{s-1} dx. Applying the Mellin transform to the functional equation:Sum_{k=1}^n k M(2^k s) =0.But this is a functional equation for the Mellin transform M(s). The only solution to this might be M(s) ≡0, implying f ≡0. However, I need to verify this.Assuming that f is such that its Mellin transform exists in some strip, and the functional equation Sum_{k=1}^n k M(2^k s) =0 holds. Let’s denote t =2^s. Then the equation becomes Sum_{k=1}^n k M(t^{2^k}) =0. But this might not lead to a straightforward solution.Alternatively, if we suppose that M(s) is analytic and satisfies this equation, then unless M(s) is zero everywhere, the relation Sum_{k=1}^n k M(2^k s) =0 must hold. If we look for non-zero solutions, perhaps M(s) must have a specific form. For example, if M(s) is an exponential function, but Sum_{k=1}^n k e^{c 2^k s} =0 for all s. However, this is impossible unless all coefficients are zero. Therefore, the only solution is M(s)=0, implying f=0.This supports the conclusion that f≡0 is the only solution.Another approach: suppose there exists a non-zero continuous solution f. Let’s take the maximum of |f(x)| on [0, ∞). Let M = sup_{x ≥0} |f(x)|. Since f is continuous and f(0)=0, if M >0, it must be attained at some x >0. Let’s assume x0 >0 is where |f(x0)|=M. Consider the equation Sum_{k=1}^n k f(x0^{2^k}) =0. Therefore, |Sum_{k=1}^n k f(x0^{2^k})| =0. But each |f(x0^{2^k})| ≤ M. Therefore, |Sum_{k=1}^n k f(x0^{2^k})| ≤ Sum_{k=1}^n k |f(x0^{2^k})| ≤ M Sum_{k=1}^n k = M n(n+1)/2. This must be zero, which implies M=0. Therefore, f≡0.Wait, this is a key argument. Suppose f is not identically zero, then let M= sup |f(x)| >0. Then, there exists x0 such that |f(x0)|=M (by continuity, since if f approaches M at infinity, but not sure). However, for maximum attained at x0 (if the function is compactly supported or attains its maximum), then the equation Sum_{k=1}^n k f(x0^{2^k}) =0. Taking absolute values, Sum_{k=1}^n k |f(x0^{2^k})| ≥ |Sum_{k=1}^n k f(x0^{2^k})| =0. But since each |f(x0^{2^k})| ≤ M, then Sum_{k=1}^n k |f(x0^{2^k})| ≤ M Sum_{k=1}^n k = M n(n+1)/2. But the sum is zero, which requires each term to be zero. Therefore, f(x0^{2^k}) =0 for all k=1,...,n. But x0^{2^k} are points obtained by squaring x0 repeatedly. If x0 ≠0, then x0^{2^k} are distinct points (unless x0=1). If f(x0^{2^k})=0 for all k=1,...,n, but we assumed |f(x0)|=M. Therefore, if x0 is such that x0^{2^k} ≠x0 for all k=1,...,n, then we have a contradiction unless M=0. If x0=1, then x0^{2^k}=1 for all k. So if x0=1, then from the equation Sum_{k=1}^n k f(1) =0. So f(1) Sum_{k=1}^n k =0 => f(1)=0. Therefore, even if x0=1, f(1)=0, contradicting |f(x0)|=M>0. Therefore, in any case, the maximum M must be zero. Hence, f≡0.This is a powerful argument. If there exists a point x0 where |f(x0)| is maximum, then the equation forces this maximum to be zero. Therefore, f must be identically zero.Therefore, regardless of n, the only continuous solution is f≡0.This seems to be the case. For any natural number n, the only continuous function satisfying the equation is the zero function.Final AnswerThe only continuous function is the zero function. Thus, the solution is boxed{0}.

✅:To find all continuous functions ( f(x) ) that satisfy the equation[0 cdot f(x) + 1 cdot f(x^2) + 2 cdot f(x^4) + cdots + (n-1) cdot fleft(x^{2^{n-1}}right) + n cdot fleft(x^{2^n}right) = 0,]we start by analyzing the structure of the equation. The equation simplifies to:[sum_{k=1}^n k cdot fleft(x^{2^k}right) = 0 quad text{for all } x geq 0.] Key Steps:1. Case ( n = 1 ): - The equation becomes ( f(x^2) = 0 ) for all ( x geq 0 ). This implies ( f(y) = 0 ) for all ( y geq 0 ). For continuity, ( f ) must be zero on the entire domain ([0, infty)).2. General Case for ( n geq 1 ): - Continuity at Zero: Setting ( x = 0 ) in the equation, we find ( f(0) = 0 ). - Recursive Analysis: For ( x in [0, 1) ), repeated squaring leads to ( x^{2^k} to 0 ). Using continuity, ( f(x^{2^k}) to f(0) = 0 ). This forces ( f ) to be zero on ([0, 1)). - Behavior for ( x geq 1 ): - If ( f ) is non-zero at some ( x geq 1 ), the equation implies a recursive relation that would require ( f ) to be zero at points ( x^{2^k} ). Continuity and the maximum modulus principle show that the maximum value of ( |f| ) must be zero, leading to ( f equiv 0 ).3. Maximum Modulus Argument: - Suppose ( f ) has a non-zero maximum value ( M ). The equation forces the sum of terms involving ( f ) evaluated at iterated squares to zero, leading to a contradiction unless ( M = 0 ). Conclusion:The only continuous function ( f(x) ) that satisfies the given equation for all ( x geq 0 ) and any natural number ( n ) is the zero function.[boxed{0}]

🔑:1. Define a function to simplify notation: [ g_{n}(x) = sum_{k=0}^{n} binom{n}{k} fleft(x^{2^{k}}right) = 0 ] This simplifies the given condition considerably by grouping and summing terms.2. Express ( g_{n-1}(x) ) using the function ( g_n(x) ): [ g_{n-1}(x) = sum_{k=0}^{n-1} binom{n-1}{k} fleft(x^{2^{k}}right) ] By changing the index and using properties of binomial coefficients, we get: [ g_{n-1}(x) = sum_{k=1}^{n} binom{n-1}{k-1} fleft(x^{2^{k-1}}right) ]3. Utilize binomial coefficient properties: Recall that: [ binom{n-1}{k} + binom{n-1}{k-1} = binom{n}{k} ] Using this identity, we see that: [ g_{n}(x) = g_{n-1}(x) + g_{n-1}left(x^{2}right) = 0 ]4. Generalize to negative values of x: [ g_{n-1}(-x) + g_{n-1}left((-x)^{2}right) = g_{n}(-x) = 0 ]5. Symmetry and even function: Since (-x)^2 = x^2, we have: [ g_{n-1}(-x) = g_{n-1}(x) ] and hence g_{n-1} must be an even function.6. Special cases - analysis at x = 0 and x = 1: If we substitute x = 0 and x = 1: [ g_{n-1}(0) + g_{n-1}(0) = 0 quad Rightarrow quad g_{n-1}(0) = 0 ] [ g_{n-1}(1) + g_{n-1}(1) = 0 quad Rightarrow quad g_{n-1}(1) = 0 ]7. Even more general cases: Using the iterative substitution, [ g_{n-1}(x) = -g_{n-1}left(x^{2}right) ] [ g_{n-1}left(x^{2}right) = -g_{n-1}left(x^{2^{2}}right) ] Continuing in this manner, [ g_{n-1}left(x^{2^{k-1}}right) = -g_{n-1}left(x^{2^{k}}right) ]8. Inductive step - repeated substitution: Via induction on k, we find: [ g_{n-1}(x) = (-1)^{k} g_{n-1}left(x^{2^{k}}right) ] As 0 leq x < 1, x^{2^k} to 0 as k to infty. Hence, the continuity of g_{n-1}(x) gives: [ g_{n-1}(x) = pm g_{n-1}(0) = 0 ]9. Generality for all x geq 0: From the relation: [ g_{n-1}(x^{2}) = -g_{n-1}(x) ] we know: [ g_{n-1}(x) = (-1)^k g_{n-1}left(x^{2^k}right) ] Since g_{n-1}(1) = 0, it follows that: [ g_{n-1}(x) = 0 quad forall , x , (text{even}) ]10. Conclusion by induction on n: Repeating this process, we establish: [ g_{n}(x) = 0 implies g_{n-1}(x) = 0 implies cdots implies g_{0}(x) = f(x) = 0 ] Thus, the only continuous function satisfying the given condition is the zero function.Conclusion:[boxed{f(x) = 0}]

🧠:Okay, so I need to figure out how many intersection points there are between the two curves given by the equations ( y = A x^{2} ) and ( y^{2} + 3 = x^{2} + 4 y ). The options are (A) to (E), and I need to determine which one is correct. Let me start by understanding each step carefully.First, the problem involves finding the intersection points of two curves. To find intersections, we can set the two equations equal to each other by substituting ( y ) from the first equation into the second one. That should give us a system to solve for ( x ) and ( y ).The first curve is a parabola that opens upward since ( A ) is positive. The second equation is a bit more complex. Let me rewrite the second equation to see if I can recognize its shape. The equation is ( y^{2} + 3 = x^{2} + 4 y ). If I rearrange the terms, it might become clearer.Let me subtract ( x^{2} + 4 y ) from both sides to get:( y^{2} + 3 - x^{2} - 4 y = 0 )Hmm, maybe completing the square for ( y ) would help. Let me group the ( y ) terms:( y^{2} - 4 y - x^{2} + 3 = 0 )Now, completing the square for ( y ):Take ( y^{2} - 4 y ). To complete the square, take half of -4, which is -2, square it to get 4. So,( y^{2} - 4 y + 4 - 4 - x^{2} + 3 = 0 )Which simplifies to:( (y - 2)^2 - x^{2} -1 = 0 )Then, bringing the constants to the other side:( (y - 2)^2 - x^{2} = 1 )Ah, this is a hyperbola. The standard form of a hyperbola is ( frac{(y - k)^2}{a^2} - frac{(x - h)^2}{b^2} = 1 ). So, this is a hyperbola centered at (0, 2), opening upwards and downwards, with ( a^2 = 1 ) and ( b^2 = 1 ). So, it's a hyperbola with vertices at (0, 2 + 1) = (0, 3) and (0, 2 - 1) = (0, 1), and asymptotes that are lines with slopes ±1 (since a = b = 1). The asymptotes would be ( y - 2 = pm x ).So, the first curve is a parabola opening upwards, and the second is a hyperbola centered at (0,2). The question is about how they intersect, depending on the value of A.Now, to find the intersection points, substitute ( y = A x^2 ) into the hyperbola equation:( (A x^2 - 2)^2 - x^2 = 1 )Let me expand that:First, expand ( (A x^2 - 2)^2 ):= ( A^2 x^4 - 4 A x^2 + 4 )So substituting back into the equation:( A^2 x^4 - 4 A x^2 + 4 - x^2 = 1 )Combine like terms:( A^2 x^4 - (4 A + 1) x^2 + 4 - 1 = 0 )Simplify:( A^2 x^4 - (4 A + 1) x^2 + 3 = 0 )So, this is a quartic equation in terms of ( x ), but since it only has even powers of ( x ), we can let ( t = x^2 ), which gives a quadratic equation:( A^2 t^2 - (4 A + 1) t + 3 = 0 )Now, solving for ( t ):Quadratic equation: ( A^2 t^2 - (4 A + 1) t + 3 = 0 )Let me compute the discriminant ( D ):( D = [ - (4 A + 1) ]^2 - 4 cdot A^2 cdot 3 )= ( (4 A + 1)^2 - 12 A^2 )Expand ( (4 A + 1)^2 ):= ( 16 A^2 + 8 A + 1 - 12 A^2 )Simplify:= ( 4 A^2 + 8 A + 1 )So, discriminant ( D = 4 A^2 + 8 A + 1 ). Since ( A ) is positive, ( D ) is definitely positive because all terms are positive. Therefore, there are two real roots for ( t ).But we need to check if these roots are positive, since ( t = x^2 ) must be non-negative.So, the quadratic equation in ( t ) is:( A^2 t^2 - (4 A + 1) t + 3 = 0 )Let me denote the roots as ( t_1 ) and ( t_2 ). Using quadratic formula:( t = frac{ (4 A + 1) pm sqrt{4 A^2 + 8 A + 1} }{ 2 A^2 } )Now, since ( D > 0 ), two distinct real roots. Now, need to check if these roots are positive.Since all coefficients of the quadratic are positive (A^2 is positive, coefficient of t is negative: -(4A +1), and the constant term is 3). Wait, hold on, coefficients:The quadratic equation is in the form ( a t^2 + b t + c = 0 ), where:- ( a = A^2 > 0 )- ( b = - (4 A + 1) < 0 )- ( c = 3 > 0 )So, quadratic with positive leading coefficient, negative linear coefficient, positive constant term.By Descartes' Rule of Signs, there are two sign changes in the coefficients (from +a to -b to +c), so two positive real roots. So, ( t_1 ) and ( t_2 ) are both positive. Therefore, each positive root ( t ) corresponds to two real solutions for ( x ) (since ( x = pm sqrt{t} )), so each root ( t ) gives two real ( x ) values. Hence, if both roots ( t_1 ) and ( t_2 ) are positive, then there are 4 real solutions for ( x ), hence 4 intersection points. However, if one of the roots is zero or negative, but since we already established that both roots are positive, then it's 4 intersection points. But wait, this seems to suggest that there are always 4 intersection points, but that contradicts the options given. The options include (A) Exactly 4, (B) Exactly 2, etc. But maybe this reasoning is flawed?Wait, let's verify this. So according to this, there are two positive roots for ( t ), so each gives two real x values, so total four intersection points. But let me check with a specific example. Let's take a value for A and see.Take A = 1. Then the quadratic equation becomes ( 1 cdot t^2 - (4*1 +1) t +3 = t^2 -5 t +3 =0 ). The roots would be [5 ± sqrt(25 -12)] /2 = [5 ± sqrt(13)]/2. Both roots are positive, so x^2 = t1 and t2, so x = ±sqrt(t1), ±sqrt(t2). So four points. So in this case, 4 intersection points. Similarly, maybe for other A?Wait, but perhaps for some A, the quadratic in t might have a double root, but since discriminant is D=4A² +8A +1 which is always positive (since A is positive), so quadratic in t will always have two distinct real roots, each positive. Thus, for any positive A, there are two positive t's, hence four x's. Therefore, answer (A) Exactly 4 points. But wait, the options include (A), (B), etc. However, the options given also have (C) which says the number varies with A, (D) says there exists A where there are 0 intersections. So why is this contradiction?Wait, maybe my reasoning is incorrect. Let me check with another value. Suppose A is very large. For example, A approaching infinity. Then the quadratic equation in t is dominated by the terms with A^2. Let's see:Quadratic equation: ( A^2 t^2 - (4 A + 1) t + 3 =0 )For large A, approximate by ignoring lower terms:( A^2 t^2 -4 A t approx 0 )Factor:( A t (A t -4 ) approx 0 )Solutions: t ≈0 and t≈4/A. So as A becomes very large, one root approaches zero and the other approaches 4/A, which also approaches zero. Wait, but if A is large, then 4/A is small. So both roots are approaching zero, but since A is large, maybe the actual roots?Wait, but even so, for very large A, the quadratic equation would be:t^2 - (4/A + 1/A²) t + 3/A² =0. Hmm, maybe scaling variables. Let me divide the equation by A²:t^2 - (4/A + 1/A²) t + 3/A² =0As A approaches infinity, this tends to t^2 =0, so double root at t=0. But since A is finite, even for very large A, we would have two roots close to zero, but still positive? Because discriminant D=4A² +8A +1, which for large A is approx 4A², so sqrt(D)=2A. Then, the roots:t = [ (4A +1) ± 2A ] / (2 A² )Which is approximately:For the positive root: [4A +1 +2A]/(2 A²) = (6A +1)/(2A²) ≈ 6A/(2A²)=3/(A)For the negative root: [4A +1 -2A]/(2 A²) = (2A +1)/(2 A²) ≈ 2A/(2A²)=1/ASo both roots are positive but approaching zero as A increases. Thus, t1 ≈3/A and t2≈1/A. Therefore, x would be ±sqrt(3/A) and ±sqrt(1/A). So four points. So even for very large A, there are four real intersection points.Wait, but maybe when A is very small? Let's take A approaching zero. Then the quadratic equation in t:( A^2 t^2 - (4A +1) t +3 =0 )As A approaches zero, this becomes:-1*t +3=0, so t=3. So the equation becomes approximately -t +3=0, so t=3. Then the other term, A^2 t^2 is negligible, so the other root would be (as A approaches zero):Using quadratic formula:t = [ (4A +1) ± sqrt(4A² +8A +1) ] / (2 A² )For A approaching zero, let's expand sqrt(4A² +8A +1). Let me set A approaching 0, so let me write sqrt(1 +8A +4A²). Using Taylor expansion: sqrt(1 + x) ≈1 + x/2 -x²/8 + ... where x=8A +4A²≈8A.Thus, sqrt(1 +8A) ≈1 + (8A)/2 - ( (8A)^2 )/8 + ... =1 +4A - 8A² +...Therefore, sqrt(4A² +8A +1) ≈1 +4A -8A² +...Therefore, the numerator for t:(4A +1) ± [1 +4A -8A²]Taking the positive sign:(4A +1) +1 +4A -8A² = 8A +2 -8A²Divide by 2A²:(8A +2 -8A²)/(2A²) = (8A)/(2A²) + 2/(2A²) -8A²/(2A²) = 4/(A) + 1/A² -4But as A approaches zero, this term goes to infinity. So one root goes to infinity.Taking the negative sign:(4A +1) - [1 +4A -8A²] =4A +1 -1 -4A +8A²=8A²Divide by 2A²: 8A²/(2A²)=4. So t=4. So as A approaches zero, one root approaches 4, the other goes to infinity? Wait, but t=4 is finite. Wait, maybe my approximation is off.Wait, let me compute the limit as A approaches zero of the roots.First, the quadratic equation is:( A^2 t^2 - (4A +1) t +3 =0 )As A approaches zero, we can write:The roots are:t = [ (4A +1) ± sqrt( (4A +1)^2 -12 A^2 ) ] / (2 A^2 )Let me compute the discriminant as A approaches zero:(4A +1)^2 -12A² =16A² +8A +1 -12A²=4A² +8A +1≈1 +8A when A is near zero.Thus, sqrt(1 +8A)≈1 +4A - 8A² as before.Thus, the numerator for t is:For the positive root:4A +1 +1 +4A -8A²= 8A +2 -8A²Divided by 2A²: (8A +2 -8A²)/(2A²)= (8A)/(2A²) + 2/(2A²) -8A²/(2A²)=4/A +1/A² -4So as A approaches zero, this tends to infinity.For the negative root:4A +1 - [1 +4A -8A²] =8A²Divided by 2A²: 8A²/(2A²)=4.So as A approaches zero, one root tends to 4, and the other tends to infinity. Therefore, for very small A, one of the roots is around 4 and the other is very large.But since t =x², x=±sqrt(t), so even if t is large, x exists. So for A approaching zero, we have two roots t=4 and t approaching infinity, but both are positive, so x=±sqrt(4)=±2 and x=±sqrt(large t), which are real. Therefore, even as A approaches zero, there are four intersection points.Wait, but this contradicts the idea that as A approaches zero, the parabola becomes flatter (since y = A x²). If A is zero, the parabola becomes the x-axis (y=0). But A is positive, so A can't be zero. However, as A approaches zero, the parabola becomes very wide and flat.But the hyperbola is fixed as (y-2)^2 -x²=1. So even if the parabola is very flat, does it intersect the hyperbola in four points?If A approaches zero, then y = A x² is almost the x-axis. The hyperbola (y-2)^2 -x²=1 is centered at (0,2), opening up and down. The vertices are at y=3 and y=1. The asymptotes are y=2±x. The x-axis (y=0) is far below the hyperbola. The closest points on the hyperbola to the x-axis are at y=1 (since the hyperbola's lower vertex is at y=1). The distance between y=0 and y=1 is 1 unit. But the parabola y=A x², when A is very small, will be very flat, so for very large x, y can reach up to 1? Wait, because if A is small, say A=0.001, then y=0.001 x². To get y=1, x²=1000, so x=±sqrt(1000)=±31.62... So the parabola intersects the hyperbola at points where y=1, which is the vertex of the hyperbola. Let's check:If y=1, substitute into hyperbola equation: (1 -2)^2 -x²=1 → 1 -x²=1 → x²=0 → x=0. So the hyperbola has a point at (0,1). But the parabola y=A x² at x=0 is y=0. So when A approaches zero, the parabola is y=0 near the origin, but for very large x, it can reach up to y=1. However, does it actually intersect the hyperbola?Wait, let me plug y=A x² into the hyperbola equation:At intersection points: (A x² -2)^2 -x²=1As A approaches zero, let's see what x is needed to satisfy this equation.Expanding:A² x^4 -4A x² +4 -x²=1 → A² x^4 - (4A +1)x² +3=0If A is very small, say A=ε approaching zero, then the equation becomes approximately:- (1) x² +3=0 → -x² +3=0 → x²=3 → x=±√3. But we neglected the A terms. Wait, but even for tiny A, there are terms with A. Let me see.Suppose A is very small, and x is large. Let me suppose that x is such that A x² is of order 1. Let's set A x² = c, where c is a constant. Then x² = c / A. Substitute into the equation:A² (c/A)^2 - (4A +1)(c/A) +3=0Simplify:A² (c² / A²) - (4A +1)(c / A) +3=0= c² - (4A c + c)/A +3=0= c² -4c - c/A +3=0But as A approaches zero, the term -c/A dominates unless c=0. But if c is non-zero, then -c/A tends to ±infinity. So this suggests that unless c approaches zero, the equation can't be satisfied. Hmm, maybe this approach isn't helpful.Alternatively, let's consider the case when A is very small. Let's look for solutions where y = A x² intersects the hyperbola. The hyperbola has a lower vertex at (0,1). So if y = A x² is to intersect the hyperbola, then there must be points where A x² >=1 (since the hyperbola's lower part is at y=1). But A is very small, so x must be large enough such that A x² >=1. Let's suppose that x is large, so x² = O(1/A). Let me set x² = k / A, where k is a constant. Then y = A*(k / A) = k. Substitute into the hyperbola equation:(k -2)^2 - (k / A) + 3 = 1Wait, no. Wait, original hyperbola equation is (y-2)^2 -x² =1. So substituting y =k and x² = k/A:(k -2)^2 - (k / A) =1But if A approaches zero, then (k / A) becomes very large unless k=0. If k is non-zero, then (k / A) term dominates, so equation becomes -k/A ≈1, which would require k negative, but k = y = A x² >=0 since A is positive. Therefore, no solution in this case.Wait, this is getting confusing. Maybe I should try a specific small A. Let's take A=0.1. Then the quadratic equation in t is:(0.1)^2 t^2 - (4*0.1 +1) t +3 =0 → 0.01 t^2 -1.4 t +3=0Multiply by 100 to eliminate decimals: t^2 -140 t +300=0Discriminant D=140² -4*1*300=19600 -1200=18400sqrt(18400)=135.6...Solutions: [140 ±135.6]/2Positive root: (140 +135.6)/2≈275.6/2≈137.8Negative root: (140 -135.6)/2≈4.4/2=2.2So t1≈137.8 and t2≈2.2. Both positive, so x=±sqrt(137.8)≈±11.74 and x=±sqrt(2.2)≈±1.48. Therefore, four intersection points. So even for A=0.1, there are four real intersection points.But when A=0.1, the parabola is y=0.1 x². For x=±11.74, y=0.1*(137.8)=13.78. But the hyperbola equation at y=13.78 would be (13.78 -2)^2 -x²=1 → (11.78)^2 -x²=1 → x²=138.77 -1=137.77, so x≈±11.74, which matches. For the other root t=2.2, x≈±1.48, y=0.1*(2.2)=0.22. Substitute into hyperbola: (0.22 -2)^2 - (1.48)^2 = ( -1.78 )^2 -2.19 ≈3.17 -2.19≈0.98, which is close to 1. Probably due to rounding errors. So yes, it does intersect.Therefore, even for small A, there are four intersection points. Similarly, for large A, we saw four points. So maybe the answer is (A) Exactly 4 points. But let me check another value.Suppose A=1/4. Then quadratic equation becomes:( (1/4)^2 )t^2 - (4*(1/4) +1 ) t +3 = (1/16)t² - (1 +1)t +3 = (1/16)t² -2t +3=0Multiply by 16: t² -32 t +48=0Solutions: [32 ±sqrt(1024 -192)]/2 = [32 ±sqrt(832)]/2 = [32 ± 28.84]/2Positive roots: (32 +28.84)/2≈60.84/2≈30.42 and (32 -28.84)/2≈3.16/2≈1.58. Both positive. So four intersection points.Alternatively, suppose A=2. Then quadratic equation:(4)t² - (8 +1)t +3=4t² -9t +3=0Discriminant=81 -48=33Roots: [9 ±sqrt(33)]/8≈[9 ±5.7446]/8Positive roots: (9 +5.7446)/8≈14.7446/8≈1.843 and (9 -5.7446)/8≈3.2554/8≈0.4069. Both positive. So x=±sqrt(1.843)≈±1.357 and x=±sqrt(0.4069)≈±0.638. Four points.So in all these cases, four intersection points. Therefore, it seems that regardless of the positive A, there are always four intersection points. Hence, answer (A). But the options given include (E) None of the above. Wait, but option (A) is "Exactly 4 points". But according to our analysis, it's always exactly 4 points for any positive A. Therefore, answer (A). But wait, the options are:(A) Exactly 4 points.(B) Exactly 2 points.(C) At least 1 point, and the number varies with A.(D) For at least one positive A, 0 intersections.(E) None of the above.But according to our analysis, (A) is correct. However, let me check when A is such that the quadratic in t has a double root. Wait, but discriminant D=4A² +8A +1, which is always positive. Therefore, quadratic in t always has two distinct real roots, both positive. Therefore, four intersection points always. Hence, answer (A). But wait, let me check the answer options again. If (A) is correct, then the answer is (A). However, maybe there is an error in my reasoning.Wait, let me think differently. Suppose we consider the two curves: the parabola and the hyperbola. The hyperbola has two branches, upper and lower. The parabola opens upward. So the upper branch of the hyperbola (y >=3) and the lower branch (y <=1). The parabola y=A x² is only in the lower half if A is very small? Wait, no. For any positive A, y=A x² is a parabola opening upwards, passing through the origin. The hyperbola's lower branch is at y <=1. So the parabola might intersect both the upper and lower branches?Wait, the hyperbola equation is (y-2)^2 -x²=1. So solving for y, we get y=2 ± sqrt(x² +1). Therefore, the upper branch is y=2 + sqrt(x² +1) and the lower branch is y=2 - sqrt(x² +1). The lower branch has maximum y=2 -1=1 (at x=0), and it decreases as |x| increases. The upper branch has minimum y=2 +1=3 at x=0 and increases as |x| increases.The parabola y=A x² will intersect the upper branch when A x² >=3. For example, for A=1, x²=3, so x=±sqrt(3). But also, for the lower branch, when A x² <=1. For A=1, x²<=1, so x in [-1,1]. But since the lower branch is y=2 - sqrt(x² +1), which is less than or equal to 1, and the parabola y=x² intersects it when x² =2 - sqrt(x² +1). Let's solve that:x² =2 - sqrt(x² +1)Let me set z =x² >=0.Then z=2 - sqrt(z +1)Rearrange:sqrt(z +1)=2 -zSquare both sides:z +1= (2 -z)^2=4 -4 z +z²Bring all terms to left:z +1 -4 +4 z -z²=0-z² +5 z -3=0 → z² -5 z +3=0Solutions z=(5 ±sqrt(25 -12))/2=(5 ±sqrt(13))/2≈(5 ±3.605)/2≈4.303 or 0.697. Since z =x² >=0, both solutions are valid. Therefore, two points on the lower branch (since z>0, so x=±sqrt(z)) and two points on the upper branch. Thus, total four points.Wait, this is for A=1. So this corroborates our previous result. So in this case, the parabola intersects both branches of the hyperbola twice each. Hence, four points.But maybe for some A, the parabola is tangent to the hyperbola, leading to multiple roots? For that, the discriminant of the quadratic in t would need to be zero. However, discriminant D=4A² +8A +1, which is always positive, as A>0. Therefore, the quadratic in t always has two distinct real roots, leading to four distinct x's. Therefore, exactly four intersection points for any positive A. Hence, answer (A).But wait, the options given are:(A) Exactly 4 points.(B) Exactly 2 points.(C) At least 1 point, and the number varies with different values of A.(D) For at least one positive value of A, the number of intersection points is 0.(E) None of the above statements are correct.But according to this, the correct answer would be (A). However, the original problem might have a trick. Let me check once more with a graph.Imagine the hyperbola: it has an upper branch and a lower branch. The parabola opens upward. For very large A, the parabola is very narrow, so it might intersect the upper branch in two points and the lower branch in two points. For very small A, the parabola is wide and might intersect the upper branch in two points and the lower branch in two points as well. But if the parabola is too high or too low, maybe there are fewer intersections.Wait, but since the hyperbola's lower branch is at y<=1, and the parabola y=A x² is at y>=0. So when A is very small, the parabola is very flat and might intersect the lower branch near y=1. As A increases, the parabola becomes steeper. Wait, but no matter how steep, it still passes through the origin and extends upwards. The upper branch of the hyperbola is always present, so the parabola should intersect the upper branch for large x. Wait, for any A>0, y=A x² will grow without bound as x increases, so it will eventually surpass the upper branch of the hyperbola. However, the upper branch of the hyperbola also grows, but how?The upper branch of the hyperbola is y=2 + sqrt(x² +1). As x increases, sqrt(x² +1) ≈|x|, so y≈2 + |x|. The parabola y=A x². For large x, A x² grows faster than linear, so the parabola will eventually surpass the upper branch. Therefore, the parabola and the upper branch will intersect at two points. Similarly, near the origin, the parabola may intersect the lower branch.Wait, let's analyze the intersection with the upper branch. For large x, the parabola y=A x² and the upper hyperbola y≈2 +x. Setting A x² ≈2 +x. For large x, A x² -x -2=0. This quadratic in x will have two real roots for large enough x, regardless of A. Wait, but A x² -x -2=0 has discriminant 1 +8A, which is always positive, so two real roots. Therefore, the parabola intersects the upper branch at two points for any A>0.Similarly, for the lower branch y=2 - sqrt(x² +1). The maximum y on the lower branch is 1 at x=0. The parabola y=A x² at x=0 is y=0. So near x=0, the parabola is below the lower branch. But as x increases, the parabola increases while the lower branch decreases. Therefore, there must be a point where they cross. For example, at x=0, parabola is at (0,0), while the hyperbola's lower branch is at (0,1). As x increases, the parabola goes up, and the hyperbola's lower branch goes down. Therefore, they must intersect at two points (positive and negative x). Hence, in total, four intersection points.Therefore, regardless of the value of A>0, there are always four intersection points. Therefore, the answer should be (A). However, the options include (E) None of the above. Is there a mistake here?Wait, the options are:(A) Exactly 4 points.(B) Exactly 2 points.(C) At least 1 point, and the number varies with different values of ( A ).(D) For at least one positive value of ( A ), the number of intersection points is 0.(E) None of the above statements are correct.But according to our analysis, (A) is correct. However, the system might have constraints we haven't considered. Wait, when we substituted y=A x² into the hyperbola equation, we obtained quartic equation which we reduced to quadratic in t=x², and found two positive roots, leading to four solutions. But maybe some of these solutions are complex? But no, since the quadratic in t has two positive roots, and x=±sqrt(t) gives real solutions. So four real intersections.Wait, but maybe the original system has constraints. For instance, substituting y=A x² into the hyperbola equation gives (A x² -2)^2 -x²=1. Suppose we find real x and y, but maybe some solutions are extraneous? Let me verify with a specific example.Take A=1, x=√3. Then y=1*(√3)^2=3. Substitute into hyperbola equation: (3-2)^2 - (√3)^2=1 -3= -2≠1. Wait, that's not correct. Wait, this suggests an error in my previous reasoning.Wait, what's wrong here? If A=1, x=√3, then y=3. Plugging into hyperbola equation: y² +3 =x² +4y → 9 +3=3 +12 →12=15? Which is false. Wait, this is a problem. Wait, I must have made a mistake earlier.Wait, wait. The hyperbola equation is y² +3 =x² +4y. Let me recompute for A=1 and x=√3, y=3.Left side: 3² +3=9+3=12.Right side: (√3)^2 +4*3=3+12=15.12≠15. So this is not a solution. But according to the previous substitution, x=±sqrt(t), where t= [5 ±sqrt(13)]/2≈ (5 ±3.605)/2≈4.303 or 0.697. So x=±sqrt(4.303)≈±2.074 and x=±sqrt(0.697)≈±0.835. Let's check these.Take x=2.074, A=1, so y=1*(2.074)^2≈4.303.Left side: y² +3≈(4.303)^2 +3≈18.5 +3=21.5Right side: x² +4y≈4.303 +4*4.303≈4.303 +17.212≈21.515. Close enough, considering rounding.For x=0.835, y=0.697.Left side: y² +3≈0.486 +3=3.486Right side: x² +4y≈0.697 +4*0.697≈0.697 +2.788≈3.485. Again, close.Therefore, the solutions are correct. Earlier mistake was because I used the upper branch substitution y=2 +sqrt(x² +1), but forgot that substituting y=A x² into the hyperbola equation directly gives the correct solutions. The previous error when I took x=√3 and y=3 was because I incorrectly assumed that was a solution, but in reality, the correct solutions are x≈±2.074 and ±0.835.Therefore, the earlier analysis stands: four intersection points for any positive A. Therefore, answer (A). But the original problem lists option (E) as "None of the above". However, according to the options given, if (A) is correct, then (E) is not. But according to the problem statement in China, sometimes options might be tricky. Wait, maybe the original problem is in Chinese and there might be a translation error, but assuming the translation is accurate.Wait, but according to the detailed analysis, for any positive A, there are exactly four intersection points. Thus, option (A) is correct, so the answer should be (A). However, the presence of option (E) suggests that perhaps there is a mistake.Wait, let me consider A=1/16. Then the quadratic in t becomes:(A²)t² - (4A +1)t +3 = (1/256)t² - (0.25 +1)t +3 = (1/256)t² -1.25 t +3=0Multiply by 256: t² -320 t +768=0Discriminant=320² -4*768=102400 -3072=99328sqrt(99328)=315.16Roots=(320 ±315.16)/2Positive roots: (320 +315.16)/2≈635.16/2≈317.58 and (320 -315.16)/2≈4.84/2≈2.42. Both positive. Thus, x=±sqrt(317.58)≈±17.82 and x=±sqrt(2.42)≈±1.55. So four points. Hence, even for very small A, there are four points.Therefore, it seems that regardless of A>0, there are always four intersection points. Therefore, answer (A). But the options include (E) None of the above. Is there a possibility that the problem considers multiplicity of roots? For example, if the quartic equation has a double root, would that count as two points instead of four? But since discriminant D=4A² +8A +1 is always positive, quadratic in t has two distinct roots, each positive, leading to four distinct x's. Hence, four distinct intersection points. Therefore, answer (A). Therefore, the correct option is (A), so why does option (E) exist? Maybe there is a mistake in the problem options.But according to the strict analysis, answer (A) is correct, so the answer should be (A). However, the original problem might have intended different, but based on the mathematical analysis, (A) is correct.But wait, let me check once more for A=1/4. We did earlier and found four points. Another test case: A=0.5.Quadratic equation: (0.5)^2 t² - (4*0.5 +1)t +3=0.25 t² -3t +3=0Discriminant=9 -3=6.Roots=(3 ±sqrt(6))/0.5. Wait, no:Wait, quadratic formula:t=[3 ±sqrt(9 -3)]/(2*0.25)= [3 ±sqrt(6)]/0.5= 2*(3 ±2.449)=6 ±4.898Thus, t1≈10.898, t2≈1.102. Both positive. Hence, four intersection points.Therefore, regardless of A>0, four points. Hence, option (A) is correct.But the given options do include (A). Therefore, unless there is an error in the problem, the answer should be (A). However, the original problem's answer might be (E) if there's a different interpretation.Wait, another possibility: maybe the hyperbola and the parabola are both considered in the real plane, but perhaps for some A, the intersection points have complex coordinates. But we already established that t is always positive, so x is real, hence y=A x² is real. Therefore, all intersection points are real.Therefore, unless the problem has a typo, the correct answer is (A). Hence, option (A) is correct. However, since the options provided include (E), and in some cases, problems might have been designed with different parameters, but according to the thorough analysis, (A) is correct.But let me verify one last thing: the original equations.First equation: y = A x².Second equation: y² +3 =x² +4y.Substituting y=A x² into the second equation:(A x²)^2 +3 =x² +4(A x²)Simplify:A² x^4 +3 =x² +4A x²Bring all terms to left:A² x^4 - (1 +4A) x² +3=0Yes, which is the same as before. Then we set t=x²:A² t² - (1 +4A) t +3=0Discriminant D=(1 +4A)^2 -12A²=1 +8A +16A² -12A²=1 +8A +4A²=4A² +8A +1>0.Hence, two distinct real roots for t, both positive. Therefore, four real intersection points. Therefore, answer (A).Therefore, the correct answer should be (A). But the options given include (E). Maybe there's a mistake in the problem options, but according to the analysis, (A) is correct. However, the original problem might have a different intended answer. Alternatively, perhaps when A is too large or too small, the number of intersection points changes, but according to our analysis, it does not.Wait, but let's consider the geometry. The hyperbola has two branches: upper and lower. The parabola, opening upward, will intersect the upper branch in two points (since it grows faster) and the lower branch in two points (since it starts at the origin and rises, crossing the lower branch which descends from y=1). Therefore, four points in total. Regardless of A>0. Therefore, answer (A). Hence, the answer is (A).But the original options have (E). Unless there is a miscalculation in my analysis. Let me check discriminant D again:D=4A² +8A +1. Since A>0, D is always positive. Therefore, two distinct real roots for t, each positive. Hence, four real solutions for x, hence four points.Thus, answer (A). Therefore, the correct option is (A).Final Answerboxed{A}

✅:To determine the number of intersection points between the curves ( y = A x^2 ) and ( y^2 + 3 = x^2 + 4 y ), we start by substituting ( y = A x^2 ) into the second equation:[(A x^2 - 2)^2 - x^2 = 1]Expanding and simplifying this equation, we get:[A^2 x^4 - (4A + 1) x^2 + 3 = 0]This is a quartic equation in ( x ), but substituting ( t = x^2 ) converts it into a quadratic equation:[A^2 t^2 - (4A + 1) t + 3 = 0]The discriminant ( D ) of this quadratic equation is:[D = (4A + 1)^2 - 12A^2 = 4A^2 + 8A + 1]Since ( A ) is positive, ( D ) is always positive, ensuring two distinct real roots for ( t ). Both roots are positive because the quadratic equation has positive coefficients for the quadratic term and constant term, and a negative coefficient for the linear term (by Descartes' Rule of Signs). Each positive root ( t ) corresponds to two real solutions for ( x ) (since ( x = pm sqrt{t} )), leading to four intersection points.Testing various values of ( A ) (both small and large) confirms that there are always four intersection points. The analysis of the geometry of the curves (parabola and hyperbola) also supports this conclusion, as the parabola intersects both branches of the hyperbola.Thus, the correct answer is:[boxed{A}]

🔑:Step 1: Given the equations are ( y = A x^2 ) and ( y^2 + 3 = x^2 + 4y ), where ( A ) is a positive constant, and ( x ) and ( y ) are real variables. We need to find the number of intersection points of these two curves.Step 2: Substitute ( y = A x^2 ) into the second equation ( y^2 + 3 = x^2 + 4y ):[ y = A x^2 implies x^2 = frac{y}{A} ]Recall the second equation:[ y^2 + 3 = x^2 + 4y ]Now, substitute ( x^2 = frac{y}{A} ) into this equation:[ y^2 + 3 = frac{y}{A} + 4y ]Step 3: Rearrange the equation to form a standard quadratic equation in ( y ):[ y^2 - left(4 + frac{1}{A}right) y + 3 = 0 ]Step 4: Determine the discriminant ((Delta)) of the quadratic equation to find the number of real roots:[ Delta = left(4 + frac{1}{A}right)^2 - 4 cdot 1 cdot 3 ][ Delta = left(4 + frac{1}{A}right)^2 - 12 ]Step 5: Since ( A > 0 ), the term (left(4 + frac{1}{A}right)^2) is always positive, therefore:[ Delta > 0 ]Thus, the quadratic equation ( y^2 - left(4 + frac{1}{A}right) y + 3 = 0 ) has two distinct real roots. Let these roots be ( y_1 ) and ( y_2 ).Step 6: For each root ( y_1 ) and ( y_2 ), solve for ( x ) from the equation ( y = A x^2 ):[ x^2 = frac{y_1}{A} implies x = pm sqrt{frac{y_1}{A}} ][ x^2 = frac{y_2}{A} implies x = pm sqrt{frac{y_2}{A}} ]Conclusion: Each pair ((y_1, x)) and ((y_2, x)) gives two values for ( x ), leading to four intersection points in total. Thus, the number of intersection points between the two curves is exactly 4. Therefore, the correct answer is:[ boxed{A} ]

🧠:Okay, so I need to find all nonzero polynomials P(x) with integer coefficients such that whenever a and b are relatively prime integers, then P(a) and P(b) are also relatively prime. Hmm, interesting problem. Let me try to break this down step by step.First, let's understand the problem statement clearly. We need polynomials P(x) in ℤ[x] (so coefficients are integers) that are nonzero, and for any two coprime integers a and b, the values P(a) and P(b) must also be coprime. So, gcd(P(a), P(b)) = 1 whenever gcd(a, b) = 1. That seems like a strong condition. I need to figure out which polynomials satisfy this.Let me start by considering some simple polynomials and see if they meet this criterion.First, constant polynomials. If P(x) = c where c is a non-zero integer. Then, for any a and b coprime, gcd(P(a), P(b)) = gcd(c, c) = |c|. For this to be 1, c must be ±1. So constant polynomials P(x) = 1 or P(x) = -1 satisfy the condition. That's straightforward.Next, linear polynomials. Let's take P(x) = x + k for some integer k. Let's see. Suppose a and b are coprime. Then, gcd(P(a), P(b)) = gcd(a + k, b + k). Since a and b are coprime, does this imply that a + k and b + k are coprime? Not necessarily. Let's test with an example.Take k = 0, so P(x) = x. Then, gcd(a, b) = 1 implies gcd(a, b) = 1, which is exactly the hypothesis. So P(x) = x works. Wait, but if a and b are coprime, then gcd(a, b) = 1, so yes, the condition is satisfied. Therefore, P(x) = x is a solution. Similarly, P(x) = -x would also work since gcd(-a, -b) = gcd(a, b) = 1. So linear polynomials of the form P(x) = ±x + c? Wait, but if there's a constant term, does that affect it?Let me check with P(x) = x + 1. Let’s take a = 2 and b = 3. These are coprime. Then P(a) = 3, P(b) = 4. gcd(3, 4) = 1. Okay, that works. Another example: a = 1 and b = 2. P(a) = 2, P(b) = 3. gcd(2, 3) = 1. Another example: a = 3 and b = 4. P(a) = 4, P(b) = 5. gcd(4,5)=1. Hmm, seems okay. Wait, but maybe try a case where k is different.Take P(x) = x + 2. Let a = 1 and b = 2 (coprime). P(a)=3, P(b)=4. gcd(3,4)=1. Another example: a=3, b=4. P(a)=5, P(b)=6. gcd(5,6)=1. Hmm, still works. Wait, maybe all linear polynomials of the form x + c satisfy the condition? Let me check with a counterexample.Suppose P(x) = x + 3. Let’s take a = 1 and b = 4 (coprime). P(a) = 4, P(b) = 7. gcd(4,7)=1. Another test: a = 2, b = 3. P(a)=5, P(b)=6. gcd(5,6)=1. Hmm, still okay. Wait, is there any linear polynomial x + c where this fails?Wait, maybe if c is even? Let me try P(x) = x + 4. Take a = 1 and b = 3. P(a)=5, P(b)=7. gcd(5,7)=1. Another test: a=2, b=5 (coprime). P(a)=6, P(b)=9. gcd(6,9)=3 ≠1. Oh! Here we go. So if a=2 and b=5, which are coprime, then P(a)=6 and P(b)=9 have gcd 3, which is not 1. So P(x)=x + 4 does NOT satisfy the condition. Therefore, my previous assumption that all linear polynomials work is incorrect.Wait, so what's the difference between P(x)=x +1 and P(x)=x +4 here? In the case of P(x)=x +1, when a and b are coprime, is there a reason why a +1 and b +1 would be coprime? But in the case of P(x)=x +4, we found a pair where a=2 and b=5, which are coprime, but a+4=6 and b+4=9 have a common factor 3. So that polynomial fails.So why does P(x)=x +1 work? Let's see. Suppose a and b are coprime. Then, is a +1 and b +1 always coprime? Let me test another example with P(x)=x +1. Take a=2, b=3. Then P(a)=3, P(b)=4. gcd(3,4)=1. a=3, b=4. P(a)=4, P(b)=5. gcd(4,5)=1. a=4, b=5. P(a)=5, P(b)=6. gcd(5,6)=1. Hmm, seems okay. Wait, but how about a=1, b=2. P(a)=2, P(b)=3. gcd(2,3)=1. Another example: a=5, b=6. P(a)=6, P(b)=7. gcd(6,7)=1. Hmm, so maybe P(x)=x +1 does work. But why?Wait, maybe it's because x and x +1 are coprime? Wait, but here we are looking at values, not the polynomials themselves. Let me think. Suppose a and b are coprime. Then, is a + c and b + c coprime for some c?But in the case of c=1, is there a reason? For example, suppose p is a prime that divides both a +1 and b +1. Then p divides (a +1 - (b +1)) = a - b. So p divides a - b. But since a and b are coprime, can we derive a contradiction? Let's see. If p divides a - b and p divides a +1, then p divides (a +1) - (a - b) = 1 + b. Similarly, p divides a - b and p divides b +1, so p divides (b +1) + (a - b) = a +1. So p divides both a +1 and b +1. But if p divides a - b and p divides a +1, then p divides (a +1) - (a - b) = 1 + b. Therefore, p divides b +1 and a +1, as well as a - b. So if p is a prime dividing both a +1 and b +1, then p divides (a - b) and p divides (a +1). So p divides (a +1) and p divides (a - b). Therefore, p divides (a +1) + (a - b) = 2a +1 - b. Not sure if that helps.Alternatively, suppose p divides both a +1 and b +1. Then, since p divides a +1 and b +1, it divides their difference: (a +1) - (b +1) = a - b. But p divides a - b. Since a and b are coprime, then p divides a - b and p divides a, then p divides b as well, which would contradict gcd(a, b)=1. Wait, let's formalize that.Suppose p divides a +1 and p divides b +1. Then p divides (a +1) - (b +1) = a - b. Now, if p divides a - b and p divides a, then p divides b. But since gcd(a, b) =1, p must be 1, which is a contradiction. Wait, but p is a prime. So unless p divides both a and b, which are coprime, so p cannot divide both. So if p divides a - b and p divides a, then p divides b. But since a and b are coprime, p can't divide both. Therefore, if p divides a - b, then p can't divide both a and b. Wait, this is getting a bit convoluted. Maybe there's a simpler way.Alternatively, take a=1 and b=2. Then a +1=2, b +1=3, gcd(2,3)=1. Take a=2, b=3: gcd(3,4)=1. a=3, b=4: gcd(4,5)=1. a=4, b=5: gcd(5,6)=1. It seems to hold for these, but maybe there's a case where it doesn't. Wait, let's see a=5, b=7. gcd(6,8)=2. Wait, hold on! If a=5 and b=7, which are coprime, then P(a)=6 and P(b)=8. gcd(6,8)=2 ≠1. So here's a contradiction. Wait, but P(x)=x +1, when a=5 and b=7, which are coprime, gives P(a)=6 and P(b)=8. gcd(6,8)=2. That's not 1. Wait, so P(x)=x +1 doesn't satisfy the condition either. But earlier examples worked. So maybe my initial reasoning was wrong.Wait, so in that case, P(x)=x +1 is not a solution. Then, what's going on here? Maybe only specific linear polynomials work?Wait, so let's check this again. Take P(x) = x +1. Then, take a=5 and b=7. They are coprime. P(a)=6, P(b)=8. gcd(6,8)=2. So this polynomial doesn't satisfy the condition. Then, my initial thought that P(x)=x +1 works is wrong. So maybe even linear polynomials other than x and -x don't work. Wait, but earlier when I tested P(x)=x +1 with a=2 and b=3, it worked. So it's inconsistent? Wait, no, it depends on the choice of a and b. So, if there exists at least one pair of coprime integers a and b such that P(a) and P(b) are not coprime, then the polynomial is excluded. Therefore, even if some pairs work, but if there's at least one pair that doesn't, the polynomial is invalid.Therefore, P(x)=x +1 is invalid because we found a=5 and b=7 where gcd(P(a), P(b))=2≠1. Similarly, P(x)=x +4 is invalid. So maybe only the linear polynomials of the form P(x)=±x work? Let's check.Take P(x)=x. If a and b are coprime, then gcd(a, b)=1, so gcd(P(a), P(b))=gcd(a, b)=1. So that works. Similarly, P(x)=-x. Then gcd(-a, -b)=gcd(a, b)=1. So that also works. What about P(x)=2x? Let's test. Take a=1 and b=2 (coprime). P(a)=2, P(b)=4. gcd(2,4)=2≠1. So P(x)=2x fails. So scaling by a constant other than ±1 breaks the condition. Therefore, only P(x)=±x satisfy the condition among linear polynomials.So far, we have constant polynomials ±1 and linear polynomials ±x. Let's check quadratic polynomials. Suppose P(x) = x². Let's see. If a and b are coprime, then gcd(a², b²) = (gcd(a,b))² = 1² =1. So that works. Similarly, P(x)=-x². Then gcd(-a², -b²)=1. So that works too. What about P(x)=x² +1? Let's test with a=1 and b=2. P(a)=2, P(b)=5. gcd(2,5)=1. a=2, b=3. P(a)=5, P(b)=10. gcd(5,10)=5≠1. So P(x)=x² +1 fails. How about P(x)=x² +x +1? Let's test a=1, b=2. P(1)=3, P(2)=7. gcd(3,7)=1. a=2, b=3. P(2)=7, P(3)=13. gcd(7,13)=1. Hmm, maybe works? Wait, test another pair. a=1, b=3. P(1)=3, P(3)=13. gcd(3,13)=1. a=1, b=4. P(1)=3, P(4)=21. gcd(3,21)=3≠1. So here, a=1 and b=4 are coprime, but P(a)=3 and P(b)=21 have gcd 3. So P(x)=x² +x +1 doesn't work.Therefore, maybe higher-degree polynomials also have similar issues. What about P(x)=x²? Wait, we saw that works. Let me verify. Take a=2 and b=3. P(a)=4, P(b)=9. gcd(4,9)=1. Another example: a=2, b=5. gcd(4,25)=1. a=3, b=4. gcd(9,16)=1. Seems to work. Let me check with a=1 and b=2. gcd(1,4)=1. a=6, b=5. gcd(36,25)=1. How about a=6 and b=7. gcd(36,49)=1. It seems to hold. So P(x)=x² works. Similarly, P(x)=x^n for any positive integer n? Let's check.Suppose P(x)=x^3. Take a=2, b=3. gcd(8,27)=1. a=2, b=5. gcd(8,125)=1. a=3, b=4. gcd(27,64)=1. So that seems to work. Similarly, negative coefficients: P(x)=-x^3. Then gcd(-8, -27)=1. So that works too. So perhaps any monomial of the form ±x^k, where k is a non-negative integer, works? Let me test with k=0. P(x)=1, which is the constant polynomial, which we saw works. k=1, P(x)=x, works. k=2, x², works. So yes. So monomials of the form ±x^k.Wait, let's check a higher power. P(x)=x^4. Take a=2, b=3. gcd(16,81)=1. a=2, b=5. gcd(16,625)=1. a=3, b=4. gcd(81,256)=1. So seems okay. So in general, if P(x) = x^n, then for coprime a and b, a^n and b^n are coprime. Because if a prime p divides a^n and b^n, then p divides a and b, which contradicts gcd(a, b)=1. So yes, x^n works.Similarly, if P(x) = -x^n, since gcd(-a^n, -b^n) = gcd(a^n, b^n) =1. So that works.So monomials ±x^n are okay. What about products of such monomials? For example, P(x)=x(x+1). Wait, but that's a quadratic polynomial. Let's test. Take a=1 and b=2. Then P(1)=2, P(2)=6. gcd(2,6)=2≠1. So that doesn't work. Similarly, any polynomial with multiple factors would potentially fail. Wait, but x^n itself is a monomial. If we have a polynomial with multiple terms, even if they are monomials, like x^n + x^m, does that work? Let's test.Take P(x)=x² + x. Let a=1, b=2. P(a)=2, P(b)=6. gcd(2,6)=2≠1. So no. How about P(x)=x³ + x² + x +1. Test with a=1, b=2. P(1)=4, P(2)=15. gcd(4,15)=1. a=2, b=3. P(2)=15, P(3)=40. gcd(15,40)=5≠1. So that fails. So seems like any polynomial with more than one term may not satisfy the condition. Unless it's of the form ±x^n.Wait, what about P(x) = x + x^3. Let's test. Take a=1, b=2. P(1)=2, P(2)=10. gcd(2,10)=2≠1. So no. Hmmm. So maybe only monomials of the form ±x^n work.But wait, let's test another case. Suppose P(x) = x^2. Let me try a=2, b=3. Then gcd(4,9)=1. If I take a=2 and b=5, gcd(4,25)=1. If I take a=2 and b=7, gcd(4,49)=1. Similarly, a=3 and b=4, gcd(9,16)=1. Seems to hold. Now, what if I take a=1 and b= any number. gcd(1, b^n)=1, which is true. So perhaps for any monic monomial, this works.Therefore, the conjecture is that all polynomials satisfying the condition are the monomials ±x^n for non-negative integers n, and the constant polynomials ±1. Wait, but the constant polynomials are included when n=0: x^0=1.Wait, but in the problem statement, it says "nonzero polynomials". So constants ±1 are allowed, and monomials ±x^n. Let me check if these are the only solutions.Suppose P(x) is a polynomial with integer coefficients such that for any coprime integers a and b, P(a) and P(b) are coprime. Then, we need to prove that P(x) must be of the form ±x^n or ±1.Alternatively, suppose that P(x) has more than one term. Then, we can try to find a and b coprime such that P(a) and P(b) share a common prime factor.Alternatively, maybe if P(x) is not of the form ±x^n or ±1, then there exists some prime p that divides P(a) for multiple values of a, which would allow us to find a and b coprime with p dividing both P(a) and P(b).This seems similar to the concept of "prime producing" polynomials or polynomials that are coprime over integers. Wait, but in general, non-constant polynomials can take composite values, but here the requirement is stronger: whenever inputs are coprime, outputs are coprime.Suppose that P(x) is not of the form ±x^n or ±1. Then, P(x) must have either a non-trivial constant term or multiple terms. Let's analyze both cases.Case 1: P(x) has a non-trivial constant term. For example, P(x) = x + c where c ≠0. Then, as we saw earlier, taking a = k and b = k +1, which are coprime, might lead to P(a) and P(b) differing by 1, but that's not necessarily always the case. Wait, but earlier example showed that even with a constant term, sometimes they can fail. For example, P(x) = x +1 fails when a=5 and b=7. So perhaps any non-monomial polynomial (i.e., with more than one term) would have such failures.Alternatively, consider that if P(x) is not a monomial, then there exists some integer k where P(k) = 0. Wait, no. For example, P(x) = x +1 is not a monomial but doesn't have integer roots except x=-1. Hmm, maybe not the right approach.Another angle: Suppose P(x) is not a monomial. Then, it has at least two terms. Let's write P(x) = a_n x^n + ... + a_m x^m where n > m ≥0 and a_n, a_m ≠0. If there are at least two terms, then perhaps we can find some prime p that divides P(a) for multiple a's, which would allow us to choose coprime a and b such that p divides both P(a) and P(b).This is similar to the idea that a non-constant polynomial can have infinitely many primes dividing its values, but if the polynomial is not of the form ±x^n, then there exists a prime p that divides P(a) for some a, and by shifting a by multiples of p, we can get another b ≡ a mod p such that p divides P(b). However, if a and b are chosen such that they are congruent modulo p but are coprime, then their images under P would share the common factor p.Wait, maybe using the Chinese Remainder Theorem? Let me think.Suppose that P(x) is not a monomial. Then, there exists a prime p that divides P(0) if the constant term is non-zero. For example, if P(0) = c ≠0, ±1, then there's a prime p dividing c. Then, take a=0 and b=p. But wait, a=0 and b=p are not coprime unless p=1, which is not prime. So maybe that's not helpful.Alternatively, suppose that P(x) has a term with x^k where k ≥1 and a constant term. Let’s assume P(x) = x^k + c. Then, for a=1 and b=1 + c, if we can choose b such that gcd(1, b)=1, which would require b=1 + c to be coprime to 1, which is always true. Then, P(1) = 1 + c and P(b) = (1 + c)^k + c. Then, gcd(1 + c, (1 + c)^k + c). Let's compute this. Let d = gcd(1 + c, (1 + c)^k + c). Then d divides (1 + c) and (1 + c)^k + c. Therefore, d divides c. But d also divides 1 + c. Therefore, d divides 1 + c and c, so d divides 1. Therefore, gcd is 1. Hmm, so in this case, maybe P(x)=x^k + c could satisfy the condition? But earlier examples contradict this. For example, P(x) = x + 1 failed when a=5 and b=7. Wait, but according to this reasoning, gcd(P(a), P(b)) =1. But in reality, with a=5 and b=7, P(a)=6 and P(b)=8, which have gcd 2. So my reasoning here must be flawed.Wait, let's check with P(x)=x +1. Then, according to the above, if we take a=1 and b=1 +1=2. Then, P(1)=2 and P(2)=3. gcd(2,3)=1. But when a=5 and b=7, which are coprime, P(a)=6 and P(b)=8 have gcd 2. So even though for some a and b, the gcd is 1, there exist other a and b where gcd is not 1. Therefore, my previous approach was incomplete.So, maybe the key is that if a polynomial is not of the form ±x^n, then there exists some prime p that divides P(a) for some a, and by varying a, we can find another a’ such that p divides P(a’) and a’ is coprime to a. Then, gcd(a, a’)=1 but gcd(P(a), P(a’)) ≥ p >1, which would violate the condition.Therefore, to formalize this: suppose that P(x) is not of the form ±x^n. Then, there exists a prime p that divides P(a) for some integer a. If we can find an integer a’ such that p divides P(a’) and gcd(a, a’)=1, then we have a contradiction.Therefore, the problem reduces to showing that for any polynomial not of the form ±x^n, there exists a prime p and integers a, a’ with gcd(a, a’)=1 such that p divides both P(a) and P(a’).Therefore, to prove that only ±x^n and ±1 work, we need to show that for any other polynomial, such p, a, a’ exist.Let me try to construct such a prime p and integers a, a’.Assume P(x) is not of the form ±x^n or ±1. Then, either:1. P(x) has a non-zero constant term, or2. P(x) is a non-constant monomial but with coefficient other than ±1, but wait, non-constant monomials with coefficients other than ±1 would be like 2x^n, which we saw fail because, for example, a=1 and b=2, P(1)=2 and P(2)=2*2^n, so gcd(2, 2*2^n)=2≠1. Therefore, such polynomials are already excluded except for ±x^n.So, the only possibilities are polynomials of the form ±x^n or ±1, and polynomials with additional terms.So, suppose P(x) is not of the form ±x^n or ±1. Then, either it has a non-zero constant term or it has multiple terms with exponents.Case 1: P(x) has a non-zero constant term.Let’s write P(x) = Q(x) + c, where c ≠0 is the constant term, and Q(x) is a non-constant polynomial with integer coefficients.If we can choose a such that Q(a) ≡ -c mod p for some prime p, then p divides P(a). If we can also find a’ such that Q(a’) ≡ -c mod p and gcd(a, a’)=1, then p divides both P(a) and P(a’), and thus gcd(P(a), P(a’)) ≥ p >1, even though gcd(a, a’)=1. Therefore, such a polynomial would not satisfy the condition.To make this concrete, let's pick a prime p dividing c (if c ≠±1). Then, P(0) = Q(0) + c = c. So, if p divides c, then p divides P(0). Now, if we can find another integer a’ such that p divides P(a’) and gcd(0, a’)=1. But gcd(0, a’) = |a’|, so to have gcd(0, a’)=1, a’ must be ±1. Therefore, if p divides P(0) and p divides P(1) or P(-1), then we have a problem. For example, if p divides c and p divides P(1)=Q(1) + c, then p divides Q(1). Therefore, Q(1) must be ≡0 mod p. But Q(1) is an integer. So unless Q(1) ≡0 mod p, which depends on the coefficients.Alternatively, suppose c ≠ ±1, so there exists a prime p dividing c. Then, p divides P(0). Now, consider a=0 and a’=1. If p divides P(1), then gcd(P(0), P(1)) ≥p >1, but a=0 and a’=1 are coprime (gcd(0,1)=1). Therefore, if p divides both P(0) and P(1), then the condition is violated. Therefore, for the polynomial to satisfy the condition, it must be that no prime p divides both P(0) and P(1).But P(0)=c and P(1)=Q(1)+c. Therefore, gcd(c, Q(1) +c) = gcd(c, Q(1)). If Q(x) is a non-constant polynomial, Q(1) is some integer. So unless gcd(c, Q(1))=1, which is not necessarily the case. For example, take P(x) = x + c. Then Q(x) =x, so Q(1)=1. Then gcd(c, 1)=1. Therefore, in this case, gcd(c, Q(1))=1, so p divides c and p divides Q(1) implies p divides 1, which is impossible. Hence, for P(x)=x +c, gcd(P(0), P(1))=gcd(c, 1 +c)=gcd(c,1)=1. So even though p divides c, it does not divide P(1). Therefore, in this specific case, even with a constant term, if the polynomial is linear, then P(0)=c and P(1)=1 +c are coprime.But as we saw earlier, even though gcd(P(0), P(1))=1, there may exist other pairs a, b where gcd(P(a), P(b))≠1. For example, in the case P(x)=x +1, taking a=5 and b=7 gives P(a)=6 and P(b)=8, which have gcd 2. So even if P(0) and P(1) are coprime, there can be other pairs that fail.Therefore, this approach might not cover all cases. Maybe a different method is needed.Another approach: Suppose that P(x) is not of the form ±x^n. Then, there exists some integer k such that P(k) is not ±1 or 0, and has a prime factor p. Then, by the polynomial congruence theorem, P(k + np) ≡ P(k) ≡0 mod p for any integer n. So, P(k + np) is divisible by p. Now, choose n such that k + np is coprime to k. Since gcd(k, k + np) = gcd(k, np). If we choose n such that gcd(n, k)=1, then gcd(k, np)=gcd(k, p). If p divides k, then gcd(k, p)=p. Otherwise, gcd(k, p)=1. Wait, this is getting complicated.Alternatively, choose n=1. Then, gcd(k, k + p)=gcd(k, p). If p does not divide k, then gcd(k, p)=1. So, if we can ensure that p does not divide k, then k and k + p are coprime. Therefore, if we have a prime p dividing P(k), and if p does not divide k, then k and k + p are coprime, and p divides both P(k) and P(k + p). Therefore, gcd(P(k), P(k + p)) ≥p >1, violating the condition.Therefore, to avoid this, we must have that for every prime p dividing P(k), p divides k. So, if every prime divisor of P(k) divides k, then P(k) is a power of primes dividing k.But this needs to hold for all integers k. Wait, no. The condition is that whenever a and b are coprime, P(a) and P(b) are coprime. So, to violate it, we just need existence of some a and b coprime with P(a) and P(b) not coprime. So, if for some k, P(k) has a prime divisor p not dividing k, then taking a=k and b=k + p (which are coprime if p does not divide k), we have p divides P(k) and p divides P(k + p) (by the earlier congruence), so gcd(P(a), P(b)) ≥p >1. Thus, violating the condition.Therefore, to satisfy the condition, it must be that for every integer k, every prime divisor of P(k) divides k. That is, P(k) is a product of primes dividing k. In other words, P(k) divides some power of k for every integer k.But how can a polynomial P(x) with integer coefficients divide a power of x for all integer x? That is, for each integer x, P(x) divides x^m for some m. This seems restrictive.For example, the polynomials ±x^n satisfy this because P(x)=x^n divides x^n. Similarly, P(x)=1 divides any x^0=1. But other polynomials would have to divide some power of x for all x, which seems unlikely unless they are monomials.Indeed, suppose P(x) is a non-monomial polynomial. Then, for example, take P(x)=x +1. Then, P(2)=3, which does not divide any power of 2. Similarly, P(3)=4, which does not divide any power of 3. Therefore, this condition fails. Therefore, only monomials ±x^n and ±1 would satisfy that P(x) divides x^m for some m for all x.Wait, but even for P(x)=x^2, P(2)=4 divides 2^2=4. P(3)=9 divides 3^2=9. So yes. For P(x)=x^n, P(k)=k^n divides k^n. For P(x)=1, P(k)=1 divides k^0=1. For P(x)=-1, similarly. For P(x)=-x^n, P(k)=-k^n divides k^n. Since divisibility ignores signs, this works.But any other polynomial would fail. For example, P(x)=x+1 evaluated at x=2 is 3, which does not divide any power of 2. Therefore, such a polynomial cannot satisfy the condition that every prime divisor of P(k) divides k. Hence, only monomials of the form ±x^n and constants ±1 work.Therefore, the only polynomials that satisfy the condition are P(x)=±x^n for some non-negative integer n, or P(x)=±1. However, the problem states "nonzero polynomials", so constants ±1 are allowed. The problem also specifies "integer coefficients", which these polynomials satisfy.Therefore, the solution should be all polynomials of the form ±x^n where n is a non-negative integer, including n=0 (which gives ±1). To confirm, let's check if these are the only ones.Suppose P(x) is such a polynomial. If it's constant ±1, then obviously for any a, b, gcd(P(a), P(b))=gcd(±1, ±1)=1. If it's ±x^n, then for coprime a, b, gcd(a^n, b^n)=1, since a and b are coprime. Hence, these polynomials satisfy the condition.Conversely, suppose P(x) is not of this form. Then, as shown earlier, there exist coprime integers a, b such that gcd(P(a), P(b)) >1. Hence, such polynomials cannot exist.Therefore, the only solutions are P(x)=±x^n for some non-negative integer n, including n=0 (i.e., the constants ±1). The problem says "nonzero polynomials", so these are all valid.Hence, the answer should be all constant polynomials ±1 and all monomials ±x^n for n ≥1.Final AnswerThe nonzero polynomials are exactly the constants (pm 1) and the monomials (pm x^n) for non-negative integers (n). Thus, the solutions are (boxed{pm x^n}) where (n) is a non-negative integer, including the constant polynomials (boxed{pm 1}).However, following the standard convention of presenting answers in boxes and considering that the constant polynomials can be represented as ( pm x^0 ), we can succinctly present all solutions as:boxed{pm x^n} for some non-negative integer ( n ).

✅:To find all nonzero polynomials ( P(x) ) with integer coefficients such that whenever ( a ) and ( b ) are relatively prime integers, ( P(a) ) and ( P(b) ) are also relatively prime, we proceed as follows:1. Constant Polynomials: If ( P(x) = c ) where ( c ) is a non-zero integer, then for ( P(a) ) and ( P(b) ) to be coprime for all coprime ( a ) and ( b ), ( c ) must be ( pm 1 ). Thus, the constant polynomials ( pm 1 ) are solutions.2. Linear Polynomials: For linear polynomials ( P(x) = pm x + c ), we found that only ( P(x) = pm x ) work. Any constant term ( c neq 0 ) leads to cases where ( P(a) ) and ( P(b) ) are not coprime for some coprime ( a ) and ( b ).3. Higher-Degree Polynomials: For polynomials of the form ( P(x) = pm x^n ), we verified that if ( a ) and ( b ) are coprime, then ( a^n ) and ( b^n ) are also coprime. This is because any prime dividing both ( a^n ) and ( b^n ) must divide ( a ) and ( b ), contradicting their coprimality.4. Non-Monomial Polynomials: Polynomials with more than one term or non-zero constant terms were shown to fail. For example, ( P(x) = x + 1 ) fails because there exist coprime integers ( a ) and ( b ) such that ( P(a) ) and ( P(b) ) share a common prime factor.5. General Argument: For any polynomial not of the form ( pm x^n ) or ( pm 1 ), there exists a prime ( p ) and integers ( a ) and ( a' ) such that ( p ) divides both ( P(a) ) and ( P(a') ) while ( a ) and ( a' ) are coprime. This contradicts the requirement that ( P(a) ) and ( P(a') ) be coprime.Thus, the only polynomials that satisfy the condition are the monomials ( pm x^n ) for non-negative integers ( n ) (including ( n = 0 ), which gives the constants ( pm 1 )). Final AnswerThe nonzero polynomials ( P(x) ) with integer coefficients that satisfy the given property are precisely the monomials of the form (boxed{pm x^n}) where ( n ) is a non-negative integer. This includes the constant polynomials (boxed{pm 1}) when ( n = 0 ).

🔑:To find all nonzero polynomials ( P(x) ) with integer coefficients that satisfy the property that whenever ( a ) and ( b ) are relatively prime integers, ( P(a) ) and ( P(b) ) are also relatively prime, follow these steps:1. Finding Possible Polynomials: First consider the polynomials of the form ( P(x) = pm x^n ), where ( n geq 0 ) is an integer. - When ( P(x) = x^n ), for any two relatively prime integers ( a ) and ( b ), the values ( P(a) = a^n ) and ( P(b) = b^n ) are relatively prime. - This is because the only prime factors of ( a^n ) are those of ( a ) and the only prime factors of ( b^n ) are those of ( b ). Since ( a ) and ( b ) are relatively prime, they share no common prime factors, so neither do ( a^n ) and ( b^n ). - Similarly, for ( P(x) = -x^n ), the same logic follows since if ( a ) and ( b ) are relatively prime, so are ( -a ) and ( -b ).2. Demonstrating that Other Polynomials Fail: Now consider any polynomial ( P(x) ) not of the form ( pm x^n ). Write it as follows: [ P(x) = c_n x^n + c_{n-1} x^{n-1} + cdots + c_0, ] where ( c_n ) is the leading coefficient and ( c_n neq 0 ). - Assume without loss of generality that ( c_n > 0 ) (we can replace ( P(x) ) with ( -P(x) ) if necessary). - Suppose ( c_n = 1 ) and the next nonzero coefficient (say ( c_k )) is negative. This implies that for sufficiently large ( x ), the polynomial ( P(x) ) lies between two consecutive powers of ( x ): [ x^n < P(x) < x^{n+1}. ] - Consequently, if ( P ) has a form other than ( x^n ), for sufficiently large prime ( q ), ( P(q) ) will be a positive integer between ( q^n ) and ( q^{n+1} ). Therefore, ( P(q) ) cannot be a power of ( q ).3. Prime Factor Argument: If ( P(q) ) has some prime factor ( r neq q ), then for relatively prime integers ( q ) and ( q + r ): [ r = (q + r) - q, ] thus ( r mid P(q + r) - P(q) ). Both ( P(q) ) and ( P(q + r) ) are divisible by ( r ), meaning they cannot both be relatively prime.4. Final Argument: Given that ( P ) cannot be of any other form besides ( pm x^n ), it confirms that ( P(x) = pm x^n ) are the only possible polynomials satisfying the condition.Conclusion: The only nonzero polynomials ( P(x) ) with integer coefficients that make ( P(a) ) and ( P(b) ) relatively prime whenever ( a ) and ( b ) are relatively prime are:[boxed{P(x) = pm x^n text{ for integer } n geq 0.}]

🧠:Okay, so I need to prove that there's a straight line g that splits the plane into two half-planes, each containing exactly half of the points from both M1 and M2. The sets M1 and M2 are disjoint, have 2n1 and 2n2 points respectively, and no three points are collinear. Hmm, let me think about how to approach this.First, I remember something about the ham-sandwich theorem, which in two dimensions says that for any two measurable "objects" (like areas or finite sets of points), there's a line that simultaneously bisects both. Since we have two finite point sets here, maybe this theorem applies? But the problem states that each half-plane must contain exactly half of each set. The ham-sandwich theorem does talk about bisecting each measure, so maybe that's the key. But I need to make sure that the line doesn't pass through any points since the problem says the line isn't included in either half-plane, and all points must be strictly on one side or the other.But wait, the ham-sandwich theorem usually applies to continuous measures, but here we have discrete points. However, since the points are in general position (no three collinear), maybe we can adapt the theorem. Let me check my understanding. For two point sets in the plane with no three collinear, there exists a line that bisects both sets, with at most half of each set on either side. But the problem here requires exactly half in each half-plane. Since both M1 and M2 have even numbers of points (2n1 and 2n2), exactly half would be n1 and n2 points in each half-plane. That makes sense.So maybe by using a continuity argument or rotating a line and counting the points on each side. Let's try to formalize this.Imagine starting with a vertical line far to the left of all points. All points are to the right of this line, so zero points on the left, all on the right. Then slowly move the line to the right. As we pass points, the count on the left increases. When the line has passed n1 points of M1 and n2 points of M2, we stop. But how do we ensure that such a position exists where both counts are exactly n1 and n2?Alternatively, maybe use a rotating line. Start at some angle and rotate it 180 degrees. For each angle θ, define a line at angle θ that bisects both sets. As the line rotates, the number of points on each side might change, but since the numbers are discrete, we need to find an angle where both counts are exactly n1 and n2. Maybe using the intermediate value theorem?Wait, here's an approach: For each direction (angle), there's a line that bisects M1. Similarly, for the same direction, a line that bisects M2. But we need a single line that bisects both. Maybe by considering the difference between these two lines and showing that there's a direction where they coincide?Alternatively, parameterize the problem. Let's consider lines in the plane. Each line can be represented by an equation ax + by + c = 0. The condition is that exactly n1 points of M1 and n2 points of M2 lie on each side. Since the points are in general position, moving the line continuously won't cause points to cross the line except one at a time. So maybe we can use some kind of sweeping argument.Alternatively, think about duality. In computational geometry, problems about lines and points can sometimes be transformed into dual problems. But I'm not sure if that's helpful here.Wait, another idea: Let's pair up the points in M1. Since there are 2n1 points, we can form n1 pairs. Similarly, pair up M2 into n2 pairs. Then, find a line that separates all the pairs, with one point of each pair on either side. But I'm not sure if that directly helps.Alternatively, use induction. Suppose for smaller numbers of points, the statement holds. But since the numbers are even, maybe splitting into smaller even numbers? Not sure.Wait, here's a more concrete idea. Let's parameterize lines in the plane by their angle θ and their offset d. For a given θ, we can sort the projections of all points in M1 and M2 onto the direction perpendicular to θ. The median of these projections would give a line that bisects the set. But since we have two sets, we need a line that is a median for both. Since both M1 and M2 have even numbers of points, the median would be between two points. So, maybe there's a θ where the median for M1 and the median for M2 coincide?This seems similar to the ham-sandwich cut. In two dimensions, the ham-sandwich theorem states that given two finite point sets in the plane, there exists a line that simultaneously bisects both, with each half-plane containing at most half of each set. But since our sets have even cardinality, we can actually have exactly half in each. The theorem should hold here because the points are in general position, so no line will pass through more than two points. But wait, the general position here only specifies no three collinear, so a line could pass through two points, but since the line is excluded from the half-planes, those two points would have to be on opposite sides? Wait, no—if a line passes through a point, the point isn't included in either half-plane. But the problem states that the line isn't included, so the points on the line are excluded. But our sets M1 and M2 are disjoint, so maybe the line can't pass through any points? Wait, the problem says "no three points of the union are collinear," so two points can be collinear. So maybe there can be a line passing through two points, but not three. However, if such a line passes through two points, then those two points are excluded from both half-planes. But the problem requires that each half-plane contains exactly half of each set. So if a line passes through two points of M1, then each half-plane must contain (2n1 - 2)/2 = n1 - 1 points. But the problem states that each half-plane should contain exactly half, which is n1 points. Therefore, the line cannot pass through any points. Therefore, the line must avoid all points. Since the points are in general position (no three collinear), but pairs can lie on a line. However, we need a line that doesn't contain any points of M1 ∪ M2.So in our case, the ham-sandwich line must not pass through any points. Since the sets are in general position (no three collinear), but there could be lines that pass through two points. However, since the total number of such lines is finite (only a finite number of pairs), and we can choose a line that doesn't pass through any points. So the ham-sandwich theorem should still apply here, giving us a line that bisects both sets without passing through any points. Therefore, each half-plane would contain exactly n1 and n2 points of M1 and M2 respectively.But maybe I need to elaborate more rigorously. Let me recall the proof technique for the ham-sandwich theorem in two dimensions. One method involves parameterizing the direction of the line, and using the intermediate value theorem.Here's an outline: For each angle θ, consider lines in direction θ. Sort the points of M1 and M2 by their projection onto a line perpendicular to θ. The median of each set would give a bisecting line. However, the medians for M1 and M2 might not coincide. But by rotating θ from 0 to π, we can find a direction where the bisectors for M1 and M2 coincide. This is because the difference between the two medians changes continuously and must cross zero.Wait, more precisely: For each direction θ, we can define a line that bisects M1. Let's call this line L1(θ). Similarly, define L2(θ) that bisects M2. If we can show that for some θ, L1(θ) = L2(θ), then that line would bisect both. To show this, we can parameterize θ from 0 to π, and consider the difference between the positions of L1(θ) and L2(θ). As θ varies, this difference will change continuously. If we can show that the difference changes sign as θ goes from 0 to π, then by the intermediate value theorem, there must be some θ where the difference is zero, meaning L1(θ) = L2(θ).Let me make this more precise. For a direction θ, we can represent the lines as points in a dual space. For each θ, the bisecting line for M1 is uniquely determined by the median of the projections. Similarly for M2. Since the projections vary continuously with θ, the positions of L1(θ) and L2(θ) also vary continuously. Now, consider the difference in the offsets of L1(θ) and L2(θ). If we rotate θ by 180 degrees, the lines L1(θ + π) and L2(θ + π) would be the same as L1(θ) and L2(θ) but shifted in the opposite direction. Hence, the difference in offsets would reverse sign. Therefore, there must be some θ where the difference is zero, hence L1(θ) = L2(θ). That line would bisect both M1 and M2.But since we have even numbers of points, the median is not a single point but between two points. Wait, but projections onto a line would be points on a line, so the median would be a point such that half the projections are on one side and half on the other. Since the number of points is even, the median is between two projections. However, by slightly perturbing the line, we can ensure that exactly half are on each side. Wait, but in our case, the problem allows the line to be anywhere, as long as it doesn't pass through any points. So as we rotate the direction θ, the bisecting line can be moved such that it doesn't pass through any points, due to the general position assumption. Therefore, there exists a θ where the bisecting line for M1 and M2 coincide, and this line doesn't pass through any points. Therefore, each half-plane would contain exactly half of each set.Alternatively, here's another way to think about it. Consider all possible oriented directions (angles) for lines. For each direction, we can find a line that bisects M1, i.e., has exactly n1 points on each side. Similarly, for each direction, there's a line that bisects M2. Now, if we can find a direction where the same line bisects both, we are done. To find such a direction, we can model the problem as a continuous function from the circle (directions) to the real line, representing the offset between the two bisecting lines. If we can show this function crosses zero, then we have our desired line.Alternatively, think of the bisector for M1 as a function f(θ) and the bisector for M2 as g(θ). As θ varies from 0 to π, the difference f(θ) - g(θ) must cross zero. Since when θ is rotated by 180 degrees, the bisectors invert, leading to a sign change in the difference. Therefore, by continuity, there must be a θ where the difference is zero.This seems plausible. To make this rigorous, maybe consider the following:For each angle θ, define a coordinate system where the x-axis is in direction θ. Sort all points in M1 ∪ M2 by their x-coordinate. Let’s define a line perpendicular to θ that separates the points into two equal halves for each set. Since both sets have even cardinality, there exists a line between the n1-th and (n1+1)-th point of M1 and similarly for M2. However, these two lines might not coincide. The key is to find a θ where these two separating lines are the same.Define a function that measures the difference between the positions of the two separating lines as θ varies. If this function changes sign as θ goes from 0 to π, then by the intermediate value theorem, there exists a θ where the difference is zero, meaning the lines coincide.To see why the function changes sign, consider rotating θ by 180 degrees. The order of the projections reverses, so the separating line for M1 that was at position x is now at position -x. Similarly for M2. Therefore, the difference between the two lines also reverses sign. Hence, by continuity, there must be some angle where the difference is zero.Therefore, such a line exists. Since the points are in general position (no three collinear), the separating line won't pass through any points, ensuring that each half-plane contains exactly half of each set.Alternatively, here's a different approach using graph theory. Consider the arrangement of all lines formed by pairs of points in M1 and M2. Since there are a finite number of such lines, the plane is divided into regions by these lines. Any line that doesn't cross any of these regions can be continuously moved without changing the number of points on each side. But as we rotate a line, the number of points on each side changes only when the line crosses a point. However, since no three points are collinear, a rotating line can only cross one point at a time. Therefore, as we rotate the line from angle 0 to π, the counts of points on each side for M1 and M2 change by one each time we cross a point. Starting with all points on one side, and ending with all on the other, we must pass through a state where exactly half are on each side. But since we have two sets, we need to synchronize this for both M1 and M2. However, since the parity is even for both, and we can adjust the rotation to hit the exact halfway points for both. This seems similar to the proof of the ham-sandwich theorem via rotating a line.Wait, maybe another way: Consider the set of all directed lines. For each directed line, we can define a pair (a, b) where a is the number of points of M1 on the left side and b the number of points of M2 on the left side. As we rotate the line continuously, a and b change by at most 1 when crossing a point. Starting from a line where all points are on the right (a=0, b=0) and rotating 180 degrees to where all are on the left (a=2n1, b=2n2). We need to find a line where a = n1 and b = n2. Imagine a grid where one axis is a and the other is b. Starting at (0,0) and ending at (2n1, 2n2), each step moves either right or up. By continuity, the path must cross the point (n1, n2). Hence, such a line exists.This is similar to a combinatorial argument in the proof of the ham-sandwich theorem. Since the numbers of points are even, and as we rotate the line, the counts change by 1 each time we cross a point. Since we need to reach exactly half, and the parity allows it, there must be a position where both counts are exactly half.Therefore, combining these ideas, the existence of such a line g is guaranteed by the ham-sandwich theorem or a continuity argument with the intermediate value theorem, leveraging the evenness of the number of points in each set and the general position assumption ensuring no three points are collinear, which avoids complications with overlapping counts.So, putting it all together, the key steps are:1. Recognize the problem as a discrete version of the ham-sandwich theorem for two sets with even cardinality.2. Use a continuity argument by rotating a line and observing that the counts of points on each side must cross the desired halfway point.3. Apply the intermediate value theorem to ensure that there exists a position where both counts are exactly half.4. Use the general position assumption to ensure the line doesn't pass through any points, thus strictly separating them into the required halves.Therefore, such a line g must exist.

✅:To prove the existence of a straight line ( g ) that bisects both sets ( M_1 ) and ( M_2 ) into equal halves in each half-plane, we employ a continuity argument inspired by the ham-sandwich theorem. Here's the step-by-step reasoning:1. General Position and Even Cardinality: The sets ( M_1 ) and ( M_2 ) have ( 2n_1 ) and ( 2n_2 ) points, respectively, with no three points collinear. This ensures that any line can pass through at most two points, but we seek a line avoiding all points.2. Parameterizing Directions: For any angle ( theta ), define a coordinate system with the ( x )-axis aligned along ( theta ). Project all points orthogonally onto the ( x )-axis and sort them. A line perpendicular to ( theta ) can bisect ( M_1 ) and ( M_2 ) by separating the first ( n_1 ) (resp. ( n_2 )) points from the rest.3. Continuous Rotation Argument: Rotate ( theta ) from ( 0 ) to ( pi ). For each ( theta ), let ( L_1(theta) ) and ( L_2(theta) ) denote the bisecting lines for ( M_1 ) and ( M_2 ), respectively. These lines vary continuously with ( theta ).4. Intermediate Value Theorem: Define the vertical offset difference between ( L_1(theta) ) and ( L_2(theta) ). As ( theta ) rotates by ( pi ), the roles of left and right reverse, causing the offset difference to change sign. By continuity, there exists some ( theta ) where ( L_1(theta) = L_2(theta) ), forming a common bisector ( g ).5. Avoiding Points: Since no three points are collinear, ( g ) cannot pass through any points of ( M_1 cup M_2 ). Thus, each half-plane contains exactly ( n_1 ) points of ( M_1 ) and ( n_2 ) points of ( M_2 ).Conclusion: By the intermediate value theorem and the general position assumption, there exists a line ( g ) that bisects both ( M_1 ) and ( M_2 ) as required. This line ensures each half-plane contains exactly half of each set. (boxed{text{Such a line } g text{ exists.}})

🔑:1. Define Half-Line and Rotate: Choose a half-line ( s ) starting from a point ( O ) in the plane ( E ). For every angle ( alpha ) in the interval ([0, 180^circ]), denote by ( s(alpha) ) the line obtained by rotating ( s ) counterclockwise about ( O ) by ( alpha ). Let ( g(alpha) ) be the oriented line containing ( s(alpha) ), on which ( s(alpha) ) defines the positive direction.2. Projection and Distance Notation: For each point ( P ) in the set ( M_i ) (where ( i ) is either 1 or 2), let ( P(alpha) ) be the foot of the perpendicular from ( P ) to ( g(alpha) ), and denote by ( l_P(alpha) ) the oriented distance from ( O ) to ( P(alpha) ) along ( g(alpha) ).3. Order Distances: Consider the set ( {l_P(alpha) mid P in M_i} ). We can arrange the distances ( l_P(alpha) ) in ascending order: [ l_1(alpha), l_2(alpha), ldots, l_{2n_i}(alpha) ] for ( i = 1, 2 ).4. Define Interval for Median Cut: Define the interval ( J_i(alpha) = [l_{n_i}(alpha), l_{n_i+1}(alpha)] ). This interval represents the ( n_i )-th and ((n_i+1))-th ordered distances along ( g(alpha) ).5. Finding a Common Point: To prove the existence of a line ( g ) with the desired property, it remains to show that there is some ( alpha ) such that the interiors of the intervals ( J_1(alpha) ) and ( J_2(alpha) ) have a common point. 6. Initial Overlap Check: If such a common point exists for ( alpha = 0 ), then we are done. Suppose without loss of generality that ( J_1(0) ) lies entirely to the left of ( J_2(0) ) on the line ( g(0) ). 7. Position Shift Analysis: By the time ( alpha ) grows to ( 180^circ ), the intervals ( J_1 ) and ( J_2 ) will have shifted such that ( J_1(180^circ) ) now lies entirely to the right of ( J_2(180^circ) ). 8. Continuity and Intersection: Note that ( J_1 ) and ( J_2 ) cannot simultaneously degenerate to a point (otherwise, there would be four collinear points in ( M_1 cup M_2 ), which contradicts the problem's stipulations). Each interval can only degenerate to a point for finitely many values of ( alpha ). Since ( J_1(alpha) ) and ( J_2(alpha) ) move continuously as ( alpha ) varies from ( 0 ) to ( 180^circ ), there must exist a subinterval ( I ) of ( [0, 180^circ] ) on which these intervals are not disjoint.9. Conclusion: Therefore, at some point ( alpha ) within this subinterval ( I ), the intervals ( J_1(alpha) ) and ( J_2(alpha) ) will both be non-degenerate and will have a common interior point. Hence, the halves of the plane defined by ( g(alpha) ) will each contain exactly half of the points of ( M_1 ) and ( M_2 ). [boxed{text{A line } g text{ exists such that each half-plane contains exactly half of the points from } M_1 text{ and } M_2.}]