🧠:Alright, so I need to prove that if we have three real polynomials P, Q, and R such that P² + Q² = R², and at least one of them is quadratic and one is cubic, then at least one of the cubic polynomials (I guess there's at least one cubic, so that one) has all real roots. Hmm, okay. Let me try to break this down step by step.First, let's make sure I understand the problem correctly. We have three polynomials: P, Q, R. All are real polynomials. The equation they satisfy is P² + Q² = R². So, similar to the Pythagorean theorem but with polynomials. The degrees: at least one of them is quadratic (degree 2) and one is cubic (degree 3). The conclusion is that at least one of the cubic polynomials (so if there is a cubic among P, Q, R, then that cubic must have all real roots). Wait, but the problem says "at least one of the cubic polynomials has all real roots." But since the problem states that there is at least one cubic polynomial among P, Q, R, then the conclusion is that this cubic polynomial must have all real roots.Wait, but could there be more than one cubic polynomial here? Let me check. If P is quadratic, Q is cubic, and R is cubic, then both Q and R are cubic. So in such a case, the conclusion would be that at least one of Q or R has all real roots. But maybe it's possible that only one of them is cubic. Let me clarify.The problem states that "at least one of them being a quadratic polynomial and one being a cubic polynomial." So, among P, Q, R, there is at least one quadratic (degree 2) and at least one cubic (degree 3). The other one can be of any degree, but presumably, since we have an equation P² + Q² = R², the degrees must align. Let's think about the degrees.Suppose P is quadratic, Q is cubic. Then P² is degree 4, Q² is degree 6. Adding them together, the highest degree term would be from Q², which is degree 6. Therefore, R² must be degree 6, so R must be degree 3. So in this case, R is cubic, so Q and R are both cubic. Wait, but Q is cubic, R is cubic. So the two cubics here are Q and R. Then according to the problem, at least one of them must have all real roots. Alternatively, if the third polynomial is of lower degree, say linear or constant, but the problem says at least one is quadratic and one is cubic. So the third one can be of any degree. Wait, but in the equation P² + Q² = R², the degrees have to satisfy that the maximum degree among P² and Q² is equal to the degree of R². So if P is quadratic (degree 2), then P² is degree 4; if Q is cubic (degree 3), then Q² is degree 6. Then R² is degree 6, so R is degree 3. So R is cubic. Therefore, in this case, Q and R are both cubic. So the conclusion is that at least one of Q or R must have all real roots. Alternatively, if another configuration is possible.Alternatively, maybe P is cubic, Q is quadratic, and R is cubic. Then similar reasoning: P² is degree 6, Q² is degree 4, so R² must be degree 6, so R is cubic. Then again, both P and R are cubic. So in any case, when one of P or Q is quadratic and the other is cubic, then R must be cubic. So the cubic polynomials are two of them, and the third is quadratic. So the conclusion is that among the two cubics, at least one has all real roots.Alternatively, maybe the quadratic is R. But then R² is degree 4. Then P and Q would have degrees such that P² and Q² add up to a degree 4 polynomial. So if R is quadratic (degree 2), then R² is degree 4. Then P² and Q² must each be of degree at most 4, but their squares add up to degree 4. Therefore, the maximum degree between P and Q is 2. But the problem states that at least one of them is cubic, so this scenario is impossible. Therefore, R cannot be quadratic. Therefore, the quadratic must be either P or Q, and the cubic must be either P, Q, or R, but given the equation, R must be cubic if one of P or Q is cubic. Wait, let's formalize this.Let me think about the degrees. Let’s denote deg(P) = p, deg(Q) = q, deg(R) = r. Then, since P² + Q² = R², the degree of R² must be equal to the maximum of deg(P²), deg(Q²). So, 2r = max{2p, 2q}, so r = max{p, q}.The problem states that at least one is quadratic (so p, q, or r is 2) and at least one is cubic (so p, q, or r is 3). However, since r = max{p, q}, then if either p or q is 3, then r must be 3. If the cubic is r, then r = 3, so max{p, q} = 3, so either p or q is 3. Alternatively, if the cubic is among P or Q, then r must be 3 as well. So in any case, there must be at least two polynomials of degree 3. Wait, no. Let's check different possibilities.Case 1: Suppose the quadratic is P (so p = 2). Then, if Q is cubic (q = 3), then R must be cubic (r = 3). So here, Q and R are cubic. If instead, Q is quadratic and P is cubic, then similarly R is cubic. If the quadratic is one of them and the cubic is the other, then R is cubic. Alternatively, suppose the quadratic is R, but then R is quadratic, so r = 2. Then, since R² is degree 4, P and Q must be such that their squares add up to a degree 4 polynomial. Therefore, max{2p, 2q} = 4, so max{p, q} = 2. Therefore, both P and Q must be at most quadratic. However, the problem states that at least one is cubic, which would be impossible if R is quadratic. Therefore, R cannot be quadratic. Hence, the quadratic must be either P or Q, and the cubic must be either P, Q, or R. But given the equation, if either P or Q is cubic, then R must be cubic. If the cubic is R, then since R = max{p, q}, then either p or q must be 3. Wait, but the problem only requires that at least one is cubic and one is quadratic. So if R is cubic, then either P or Q is cubic, but that's not necessary. Wait, no. If R is cubic, that just means that the maximum of p and q is 3, so at least one of P or Q is cubic. Therefore, in this case, the cubic polynomial is either P or Q, and R is also cubic. But the problem says "at least one of them being a quadratic polynomial and one being a cubic polynomial." So if R is cubic, then we need one quadratic among P or Q. If R is quadratic, that's impossible as shown earlier. So possible cases:1. P is quadratic (2), Q is cubic (3), so R is cubic (3).2. Q is quadratic (2), P is cubic (3), so R is cubic (3).3. Suppose R is cubic (3), so max{p, q} = 3, and one of P or Q is quadratic (2). Then the other (P or Q) can be of degree ≤ 3, but since we need at least one quadratic and one cubic, then in this case, one is quadratic (2) and the other is cubic (3). So this is similar to cases 1 and 2.Alternatively, maybe one of them is quadratic, another is cubic, and the third is something else. But according to the equation, R is determined by P and Q. So if P is quadratic and Q is cubic, R is cubic. If P is quadratic and Q is linear, then R would be quadratic. But the problem requires at least one cubic and one quadratic. So if in that case, we have P quadratic, Q linear, R quadratic, but then there is no cubic, which violates the problem's condition. Therefore, the only valid cases are those where one of P or Q is quadratic, the other is cubic, and R is cubic.Therefore, in all valid cases, we have two cubic polynomials (either P and R, Q and R, or P and Q if R is quadratic, but R can't be quadratic). Wait, no. Wait, in the valid cases where R is cubic, then either P or Q is cubic, and the other is quadratic. Wait, no. Let me re-examine.If P is quadratic (2), Q is cubic (3), then R is cubic (3). So Q and R are cubic. If Q is quadratic (2), P is cubic (3), then R is cubic (3). So P and R are cubic. If, hypothetically, R were cubic, and say P is cubic and Q is quadratic, then yes. So in all valid scenarios, there are two cubic polynomials: one of P or Q and R. Therefore, the two cubics are, for example, Q and R in the first case, or P and R in the second case. The third polynomial is quadratic.Therefore, the problem states that at least one of the cubic polynomials (so either of the two cubics) has all real roots. So we need to show that in such a setup, either Q and R in the first case, or P and R in the second case, at least one of them has all real roots.So perhaps I need to show that if you have two cubic polynomials, say Q and R, such that Q² = R² - P², where P is quadratic, then either Q or R must have all real roots.Alternatively, perhaps considering the identity P² + Q² = R², we can think of this as similar to Pythagorean triples but for polynomials, and analyze the roots.Let me recall that for real polynomials, if a polynomial can be written as a sum of squares of two other real polynomials, then it must be non-negative for all real x. Wait, but here R² = P² + Q², so R² is a sum of squares, hence non-negative. Therefore, R² is always non-negative, which is obvious since it's a square. But also, P² + Q² is a sum of squares, which is non-negative, so R² is non-negative. So that doesn't give us much.But perhaps considering the roots. Let me think. If a polynomial S(x) is a sum of squares of two polynomials, then S(x) is non-negative for all real x. Therefore, if R² = P² + Q², then R² is non-negative, which is true. But also, if we factor R² as (R(x))^2, then all roots of R(x) must have even multiplicity, because squares. Similarly, P² and Q² have roots with even multiplicities.But how does this help? Well, the problem is about the roots of the cubic polynomials. A cubic polynomial must have at least one real root, and can have up to three. So if a cubic polynomial has all real roots, then it can be factored as (x - a)(x - b)(x - c) where a, b, c are real. If it has one real root and two complex-conjugate roots, it factors as (x - a)(x² + bx + c) where the quadratic has no real roots.Therefore, the problem requires that at least one of the cubic polynomials in the equation must have all real roots, i.e., it cannot have a pair of complex-conjugate roots.Suppose, for contradiction, that both cubic polynomials have a pair of complex-conjugate roots. Then each cubic would have one real root and two complex roots. Let's see if this is possible under the given equation.Let me denote the cubics as Q and R (assuming P is quadratic). So Q(x) = (x - α)(x² + bx + c) where the quadratic has no real roots, and R(x) = (x - β)(x² + dx + e) similarly. Then, Q²(x) + P²(x) = R²(x). Let's see what this would imply.But perhaps another approach. Let me consider the degrees. Suppose P is quadratic, Q and R are cubic. Then, the leading terms: Let P(x) = a x² + lower degree terms, Q(x) = b x³ + ... , R(x) = c x³ + ... Then P² + Q² = R² implies that the leading term of Q² must equal the leading term of R². Because Q is cubic, Q² is degree 6, P² is degree 4, so the leading term of R² is the same as that of Q², so (b x³)^2 = (c x³)^2. Therefore, b² = c². Therefore, c = ±b. Similarly, if Q and R are cubics, their leading coefficients squared are equal. So R(x) is either Q(x) plus lower degree terms or -Q(x) plus lower degree terms. But since R² = P² + Q², R must be either Q plus something squared, but not sure.Alternatively, if we think of P, Q, R as real polynomials satisfying P² + Q² = R², then over the complex numbers, this factors as (R - iP)(R + iP) = Q². But not sure if complex factorization helps here.Alternatively, maybe using differentiation. If R² = P² + Q², then differentiating both sides, we get 2 R R' = 2 P P' + 2 Q Q', so R R' = P P' + Q Q'. But how does this help with the roots?Alternatively, think about the roots of R. Since R² = P² + Q², for any real root of R, say x₀, we have P(x₀)^2 + Q(x₀)^2 = 0, which implies P(x₀) = Q(x₀) = 0. So every real root of R is a common root of P and Q. Similarly, if x₀ is a root of R, then it's a root of R², so P(x₀)^2 + Q(x₀)^2 = 0, so P and Q both have x₀ as a root. Therefore, all real roots of R are common roots of P and Q. But since R is cubic, it has either one or three real roots (counting multiplicities). Wait, but multiplicities. If R has a real root with multiplicity, say two, then R² would have that root with multiplicity four. But in our case, R is a cubic, so R² is a degree 6 polynomial. But since R is cubic, R² is degree 6. But in our problem, R is cubic and another polynomial is quadratic, so P² + Q² is degree 6 (since Q is cubic, Q² is degree 6, P is quadratic, P² is degree 4). So R² is degree 6, which is consistent with R being cubic.But the key point is that any real root of R must be a common root of P and Q. Therefore, the real roots of R are common roots of P and Q. Now, suppose R has three real roots. Then each of these roots is a common root of P and Q. However, P is quadratic, so it can have at most two distinct roots (unless it's a double root). Therefore, if R has three distinct real roots, then P and Q would have to share all three roots, which is impossible since P is quadratic. Therefore, R cannot have three distinct real roots. Wait, but multiplicities? Let's see.Suppose R has a multiple root. For example, R could have a double root and a simple root. Then, the double root would be a root of R, hence a common root of P and Q. Similarly, the simple root would be a common root of P and Q. But P is quadratic, so it can have at most two roots (counting multiplicity). Therefore, if R has three real roots (counting multiplicities), then P would have to have three roots, which is impossible because P is quadratic.Therefore, R cannot have three real roots (counting multiplicities). Therefore, R must have exactly one real root (with multiplicity 1) and a pair of complex-conjugate roots. Wait, but cubic polynomials must have at least one real root. So if R is cubic, it has either one real root or three real roots. But according to the above, if R had three real roots, even with multiplicities, then P would have three roots, which is impossible. Therefore, R must have exactly one real root (of multiplicity 1) and two complex roots. Therefore, R cannot have all real roots. Therefore, if R is cubic, it cannot have all real roots. Wait, but the problem states that at least one of the cubic polynomials must have all real roots. So if R is cubic and cannot have all real roots, then the other cubic (either P or Q) must have all real roots. Wait, but in the case where P is quadratic and Q is cubic, then R is cubic. So in this case, Q and R are cubic. If R cannot have all real roots, then Q must have all real roots. But how?Wait, maybe there is a mistake in my previous reasoning. Let's re-examine.Suppose R has a real root at x₀. Then P(x₀) = Q(x₀) = 0. So x₀ is a common root of P and Q. If R is cubic, it must have at least one real root. Let's say it has exactly one real root x₀ (with multiplicity 1) and two complex roots. Then, x₀ is a common root of P and Q. Since P is quadratic, it can have at most two roots. If P has x₀ as a root, and since P is quadratic, it could have another root x₁. Similarly, Q is cubic, so it can have three roots. If Q has x₀ as a root, then it could have two more roots, which could be real or complex.But if Q has x₀ as a root, and is cubic, then Q(x) = (x - x₀) * S(x), where S(x) is a quadratic polynomial. If S(x) has real roots, then Q has three real roots; if S(x) has complex roots, then Q has one real root and two complex roots. Similarly, R(x) = (x - x₀) * T(x), where T(x) is a quadratic polynomial with two complex roots (since R has only one real root x₀). Therefore, T(x) is irreducible over the reals.So, given that R(x) = (x - x₀) * T(x), and Q(x) = (x - x₀) * S(x), and P(x) has x₀ as a root (since P(x₀) = 0), so P(x) = (x - x₀) * L(x), where L(x) is linear (since P is quadratic). Therefore, P(x) = (x - x₀)(a x + b).Therefore, substituting into the equation P² + Q² = R²:[(x - x₀)(a x + b)]² + [(x - x₀) S(x)]² = [(x - x₀) T(x)]²Factor out (x - x₀)^2:(x - x₀)^2 [ (a x + b)^2 + S(x)^2 ] = (x - x₀)^2 T(x)^2Cancel (x - x₀)^2 (since it's non-zero except at x = x₀):(a x + b)^2 + S(x)^2 = T(x)^2Now, this is a new equation involving the polynomials (a x + b), S(x), and T(x), where S(x) is quadratic (since Q is cubic and we factored out (x - x₀)), and T(x) is quadratic (since R is cubic and we factored out (x - x₀)). So we have:Linear^2 + Quadratic^2 = Quadratic^2Wait, but (a x + b) is linear, S(x) is quadratic, T(x) is quadratic. So we have (linear)^2 + (quadratic)^2 = (quadratic)^2. Therefore, rearranged, we get (linear)^2 = (quadratic)^2 - (quadratic)^2. Hmm, but (quadratic)^2 - (quadratic)^2 is zero? Wait, no. Wait, the original equation is (linear)^2 + (quadratic)^2 = (quadratic)^2. Therefore, subtracting, we get (linear)^2 = 0, which would imply that the linear polynomial is identically zero. But that's only possible if a = 0 and b = 0, which would make P(x) = (x - x₀)*0 = 0, but P is supposed to be quadratic. Therefore, this is a contradiction.Wait, this suggests that my assumption that R has only one real root leads to a contradiction. Therefore, R cannot have only one real root, so it must have three real roots. But earlier, I thought that R cannot have three real roots because P is quadratic and would have to share all three roots, which is impossible. Therefore, there must be a mistake in my reasoning.Wait, let's go back. If R has three real roots, each of multiplicity 1, then all three roots are common roots of P and Q. But P is quadratic, so it can have at most two distinct roots. Therefore, R cannot have three distinct real roots. If R has a multiple root, say a double root at x₀ and a simple root at x₁. Then x₀ is a common root of P and Q with multiplicity at least 2 for R, but since P is quadratic, if x₀ is a root of P, its multiplicity can be at most 2. But Q is cubic, so x₀ can have multiplicity up to 3. However, in the equation R² = P² + Q², the multiplicity of x₀ in R² is twice the multiplicity in R. Similarly, in P² and Q², it's twice the multiplicity in P and Q, respectively. But since x₀ is a root of both P and Q, let's denote the multiplicity of x₀ in P as m and in Q as n. Then, in P², it's 2m, in Q², it's 2n, and in R², it's 2*(multiplicity in R). Since R² = P² + Q², the multiplicity of x₀ in R² must be equal to the minimum of 2m and 2n, assuming 2m ≠ 2n. Wait, but if 2m = 2n, then the multiplicity in R² is 2m + 1? No, actually, when adding two functions with a root of the same multiplicity, the resulting function has a root of at least that multiplicity, but could have higher if there's cancellation.Wait, perhaps this approach is too complicated. Let's think differently.Suppose that R has a real root x₀. Then, as before, P(x₀) = Q(x₀) = 0. Therefore, x₀ is a common root of P and Q. Let’s factor out the greatest common divisor (GCD) of P and Q. Let D(x) = gcd(P, Q). Then, we can write P(x) = D(x) * P₁(x), Q(x) = D(x) * Q₁(x), where P₁ and Q₁ are coprime polynomials. Then, substituting into the equation:(D P₁)^2 + (D Q₁)^2 = R² ⇒ D² (P₁² + Q₁²) = R².Therefore, R must be divisible by D, so let R = D * R₁. Then, the equation becomes D² (P₁² + Q₁²) = D² R₁² ⇒ P₁² + Q₁² = R₁².So now, we have a new equation with P₁, Q₁, R₁, where P₁ and Q₁ are coprime. Therefore, without loss of generality, we can assume that P and Q are coprime. Otherwise, we factor out their GCD and reduce the problem to coprime polynomials.Therefore, assuming that P and Q are coprime, which can be done by the above reasoning, then they share no common roots. But earlier, we saw that any real root of R must be a common root of P and Q. But if P and Q are coprime, they have no common roots. Therefore, R cannot have any real roots. But R is a cubic polynomial, which must have at least one real root. Contradiction.Wait, this seems like a contradiction. Therefore, our assumption that P and Q are coprime must be invalid. But we factored out the GCD, so if P and Q have a common root, then D(x) is non-constant. Therefore, in the original problem, unless P and Q are coprime, which would lead to R having no real roots, but R is cubic, which must have at least one real root, hence P and Q cannot be coprime. Therefore, D(x) must be non-constant, i.e., P and Q must share at least one common real root.But then, since D(x) is the GCD of P and Q, and we have P = D P₁, Q = D Q₁, with P₁ and Q₁ coprime, then substituting into the equation gives R = D R₁, and P₁² + Q₁² = R₁². Now, since P₁ and Q₁ are coprime, then as before, R₁ cannot have any real roots, because any real root of R₁ would have to be a common root of P₁ and Q₁, which are coprime. But R₁ is a polynomial such that R₁² = P₁² + Q₁². If R₁ has a real root x₀, then P₁(x₀)^2 + Q₁(x₀)^2 = 0 ⇒ P₁(x₀) = Q₁(x₀) = 0, contradicting that they are coprime. Therefore, R₁ cannot have any real roots, so R₁² is always positive, and hence R₁ must be a constant polynomial? Wait, no. If R₁ is a non-constant polynomial with no real roots, then it must have even degree and be always positive or always negative. But since R₁² = P₁² + Q₁², which is a sum of squares, R₁ must be a real polynomial with R₁² ≥ 0 for all real x. Therefore, R₁ must be a real polynomial that is either always non-negative or always non-positive. However, since it has no real roots, it is either always positive or always negative. If R₁ is always positive, then R₁ is a positive constant? Wait, not necessarily. For example, R₁(x) could be x² + 1, which is always positive but not constant. However, in our case, R₁² = P₁² + Q₁². If R₁ is non-constant, then its degree is at least 2. However, let's check the degrees.Originally, we have:- P is quadratic: deg(P) = 2- Q is cubic: deg(Q) = 3- R is cubic: deg(R) = 3After factoring out D(x), which is the GCD of P and Q:deg(D) = d (at least 1, since P and Q must share a common root)Then,deg(P) = d + deg(P₁) = 2 ⇒ deg(P₁) = 2 - ddeg(Q) = d + deg(Q₁) = 3 ⇒ deg(Q₁) = 3 - ddeg(R) = d + deg(R₁) = 3 ⇒ deg(R₁) = 3 - dSince P₁ and Q₁ are coprime, and R₁² = P₁² + Q₁². Let's analyze possible values of d.Case 1: d = 1.Then,deg(P₁) = 2 - 1 = 1deg(Q₁) = 3 - 1 = 2deg(R₁) = 3 - 1 = 2So we have P₁ is linear, Q₁ is quadratic, R₁ is quadratic, and they satisfy P₁² + Q₁² = R₁², with P₁ and Q₁ coprime.Similarly, since R₁ is quadratic and has no real roots (as established earlier, since P₁ and Q₁ are coprime), but R₁² = P₁² + Q₁². Wait, but if R₁ is quadratic with no real roots, then R₁(x) is either always positive or always negative. But R₁² is always positive. So P₁² + Q₁² is always positive, which it is, being a sum of squares. But how does this help?In this case, R₁ is a quadratic polynomial with no real roots, so it can be written as R₁(x) = a(x)^2 + b(x)^2 for some real polynomials a(x) and b(x)? Wait, but R₁ is quadratic, so maybe it's similar to a sum of squares. But R₁ is part of the equation R₁² = P₁² + Q₁². So R₁² is equal to the sum of squares of a linear and a quadratic polynomial. Hmm.Alternatively, perhaps using the theory of Pythagorean triples for polynomials. I recall that over the real numbers, all solutions to A² + B² = C² are given by A = m² - n², B = 2 m n, C = m² + n² for some polynomials m and n. But I'm not sure if this applies here.Wait, if we parametrize P₁, Q₁, R₁ as such, then maybe we can write P₁ = m² - n², Q₁ = 2 m n, R₁ = m² + n². But since P₁ is linear and Q₁ is quadratic, this would require m and n to be polynomials such that m² - n² is linear, and 2 m n is quadratic. Let's see. Let m and n be linear polynomials. Then m² - n² is quadratic, but we need it to be linear. Therefore, this parametrization might not work. Alternatively, maybe m is a linear polynomial and n is a constant. Then m² - n² is quadratic, still not linear. Therefore, this approach might not work.Alternatively, since P₁ is linear and Q₁ is quadratic, let's denote P₁(x) = a x + b, Q₁(x) = c x² + d x + e. Then, R₁(x)^2 = (a x + b)^2 + (c x² + d x + e)^2. Expanding this, we get a quartic polynomial on the right. But R₁(x) is quadratic, so R₁(x)^2 is quartic. Therefore, equating degrees:Right-hand side: (a x + b)^2 + (c x² + d x + e)^2 = c² x⁴ + 2 c d x³ + (2 c e + d² + a²) x² + (2 d e + 2 a b) x + (e² + b²)Left-hand side: (f x² + g x + h)^2 = f² x⁴ + 2 f g x³ + (2 f h + g²) x² + 2 g h x + h²Therefore, equating coefficients:1. x⁴ term: c² = f² ⇒ f = ±c2. x³ term: 2 c d = 2 f g ⇒ c d = f g. Since f = ±c, this gives c d = ±c g ⇒ g = ±d3. x² term: 2 c e + d² + a² = 2 f h + g². Substituting f = ±c and g = ±d:If f = c and g = d: 2 c e + d² + a² = 2 c h + d² ⇒ 2 c e + a² = 2 c h ⇒ h = e + (a²)/(2 c)If f = -c and g = -d: 2 c e + d² + a² = 2 (-c) h + (-d)^2 ⇒ 2 c e + d² + a² = -2 c h + d² ⇒ 2 c e + a² = -2 c h ⇒ h = -e - (a²)/(2 c)4. x term: 2 d e + 2 a b = 2 g h. Substituting g = ±d:If g = d: 2 d e + 2 a b = 2 d h ⇒ d e + a b = d h ⇒ h = e + (a b)/dBut from step 3, if f = c, then h = e + (a²)/(2 c). Therefore, equating the two expressions for h:e + (a²)/(2 c) = e + (a b)/d ⇒ (a²)/(2 c) = (a b)/d ⇒ a/(2 c) = b/d ⇒ a d = 2 b cSimilarly, if f = -c and g = -d, then h = -e - (a²)/(2 c). Then, from the x term:2 d e + 2 a b = 2 (-d) h ⇒ d e + a b = -d h ⇒ h = - (d e + a b)/d = -e - (a b)/dBut from step 3, h = -e - (a²)/(2 c). Therefore,-e - (a b)/d = -e - (a²)/(2 c) ⇒ (a b)/d = (a²)/(2 c) ⇒ b/d = a/(2 c) ⇒ a d = 2 b c, same as before.Therefore, the condition a d = 2 b c must hold.5. Constant term: e² + b² = h². From step 3, if f = c, then h = e + (a²)/(2 c). Therefore:e² + b² = (e + (a²)/(2 c))² = e² + (a²/c) e + (a^4)/(4 c²)Subtracting e² from both sides:b² = (a²/c) e + (a^4)/(4 c²) ⇒ Multiply both sides by 4 c²:4 b² c² = 4 a² c e + a^4 ⇒ 4 c² (b² - a² e / c) = a^4 ⇒ Not sure.Alternatively, substitute e from h = e + (a²)/(2 c) and h from step 4.Wait, this is getting complicated. Let's see if there are any solutions.Suppose we choose specific values for a, b, c, d, e to satisfy the conditions. Let's attempt to construct an example.Let’s set a = 1, c = 1 for simplicity.Then from a d = 2 b c ⇒ d = 2 b.From step 3: h = e + (a²)/(2 c) = e + 1/2.From step 4: h = e + (a b)/d = e + (1 * b)/(2 b) = e + 1/2. So this is consistent.From the constant term: e² + b² = h² = (e + 1/2)^2.Expanding:e² + b² = e² + e + 1/4 ⇒ b² = e + 1/4.We need to choose e such that this holds. For example, let’s pick e = 0. Then b² = 0 + 1/4 ⇒ b = ±1/2.Then, d = 2 b = ±1.Therefore, let’s take b = 1/2, d = 1.Then, our polynomials are:P₁(x) = a x + b = x + 1/2Q₁(x) = c x² + d x + e = x² + x + 0 = x² + xR₁(x) = f x² + g x + h = c x² + d x + h = x² + x + h, where h = e + 1/2 = 0 + 1/2 = 1/2.Therefore, R₁(x) = x² + x + 1/2Check R₁² = (x² + x + 1/2)^2 = x⁴ + 2 x³ + (1 + 1/2) x² + x + 1/4 = x⁴ + 2 x³ + (3/2) x² + x + 1/4On the other hand, P₁² + Q₁² = (x + 1/2)^2 + (x² + x)^2 = (x² + x + 1/4) + (x⁴ + 2 x³ + x²) = x⁴ + 2 x³ + 2 x² + x + 1/4But R₁² = x⁴ + 2 x³ + 3/2 x² + x + 1/4. These are not equal. Therefore, this choice doesn't work. There's a discrepancy in the x² term.Wait, so there's a mistake here. Therefore, my attempt to construct such polynomials failed, suggesting that such a solution may not exist. Therefore, perhaps there are no non-trivial solutions where R₁ is a quadratic polynomial with no real roots, and P₁ is linear, Q₁ is quadratic, coprime. Therefore, this would imply that our initial assumption that R has only one real root (and hence D(x) is linear) leads to a contradiction, meaning that such polynomials cannot exist. Hence, R must have three real roots, which would require P and Q to have three common roots, but P is quadratic, so it cannot have three distinct roots. Therefore, this is impossible unless P has a multiple root.Wait, but if P is quadratic with a multiple root, say a double root at x₀, then Q must also have x₀ as a root (since every root of R is a root of P and Q). Then Q, being cubic, would have x₀ as a root, and possibly two other roots. However, if P has a double root at x₀, then Q must also have x₀ as a root, but Q is cubic. So Q could have x₀ once or multiple times. Let's see.Suppose P(x) = (x - x₀)^2, quadratic. Then Q(x) must have x₀ as a root. Let Q(x) = (x - x₀) * S(x), where S(x) is quadratic. Then R(x) = (x - x₀) * T(x), where T(x) is quadratic. Then, substituting into P² + Q² = R²:(x - x₀)^4 + [(x - x₀) S(x)]² = [(x - x₀) T(x)]² ⇒ (x - x₀)^4 + (x - x₀)^2 S(x)^2 = (x - x₀)^2 T(x)^2 ⇒ (x - x₀)^2 [ (x - x₀)^2 + S(x)^2 ] = (x - x₀)^2 T(x)^2Cancel (x - x₀)^2:(x - x₀)^2 + S(x)^2 = T(x)^2Now, this is an equation involving polynomials where S(x) is quadratic and T(x) is quadratic. Let’s denote S(x) = a x² + b x + c, T(x) = d x² + e x + f. Then:(x - x₀)^2 + (a x² + b x + c)^2 = (d x² + e x + f)^2Expand both sides:Left side: (x² - 2 x₀ x + x₀²) + (a² x⁴ + 2 a b x³ + (2 a c + b²) x² + 2 b c x + c²)= a² x⁴ + 2 a b x³ + (2 a c + b² + 1) x² + (2 b c - 2 x₀) x + (c² + x₀²)Right side: d² x⁴ + 2 d e x³ + (2 d f + e²) x² + 2 e f x + f²Equate coefficients:1. x⁴: a² = d² ⇒ d = ±a2. x³: 2 a b = 2 d e ⇒ a b = d e. Since d = ±a, this gives a b = ±a e ⇒ b = ±e3. x²: 2 a c + b² + 1 = 2 d f + e². If d = a and e = b, then:2 a c + b² + 1 = 2 a f + b² ⇒ 2 a c + 1 = 2 a f ⇒ f = c + 1/(2 a)If d = -a and e = -b, then:2 a c + b² + 1 = 2 (-a) f + (-b)^2 ⇒ 2 a c + b² + 1 = -2 a f + b² ⇒ 2 a c + 1 = -2 a f ⇒ f = - (2 a c + 1)/(2 a) = -c - 1/(2 a)4. x term: 2 b c - 2 x₀ = 2 e f. If using d = a and e = b:2 b c - 2 x₀ = 2 b f ⇒ b c - x₀ = b f ⇒ f = (b c - x₀)/b = c - x₀ / bBut from step 3, f = c + 1/(2 a). Therefore,c - x₀ / b = c + 1/(2 a) ⇒ -x₀ / b = 1/(2 a) ⇒ x₀ = - b / (2 a)Similarly, if d = -a and e = -b:2 b c - 2 x₀ = 2 (-b) f ⇒ 2 b c - 2 x₀ = -2 b f ⇒ b c - x₀ = -b f ⇒ f = (x₀ - b c)/bFrom step 3, f = -c - 1/(2 a). Therefore,(x₀ - b c)/b = -c - 1/(2 a) ⇒ x₀ / b - c = -c - 1/(2 a) ⇒ x₀ / b = -1/(2 a) ⇒ x₀ = -b/(2 a), same as before.5. Constant term: c² + x₀² = f²From step 3 and 4, if d = a and e = b:f = c + 1/(2 a)So f² = (c + 1/(2 a))² = c² + c/a + 1/(4 a²)But from the constant term:c² + x₀² = c² + (b²)/(4 a²) = c² + (b²)/(4 a²)Therefore,c² + (b²)/(4 a²) = c² + c/a + 1/(4 a²) ⇒ (b²)/(4 a²) = c/a + 1/(4 a²)Multiply both sides by 4 a²:b² = 4 a c + 1So, summarizing the conditions:From step 4: x₀ = -b/(2 a)From step 5: b² = 4 a c + 1And from step 3: f = c + 1/(2 a)Additionally, we need to ensure consistency in the coefficients.Let’s attempt to choose specific values. Let’s set a = 1 for simplicity.Then, x₀ = -b/2From step 5: b² = 4 * 1 * c + 1 ⇒ c = (b² - 1)/4Then, f = c + 1/(2 * 1) = (b² - 1)/4 + 1/2 = (b² - 1 + 2)/4 = (b² + 1)/4Now, let's construct S(x) and T(x):S(x) = a x² + b x + c = x² + b x + (b² - 1)/4T(x) = d x² + e x + f = x² + b x + (b² + 1)/4Now, check the x term coefficient in the left side:Left side x term: 2 b c - 2 x₀ = 2 b * (b² - 1)/4 - 2*(-b/2) = (b³ - b)/2 + b = (b³ - b + 2 b)/2 = (b³ + b)/2Right side x term: 2 e f = 2 b * (b² + 1)/4 = (b³ + b)/2They match. So this works.Therefore, such polynomials exist. For example, take a = 1, b = 1:Then, x₀ = -1/2c = (1 - 1)/4 = 0/4 = 0f = (1 + 1)/4 = 0.5Therefore,S(x) = x² + x + 0 = x² + xT(x) = x² + x + 0.5Thus, P(x) = (x - x₀)^2 = (x + 1/2)^2 = x² + x + 1/4Q(x) = (x - x₀) S(x) = (x + 1/2)(x² + x) = x³ + (3/2)x² + (1/2)xR(x) = (x - x₀) T(x) = (x + 1/2)(x² + x + 0.5) = x³ + (3/2)x² + (1/2)x + 0.25Check P² + Q²:P² = (x² + x + 1/4)^2 = x⁴ + 2x³ + (3/2)x² + (1/2)x + 1/16Q² = (x³ + (3/2)x² + (1/2)x)^2 = x⁶ + 3x⁵ + (13/4)x⁴ + (3/2)x³ + (1/4)x²But R² = (x³ + (3/2)x² + (1/2)x + 0.25)^2 = x⁶ + 3x⁵ + (13/4)x⁴ + 3x³ + (13/16)x² + (3/8)x + 1/16But P² + Q² = x⁶ + 3x⁵ + (13/4)x⁴ + 3x³ + (13/16)x² + (3/8)x + 1/16Wait, no. Wait, Q is cubic, so Q² is degree 6. P is quadratic, so P² is degree 4. Therefore, P² + Q² is Q² + P², which is a degree 6 polynomial. But in our case, R is cubic, so R² is degree 6. Therefore, yes, this matches.But in this example, R(x) = x³ + (3/2)x² + (1/2)x + 0.25. Does this cubic polynomial have all real roots? Let's check.We can use the rational root theorem. Possible rational roots are factors of 0.25 over factors of 1, i.e., ±1, ±0.5, ±0.25. Testing x = -1:R(-1) = -1 + 3/2 - 1/2 + 0.25 = (-1 + 3/2) + (-1/2 + 0.25) = 0.5 + (-0.25) = 0.25 ≠ 0x = -0.5:R(-0.5) = (-0.5)^3 + (3/2)(-0.5)^2 + (1/2)(-0.5) + 0.25= -0.125 + (3/2)(0.25) + (-0.25) + 0.25= -0.125 + 0.375 - 0.25 + 0.25 = 0.25 ≠ 0x = -0.25:R(-0.25) = (-0.25)^3 + (3/2)(-0.25)^2 + (1/2)(-0.25) + 0.25= -0.015625 + (3/2)(0.0625) + (-0.125) + 0.25= -0.015625 + 0.09375 - 0.125 + 0.25 ≈ 0.203125 ≠ 0So no rational roots. Let's check the derivative for the number of real roots. The derivative R’(x) = 3x² + 3x + 0.5. The discriminant is 9 - 6 = 3, which is positive, so R’(x) has two real roots, meaning R(x) has one local maximum and one local minimum. Therefore, by the Intermediate Value Theorem, since R(x) tends to ±∞ as x→±∞, and it has a local maximum and minimum, it could have three real roots. But without finding them explicitly, it's hard to say. Alternatively, compute the discriminant of the cubic.The cubic R(x) = x³ + (3/2)x² + (1/2)x + 1/4. The general form is ax³ + bx² + cx + d. The discriminant Δ = 18abcd - 4b³d + b²c² - 4ac³ - 27a²d².Plugging in a=1, b=3/2, c=1/2, d=1/4:Δ = 18*1*(3/2)*(1/2)*(1/4) - 4*(3/2)^3*(1/4) + (3/2)^2*(1/2)^2 - 4*1*(1/2)^3 - 27*1^2*(1/4)^2Compute each term:First term: 18*(3/2)*(1/2)*(1/4) = 18*(3/16) = 54/16 = 27/8Second term: -4*(27/8)*(1/4) = -4*(27/32) = -108/32 = -27/8Third term: (9/4)*(1/4) = 9/16Fourth term: -4*(1/8) = -1/2Fifth term: -27*(1/16) = -27/16Sum all terms:27/8 - 27/8 + 9/16 - 1/2 - 27/16= 0 + (9/16 - 27/16) - 1/2= (-18/16) - 8/16= -26/16 = -13/8Since Δ < 0, the cubic has one real root and two complex conjugate roots. Therefore, R(x) has only one real root, which contradicts our earlier conclusion that R must have three real roots. But in this example, we have such polynomials P, Q, R satisfying P² + Q² = R², where P is quadratic, Q and R are cubic, and R has only one real root. Therefore, my previous reasoning was incorrect.But the problem states that at least one of the cubic polynomials must have all real roots. In this example, both Q and R are cubic. Let's check Q(x) = x³ + (3/2)x² + (1/2)x.Q(x) factors as x(x² + (3/2)x + 1/2). The quadratic factor has discriminant (9/4) - 2 = (9/4 - 8/4) = 1/4, which is positive. Therefore, the quadratic factors as (x + 1)(x + 0.5), so Q(x) = x(x + 1)(x + 0.5). Therefore, Q has three real roots: 0, -1, -0.5. Therefore, even though R has only one real root, Q has all real roots. Therefore, this example satisfies the problem's conclusion: at least one of the cubic polynomials (Q in this case) has all real roots.Ah, so the key point is that even if one cubic (R) has only one real root, the other cubic (Q) must have all real roots. Therefore, the statement is proven.Therefore, the crux is that if you have such an equation P² + Q² = R² with one quadratic and two cubics, then one of the cubics must have all real roots, while the other may not. The example confirms this.Therefore, the way to prove it is to assume the contrary: that both cubics have non-real roots, i.e., each has one real root and two complex-conjugate roots. Then, analyze the implications on the equation P² + Q² = R² and reach a contradiction.Let me formalize this.Assume, for contradiction, that both cubic polynomials have only one real root each (and two complex-conjugate roots). Let's denote the cubics as Q and R (assuming P is quadratic).Then, Q(x) = (x - α) S(x), where S(x) is an irreducible quadratic over the reals, and R(x) = (x - β) T(x), where T(x) is also an irreducible quadratic over the reals.From the equation P² + Q² = R², we know that any real root of R must be a real root of both P and Q. Since R has only one real root β, then β must be a common root of P and Q. Similarly, any real root of Q must be a real root of P and R. But Q has only one real root α, so α must be a common root of P and R.But P is quadratic, so it can have at most two distinct real roots. However, both α and β must be roots of P. If α ≠ β, then P would have two distinct roots, which is allowed. If α = β, then P has a repeated root.But let's analyze the multiplicities. Suppose α ≠ β. Then P has two distinct roots α and β, so P(x) = (x - α)(x - β). However, Q(x) has a root at α and irreducible quadratic S(x), and R(x) has a root at β and irreducible quadratic T(x).Now, substituting into the equation P² + Q² = R²:[(x - α)(x - β)]² + [(x - α) S(x)]² = [(x - β) T(x)]²Factor out (x - α)^2 from the first two terms:(x - α)^2 [(x - β)^2 + S(x)^2] = (x - β)^2 T(x)^2But this equation must hold for all x. Let's analyze the roots. The left side has a factor of (x - α)^2, and the right side has a factor of (x - β)^2. Unless α = β, these factors are different, leading to a contradiction because the left side would have a root at α and the right side at β, but polynomials can only be equal if they have the same roots with the same multiplicities.Therefore, α must equal β. So P(x) = (x - α)^2, Q(x) = (x - α) S(x), R(x) = (x - α) T(x), where S(x) and T(x) are irreducible quadratics.Substituting into the equation:(x - α)^4 + (x - α)^2 S(x)^2 = (x - α)^2 T(x)^2Divide both sides by (x - α)^2 (which is non-zero for x ≠ α):(x - α)^2 + S(x)^2 = T(x)^2Now, this is an equation involving polynomials where S(x) and T(x) are irreducible quadratics. Let's write S(x) = a x² + b x + c and T(x) = d x² + e x + f. Then:(x - α)^2 + (a x² + b x + c)^2 = (d x² + e x + f)^2Expanding both sides:Left side: x² - 2 α x + α² + a² x⁴ + 2 a b x³ + (2 a c + b²) x² + 2 b c x + c²= a² x⁴ + 2 a b x³ + (2 a c + b² + 1) x² + (2 b c - 2 α) x + (c² + α²)Right side: d² x⁴ + 2 d e x³ + (2 d f + e²) x² + 2 e f x + f²Equating coefficients:1. x⁴ term: a² = d² ⇒ d = ±a2. x³ term: 2 a b = 2 d e ⇒ a b = d e. Since d = ±a, this implies e = ±b.3. x² term: 2 a c + b² + 1 = 2 d f + e². Substituting d = ±a and e = ±b:If d = a and e = b:2 a c + b² + 1 = 2 a f + b² ⇒ 2 a c + 1 = 2 a f ⇒ f = c + 1/(2 a)If d = -a and e = -b:2 a c + b² + 1 = 2 (-a) f + (-b)^2 ⇒ 2 a c + 1 = -2 a f + b² ⇒ This introduces more variables; let's proceed with the first case.4. x term: 2 b c - 2 α = 2 e f. If e = b:2 b c - 2 α = 2 b f ⇒ b c - α = b f ⇒ f = c - α / bFrom step 3, f = c + 1/(2 a). Therefore:c - α / b = c + 1/(2 a) ⇒ -α / b = 1/(2 a) ⇒ α = -b/(2 a)5. Constant term: c² + α² = f²From step 3: f = c + 1/(2 a). Therefore:c² + α² = (c + 1/(2 a))² = c² + c / a + 1/(4 a²) ⇒ α² = c / a + 1/(4 a²)But from step 4, α = -b/(2 a). Therefore:(-b/(2 a))² = c / a + 1/(4 a²) ⇒ b²/(4 a²) = c / a + 1/(4 a²) ⇒ Multiply both sides by 4 a²:b² = 4 a c + 1Therefore, we have the conditions:α = -b/(2 a)b² = 4 a c + 1Now, since S(x) = a x² + b x + c is irreducible over the reals, its discriminant must be negative:b² - 4 a c < 0But from b² = 4 a c + 1, substitute into the discriminant:(4 a c + 1) - 4 a c = 1 < 0Which is a contradiction, since 1 is not less than 0. Therefore, our assumption that both cubics Q and R have only one real root leads to a contradiction. Therefore, at least one of the cubics must have all real roots.This completes the proof.

✅:To prove that at least one of the cubic polynomials in the equation ( P^2 + Q^2 = R^2 ) must have all real roots, we proceed by contradiction:1. Assume the Contrary: Suppose both cubic polynomials (say ( Q ) and ( R )) have only one real root each, with the remaining roots being complex conjugates. Thus, they can be factored as: [ Q(x) = (x - alpha)S(x), quad R(x) = (x - beta)T(x), ] where ( S(x) ) and ( T(x) ) are irreducible quadratics.2. Common Roots: Since ( P^2 + Q^2 = R^2 ), any real root of ( R ) must be a root of both ( P ) and ( Q ). Similarly, any real root of ( Q ) must be a root of ( P ) and ( R ). Hence, ( alpha ) and ( beta ) must be roots of ( P ). As ( P ) is quadratic, ( alpha = beta ) (otherwise ( P ) would have three roots), so ( P(x) = (x - alpha)^2 ).3. Substitute into the Equation: [ (x - alpha)^4 + [(x - alpha)S(x)]^2 = [(x - alpha)T(x)]^2. ] Dividing by ( (x - alpha)^2 ): [ (x - alpha)^2 + S(x)^2 = T(x)^2. ]4. Analyze Coefficients: Let ( S(x) = ax^2 + bx + c ) and ( T(x) = dx^2 + ex + f ). Expanding and equating coefficients leads to contradictions unless: - ( b^2 = 4ac + 1 ) (from coefficient conditions). - The discriminant of ( S(x) ), ( b^2 - 4ac ), must be negative (since ( S(x) ) is irreducible). However, substituting ( b^2 = 4ac + 1 ) yields ( 1 < 0 ), a contradiction.5. Conclusion: The assumption that both cubics have non-real roots is false. Hence, at least one cubic must have all real roots.Final Answer: At least one of the cubic polynomials in the equation ( P^2 + Q^2 = R^2 ) must have all real roots, as proven by contradiction. boxed{text{At least one cubic polynomial has all real roots.}}

🔑: Problem:Let ( P, Q, R ) be polynomials with real coefficients, where ( P ) is a quadratic polynomial and ( Q ), ( R ) are cubic polynomials satisfying the relation ( P^2 + Q^2 = R^2 ). Prove that one of the cubic polynomials must have all real roots.1. According to the problem, ( R ) and ( Q ) are cubic polynomials, while ( P ) is a quadratic polynomial. Without loss of generality, assume ( R ) and ( Q ) are cubic polynomials, and ( P ) is a quadratic polynomial.2. We start with the given equation: [ P^2 + Q^2 = R^2 ]3. Rearrange the equation as: [ P^2 = R^2 - Q^2 = (R + Q)(R - Q) ]4. Since ( R ) and ( Q ) are cubic polynomials, ( R + Q ) is also a cubic polynomial, and ( R - Q ) will be a cubic polynomial as well. Notice that ( P^2 ) is a polynomial of degree 4, leading us to conclude: [ (R + Q) text{ and } (R - Q) text{ must be polynomials whose product results in a polynomial of degree 4 (which matches } P^2 text{)} ]5. Here, ( P ) being quadratic implies that ( P^2 ) is quartic (degree 4). So, ( (R - Q) ) must be linear (degree 1) for their product to be a degree 4 polynomial since: [ P^2 = (R + Q)(R - Q) ]6. Thus, ( R - Q ) must be a first-degree polynomial (i.e., linear polynomial). Let ( R - Q = t(x - x_1) ) where ( t > 0 ) ensures the coefficient of ( x ) in ( P^2 ) is positive: [ R - Q = t(x - x_1) ]7. This implies that both ( R ) and ( Q ) are divisible by ( x - x_1 ): [ R = (x - x_1) R_1 quad text{ and } quad Q = (x - x_1) Q_1 ] Here, ( R_1 ) and ( Q_1 ) are polynomials of degree 2, with the leading coefficient positive.8. Now suppose ( P ) is given by: [ P = a(x - x_1)(x - x_2) ]9. Plugging into the original equation: [ P^2 = a^2(x - x_1)^2(x - x_2)^2 ]10. From the rearranged form ( P^2 = (R + Q)(R - Q) ): [ a^2(x - x_1)^2(x - x_2)^2 = (R_1 + Q_1)(R_1 - Q_1) ]11. Assume: [ R_1 - Q_1 = t ] and thus: [ R_1 = Q_1 + t ]12. Substituting back we get: [ a^2(x - x_2)^2 = (2Q_1 + t)t ]13. Rearrange for ( Q_1 ): [ Q_1 = frac{a^2}{2t}(x - x_2)^2 - frac{t}{2} ]14. It follows that ( Q_1 ) is a quadratic polynomial with two real roots. Consequently, ( Q ) must be cubic polynomial with three real roots as it has the factor ( x - x_1 ). Conclusion:Thus, we have shown that one of the cubic polynomials, (Q), must have all real roots.[boxed{text{Q}}]

🧠:Okay, so I need to find the sum of the series where each term is k multiplied by k factorial, right? The series is given as 1·1! + 2·2! + ... + n·n!. Hmm, let me think about how to approach this. First, maybe I should write out the first few terms to see if there's a pattern. Let's see:For n=1: 1·1! = 1·1 = 1For n=2: 1·1! + 2·2! = 1 + 4 = 5For n=3: 1·1! + 2·2! + 3·3! = 1 + 4 + 18 = 23For n=4: 1·1! + 2·2! + 3·3! + 4·4! = 1 + 4 + 18 + 96 = 119Hmm, the results are 1, 5, 23, 119... Let me see if these numbers relate to factorials. Well, 1 is 1!, 5 is 5 which is 5, not a factorial, 23 is not a factorial either. Wait, but 5 is one less than 6 (3!), 23 is one less than 24 (4!), 119 is one less than 120 (5!). Oh! Wait a minute, for n=2, the sum is 5 which is 5 = 6 -1 = 3! -1. For n=3, 23 = 24 -1 = 4! -1. For n=4, 119 = 120 -1 = 5! -1. So maybe the pattern is that the sum up to n is (n+1)! -1. Let me check for n=1: (1+1)! -1 = 2! -1 = 2 -1 =1, which matches. So that seems promising. But I need to confirm this. Let me test n=5. The sum would be 1·1! +2·2!+3·3!+4·4!+5·5! = 1 +4 +18 +96 +600 = 719. And (5+1)! -1 = 720 -1=719. Yep, that works. So the formula seems to hold. But how do I prove it?Maybe I can use mathematical induction. Let's try that. Base case: n=1. The sum is 1·1! =1. The formula gives (1+1)! -1=2! -1=2-1=1. So it works.Assume that for some k ≥1, the sum up to k is (k+1)! -1. Now consider the sum up to k+1. That would be [Sum from i=1 to k of i·i!] + (k+1)·(k+1)! By the induction hypothesis, this is [(k+1)! -1] + (k+1)·(k+1)! Let's factor out (k+1)!: (k+1)! [1 + (k+1)] -1 = (k+1)! (k+2) -1 = (k+2)! -1. Which is the formula with n=k+1. Therefore, by induction, the formula holds for all n≥1.But wait, is there another way to see why this works? Maybe by telescoping the series or manipulating the terms. Let's see. Each term is k·k!. Notice that k·k! = (k+1)! -k!. Let me verify this. For example, take k=2: 2·2! =4. (2+1)! -2! =6 -2=4. Yes, that works. Similarly, k=3: 3·6=18, 4! -3! =24 -6=18. So indeed, k·k! = (k+1)! -k!.So then, the sum from k=1 to n of k·k! is the same as the sum from k=1 to n of [(k+1)! -k!]. Let's write that out:[(2! -1!) + (3! -2!) + (4! -3!) + ... + (n+1)! -n!)]When we add these up, most terms should cancel out. Let's check:-1! +2! -2! +3! -3! + ... -n! + (n+1)! All the intermediate terms cancel, leaving -1! + (n+1)! = (n+1)! -1. Which is exactly the formula we had before. So that's a telescoping series approach, which also confirms that the sum is (n+1)! -1. Therefore, the answer should be (n+1)! -1. Let me just make sure there's no mistake here. For n=2, (2+1)! -1=6-1=5, which matches. For n=3, 24-1=23. Yep, seems right. Another way to think about it: if someone is trying to find a recursive relationship. Suppose S(n) is the sum up to n. Then S(n) = S(n-1) +n·n!. But from the formula, S(n) = (n+1)! -1, so S(n-1) =n! -1. Then, S(n) =n! -1 +n·n! =n!(1 +n) -1 =(n+1)! -1. Which again gives the same result. So all these methods confirm that the sum is (n+1)! -1. Therefore, the final answer should be that.

✅:The sum (1 cdot 1! + 2 cdot 2! + cdots + n cdot n!) can be evaluated using a telescoping series approach. Step 1: Express each term (k cdot k!) in a telescoping form. Notice that: [k cdot k! = (k + 1)! - k!] This holds because ((k + 1)! = (k + 1) cdot k!), so: [(k + 1)! - k! = k! cdot (k + 1 - 1) = k cdot k!]Step 2: Rewrite the original sum using the telescoping identity. The sum becomes: [sum_{k=1}^n k cdot k! = sum_{k=1}^n left[(k + 1)! - k!right]]Step 3: Expand the telescoping series and simplify. [begin{align*}sum_{k=1}^n left[(k + 1)! - k!right] &= left(2! - 1!right) + left(3! - 2!right) + cdots + left((n + 1)! - n!right) &= -1! + (n + 1)! quad text{(all intermediate terms cancel)} &= (n + 1)! - 1.end{align*}]Conclusion: The sum simplifies to ((n + 1)! - 1).Answer: [boxed{(n + 1)! - 1}]

🔑:To solve the problem (1 cdot 1! + 2 cdot 2! + ldots + n cdot n!), we will use the principle of mathematical induction to prove the formula:[ S(n) = 1 cdot 1! + 2 cdot 2! + ldots + n cdot n! = (n+1)! - 1. ]1. Base Case: We start by checking the base case where (n = 1): [ S(1) = 1 cdot 1! = 1. ] According to our formula, for (n = 1): [ (1+1)! - 1 = 2! - 1 = 2 - 1 = 1. ] Hence, the formula (S(n) = (n+1)! - 1) holds for (n = 1).2. Inductive Hypothesis: Assume that the formula (S(k) = 1 cdot 1! + 2 cdot 2! + cdots + k cdot k! = (k+1)! - 1) holds for some (k geq 1). [ 1 cdot 1! + 2 cdot 2! + ldots + k cdot k! = (k+1)! - 1. ]3. Inductive Step: We need to prove that the formula holds for (n = k + 1): [ S(k+1) = 1 cdot 1! + 2 cdot 2! + ldots + k cdot k! + (k+1) cdot (k+1)!. ] By the inductive hypothesis, we can substitute (S(k)): [ S(k+1) = (k+1)! - 1 + (k+1) cdot (k+1)!. ]4. Simplifying the Expression: Combine terms involving ((k+1)!): [ S(k+1) = (k+1)! - 1 + (k+1) cdot (k+1)!. ] Factor out ((k+1)!): [ S(k+1) = (k+1)! - 1 + (k+1) cdot (k+1)! = (k+1)! [1 + (k+1)] - 1. ] Simplify inside the bracket: [ S(k+1) = (k+1)! cdot (k+2) - 1 = (k+2)! - 1. ]Hence, we have shown that if the formula holds for (n = k), it also holds for (n = k+1). By the principle of mathematical induction, the formula is true for all (n geq 1):[1 cdot 1! + 2 cdot 2! + ldots + n cdot n! = (n+1)! - 1.] Conclusion:[ boxed{(n+1)! - 1} ]

🧠:Okay, so I need to find the points on the surface of a cube that are equidistant from the endpoints of one of its space diagonals. Let me start by visualizing a cube. A cube has all sides equal, and all angles are right angles. Each space diagonal runs from one corner of the cube to the opposite corner, passing through the interior of the cube. The problem is asking for the set of points on the cube's surface that are equidistant from these two endpoints of a space diagonal.First, let me set up a coordinate system to make things easier. If I consider a cube with side length 'a', I can place it in a 3D coordinate system such that one corner is at the origin (0,0,0) and the opposite corner is at (a,a,a), which would be the endpoints of the space diagonal. The space diagonal in this case connects (0,0,0) to (a,a,a). The problem now is to find all points (x,y,z) on the cube's surface where the distance from (x,y,z) to (0,0,0) is equal to the distance from (x,y,z) to (a,a,a).The distance from a point (x,y,z) to (0,0,0) is given by the Euclidean distance formula: √(x² + y² + z²). Similarly, the distance from (x,y,z) to (a,a,a) is √[(x - a)² + (y - a)² + (z - a)²]. Since these distances are equal, I can set them equal to each other and square both sides to eliminate the square roots:x² + y² + z² = (x - a)² + (y - a)² + (z - a)²Expanding the right-hand side:(x² - 2ax + a²) + (y² - 2ay + a²) + (z² - 2az + a²)Combine like terms:x² + y² + z² - 2a(x + y + z) + 3a²So setting the left-hand side equal to this:x² + y² + z² = x² + y² + z² - 2a(x + y + z) + 3a²Subtract x² + y² + z² from both sides:0 = -2a(x + y + z) + 3a²Divide both sides by 'a' (assuming a ≠ 0):0 = -2(x + y + z) + 3aThen:2(x + y + z) = 3aDivide both sides by 2:x + y + z = (3a)/2So the set of points equidistant from (0,0,0) and (a,a,a) lie on the plane x + y + z = 3a/2. However, we need to find the intersection of this plane with the cube's surface. The cube's surface consists of all points where at least one of the coordinates x, y, or z is either 0 or a. So, the points we are looking for must satisfy both x + y + z = 3a/2 and have at least one coordinate equal to 0 or a.Let me consider the different faces of the cube and see where this plane intersects them.First, consider the face where x = 0. On this face, the equation becomes 0 + y + z = 3a/2 ⇒ y + z = 3a/2. However, on the x = 0 face, y and z can range from 0 to a. The sum y + z = 3a/2 would require y and z to be greater than a, which is impossible since their maximum value is a. Therefore, there are no points on the x = 0 face that satisfy the equation.Similarly, check the face where x = a. Substituting x = a into the equation: a + y + z = 3a/2 ⇒ y + z = a/2. On the x = a face, y and z can range from 0 to a. The equation y + z = a/2 is possible here. This is a line segment on the x = a face from (a, 0, a/2) to (a, a/2, 0). Wait, let me verify that.If y + z = a/2, then possible solutions on the x = a face are points where y ranges from 0 to a/2 and z = a/2 - y, but z must also be ≥ 0. So when y = 0, z = a/2; when y = a/2, z = 0. So the line segment connects (a, 0, a/2) to (a, a/2, 0). Similarly, but actually, since both y and z must be between 0 and a, this line lies entirely within the face x = a. So this is a valid intersection.Similarly, if we check the y = 0 face, substituting y = 0 into x + y + z = 3a/2 gives x + z = 3a/2. But on the y = 0 face, x and z can only go up to a. So x + z ≤ 2a. However, 3a/2 is less than 2a (since 3a/2 = 1.5a < 2a). Wait, no, 3a/2 is 1.5a, which is greater than a. Because x and z can each be at most a. So the sum x + z can be at most 2a. But 3a/2 is 1.5a, so x + z = 1.5a on the y = 0 face. So possible?Wait, if x and z are both ≤ a, then x + z can be up to 2a. So 1.5a is possible. For example, x = a, z = 0.5a; or x = 0.5a, z = a. So the line segment on the y = 0 face would connect (a, 0, a/2) to (a/2, 0, a). Similarly, on the z = 0 face, substituting z = 0 gives x + y = 3a/2. On the z = 0 face, x and y can go up to a, so x + y = 1.5a. So similar to the previous case, this line connects (a, a/2, 0) to (a/2, a, 0).Similarly, on the faces y = a and z = a, let's check.For the face y = a: substituting y = a into x + y + z = 3a/2 gives x + a + z = 3a/2 ⇒ x + z = a/2. So on the y = a face, x and z can range from 0 to a. So x + z = a/2 is a line from (0, a, a/2) to (a/2, a, 0).For the face z = a: substituting z = a into the equation gives x + y + a = 3a/2 ⇒ x + y = a/2. On the z = a face, x and y range from 0 to a, so x + y = a/2 is a line from (0, a/2, a) to (a/2, 0, a).Similarly, for the x = 0 face, we saw earlier that substituting x = 0 gives y + z = 3a/2, which is not possible since y and z can each be at most a, so their sum is at most 2a, but 3a/2 is 1.5a. Wait, 3a/2 is 1.5a, which is less than 2a. Wait, no, 3a/2 is 1.5a, which is less than 2a, but both y and z on the x=0 face can go up to a. Wait, but 3a/2 is 1.5a, so if y and z have to sum to 1.5a, that's possible. For example, y = a, z = 0.5a; but on the x = 0 face, z can be up to a, so y + z = 1.5a would require y and z to be in the range 0.5a to a. Wait, but when x = 0, then y and z can each be from 0 to a. So if y + z = 1.5a, then possible points are where y ranges from 0.5a to a, and z = 1.5a - y. Similarly, z ranges from 0.5a to a. So this would give a line segment from (0, a, 0.5a) to (0, 0.5a, a). But wait, but earlier I thought there were no points on the x=0 face. Wait, maybe I made a mistake earlier.Wait, let me recast. If x = 0, then y + z = 3a/2. But since y and z are each at most a, their maximum sum is 2a. 3a/2 is 1.5a, which is less than 2a. So yes, y + z = 1.5a is possible on the x = 0 face. For instance, when y = a, z = 0.5a; when y = 0.5a, z = a; and points in between. So the line segment on the x = 0 face connects (0, a, a/2) to (0, a/2, a).Wait, so my initial conclusion that there are no points on x=0 was wrong. Similarly, same applies for y=0 and z=0. Wait, earlier when I considered x=0, I thought y + z = 3a/2 was impossible, but actually, 3a/2 is 1.5a, and since y and z can go up to a each, their sum can be up to 2a. So 1.5a is feasible. Therefore, the plane x + y + z = 3a/2 intersects all six faces of the cube. Wait, but when we considered x=0, y=0, z=0 faces, we have intersections. Let's verify.Wait, for example, on the x=0 face: y + z = 1.5a. Since y and z can each be up to a, then 1.5a is possible. For instance, y = a, z = 0.5a; y = 0.5a, z = a; and all points in between where y and z sum to 1.5a. So the line segment on x=0 face is from (0, a, a/2) to (0, a/2, a). Similarly, on y=0 face: x + z = 1.5a, which would connect (a, 0, a/2) to (a/2, 0, a). On z=0 face: x + y = 1.5a, connecting (a, a/2, 0) to (a/2, a, 0). Similarly, on the opposite faces x=a, y=a, z=a, we have:On x=a face: y + z = 0.5a, connecting (a, 0, a/2) to (a, a/2, 0). Wait, earlier when I considered x=a, I thought y + z = a/2. Wait, let's check again.Wait, if we substitute x=a into x + y + z = 3a/2, we get a + y + z = 3a/2 ⇒ y + z = a/2. So on the x=a face, y and z are between 0 and a. So y + z = a/2 is a line segment from (a, 0, a/2) to (a, a/2, 0). Similarly, on y=a face: x + z = a/2, which would be from (0, a, a/2) to (a/2, a, 0). On z=a face: x + y = a/2, which is from (0, a/2, a) to (a/2, 0, a).Therefore, the intersection of the plane x + y + z = 3a/2 with the cube's surface consists of six line segments, each on a different face of the cube. Each of these line segments connects midpoints of edges. For example, on the x=0 face, the line connects (0, a, a/2) to (0, a/2, a), which are midpoints of the edges where x=0, y=a, z from 0 to a and x=0, y from 0 to a, z=a. Similarly, on the x=a face, the line connects (a, 0, a/2) to (a, a/2, 0), which are midpoints of edges on the x=a face.Wait, but actually, the points (a, 0, a/2) and (a, a/2, 0) are not midpoints of edges. The edges of the cube are the lines where two coordinates are fixed (either 0 or a) and the third varies. The midpoints of edges would have two coordinates fixed and the third at a/2. For example, the midpoint of the edge from (a,0,0) to (a,a,0) is (a, a/2, 0). Similarly, the midpoint of the edge from (a,0,0) to (a,0,a) is (a,0,a/2). So yes, these points are midpoints of edges.So each of these line segments on the cube's faces connects two such midpoints. Therefore, the intersection of the plane with the cube's surface is a hexagon, with each edge of the hexagon lying on a different face of the cube. The vertices of this hexagon are the midpoints of the cube's edges.Wait, but let me confirm. The plane x + y + z = 3a/2 intersects the cube in a regular hexagon? Because a cube's space diagonal plane can form a regular hexagon when intersected with the cube. But I need to verify the coordinates.Let me list all the intersection points:1. On x=0 face: (0, a, a/2) and (0, a/2, a)2. On y=0 face: (a, 0, a/2) and (a/2, 0, a)3. On z=0 face: (a, a/2, 0) and (a/2, a, 0)4. On x=a face: (a, 0, a/2) and (a, a/2, 0)5. On y=a face: (0, a, a/2) and (a/2, a, 0)6. On z=a face: (0, a/2, a) and (a/2, 0, a)Wait, but each edge of the hexagon is shared between two faces. For example, the point (a, 0, a/2) is on both the x=a face and the y=0 face. Similarly, (0, a, a/2) is on x=0 and y=a faces. So connecting these points, the hexagon would have vertices at:1. (a, 0, a/2)2. (a, a/2, 0)3. (a/2, a, 0)4. (0, a, a/2)5. (0, a/2, a)6. (a/2, 0, a)Connecting these points in order would form a regular hexagon. Each vertex is a midpoint of an edge of the cube. Each edge of the hexagon lies on a different face of the cube. Therefore, all points on this hexagon are equidistant from (0,0,0) and (a,a,a).But wait, the problem states "points on the surface of a cube that are equidistant from the endpoints of one of its space diagonals." So the hexagon described here is the set of all such points. Therefore, the solution is the set of points forming a regular hexagon on the cube's surface, each of whose vertices is the midpoint of an edge of the cube.To confirm, let's pick a point on this hexagon and check the distances. Take the point (a, 0, a/2). The distance from (0,0,0) is √(a² + 0 + (a/2)²) = √(a² + a²/4) = √(5a²/4) = (a/2)√5. The distance from (a,a,a) is √[(0)² + (a)² + (a/2)²] = √(a² + a²/4) = same as above. So yes, equidistant.Similarly, take the point (a/2, a, 0). Distance from (0,0,0): √[(a/2)² + a² + 0] = √(a²/4 + a²) = √(5a²/4) = (a/2)√5. Distance from (a,a,a): √[(a/2)² + 0 + a²] = same result. So indeed, all points on this hexagon are equidistant from both ends of the space diagonal.Therefore, the answer is that the set of points equidistant from the endpoints of a space diagonal of a cube lie on a regular hexagon formed by the midpoints of the cube's edges. Each edge of this hexagon lies on a face of the cube, connecting midpoints of adjacent edges.Alternatively, since the cube is symmetric, the solution would be the same regardless of which space diagonal we choose. Therefore, the locus of points is this hexagon.But let me check if there are any other points outside this hexagon that might satisfy the condition. Suppose a point is on an edge or a vertex of the cube. For example, the midpoint of an edge is part of the hexagon we found. What about a vertex? Take the vertex (a,0,0). Distance to (0,0,0) is a. Distance to (a,a,a) is √[(0)^2 + (a)^2 + (a)^2] = √(2a²) = a√2. Not equal. So vertices are not part of the solution.What about the center of a face? Take the center of the face x=a, which is (a, a/2, a/2). Distance to (0,0,0): √(a² + (a/2)^2 + (a/2)^2) = √(a² + a²/4 + a²/4) = √(1.5a²) = a√(3/2). Distance to (a,a,a): √[(0)^2 + (a/2)^2 + (a/2)^2] = √(a²/4 + a²/4) = √(a²/2) = a/√2. These are not equal. So the center of a face is not part of the solution.Therefore, only the points on the hexagon described earlier satisfy the equidistant condition.In summary, the set of points on the cube's surface equidistant from the endpoints of a space diagonal form a regular hexagon, each of whose vertices is the midpoint of an edge of the cube. This hexagon lies on the plane x + y + z = 3a/2 and intersects each face of the cube along a line segment connecting the midpoints of two adjacent edges.

✅:The points on the surface of a cube that are equidistant from the endpoints of one of its space diagonals form a regular hexagon centered within the cube. Each vertex of this hexagon is the midpoint of an edge of the cube. Here's the detailed breakdown:1. Coordinate Setup: Consider a cube with side length ( a ), aligned with a coordinate system such that one endpoint of the space diagonal is at ( (0,0,0) ) and the opposite endpoint is at ( (a,a,a) ).2. Equidistant Condition: A point ( (x,y,z) ) on the cube's surface satisfies: [ sqrt{x^2 + y^2 + z^2} = sqrt{(x-a)^2 + (y-a)^2 + (z-a)^2}. ] Squaring and simplifying yields the plane equation: [ x + y + z = frac{3a}{2}. ]3. Intersection with Cube Faces: The plane intersects the cube’s surface on six faces, forming line segments. These segments connect midpoints of edges: - On ( x = 0 ): Line from ( (0, a, frac{a}{2}) ) to ( (0, frac{a}{2}, a) ). - On ( y = 0 ): Line from ( (a, 0, frac{a}{2}) ) to ( (frac{a}{2}, 0, a) ). - On ( z = 0 ): Line from ( (a, frac{a}{2}, 0) ) to ( (frac{a}{2}, a, 0) ). - On ( x = a ): Line from ( (a, 0, frac{a}{2}) ) to ( (a, frac{a}{2}, 0) ). - On ( y = a ): Line from ( (0, a, frac{a}{2}) ) to ( (frac{a}{2}, a, 0) ). - On ( z = a ): Line from ( (0, frac{a}{2}, a) ) to ( (frac{a}{2}, 0, a) ).4. Hexagon Formation: Connecting these midpoints forms a regular hexagon. Each vertex is equidistant from both ends of the space diagonal, verified by calculating distances (e.g., midpoint ( (a, 0, frac{a}{2}) ) has equal distance ( frac{asqrt{5}}{2} ) to ( (0,0,0) ) and ( (a,a,a) )).Final Answer: The equidistant points form a regular hexagon on the cube’s surface, with vertices at the midpoints of its edges. For a cube with side length ( a ), these vertices are: [boxed{left{(a, 0, frac{a}{2}), (a, frac{a}{2}, 0), (frac{a}{2}, a, 0), (0, a, frac{a}{2}), (0, frac{a}{2}, a), (frac{a}{2}, 0, a)right}}]

🔑:1. Identify the Points: We need to find the points on the surface of a cube that are equidistant from the endpoints of one of its space diagonals. Let the cube have side length a.2. Space Diagonal Description: Any cube has space diagonals that connect opposite corners of the cube, passing through its center.3. Midpoint and Plane: Consider one space diagonal of the cube and the midpoint of this diagonal. The points we are looking for lie on the plane that is perpendicular to this space diagonal and passes through its midpoint. This plane will divide the cube into two equal halves.4. Plane Intersection with the Cube: The plane perpendicular to the space diagonal at its midpoint will slice through the cube, intersecting its surface. This intersection includes certain edges and vertices of the cube.5. Geometry of the Plane Intersection: Each face of the cube that the plane intersects will have a line segment cut out that is the intersection of this plane with the face. Since the plane is through the midpoint of the space diagonal, it also cuts through the midpoint of each side of the cube that is perpendicular to this space diagonal.6. Length of the Intersection Segments: Each of these intersection segments on a face of the cube has length frac{a sqrt{2}}{2} because they go from the midpoint of one edge to the midpoint of its opposite edge on the face of a cube.7. Resulting Shape: Because the plane cuts through the midlines of each face of the cube (which are equidistant), the resulting intersection on the surface of the cube is a hexagon. Specifically, it is a regular hexagon, as each segment created by the intersection of the plane with the cube’s faces is of equal length and each internal angle is 120^circ.8. Properties of the Hexagon: The distance from each vertex of the hexagon to the center of the cube (which is also the center of the hexagon) is frac{a sqrt{2}}{2}. Conclusion:The points on the surface of the cube that are equidistant from the endpoints of a space diagonal form a regular hexagon with side length frac{a sqrt{2}}{2}. boxed{text{The cube’s surface has a regular hexagon formed by these points.}}

🧠:Alright, so I need to prove the trigonometric identity: sin2α + sin2β + sin2γ = 4 sinα sinβ sinγ. Hmm, okay. Let me think about how to approach this. I remember there are some trigonometric identities that relate sums of sines to products. Maybe I can use those?First, let me recall the sum-to-product formulas. The sum of sines formula is sinA + sinB = 2 sin((A+B)/2) cos((A-B)/2). Maybe I can apply this to pairs of the terms here. Let's see. If I take sin2α + sin2β first, applying the formula would give 2 sin(α + β) cos(α - β). Then, add sin2γ to that result. So the left-hand side (LHS) becomes 2 sin(α + β) cos(α - β) + sin2γ. Hmm, not sure if that helps directly. Maybe there's another way.Wait, the right-hand side (RHS) is 4 sinα sinβ sinγ. That makes me think of product-to-sum identities. The product of sines can be converted into sums. Let me recall that identity. The product sinA sinB = [cos(A-B) - cos(A+B)] / 2. So applying that to sinα sinβ first would give [cos(α - β) - cos(α + β)] / 2. Then multiplying by sinγ would be [cos(α - β) - cos(α + β)] / 2 * sinγ. Maybe expand that further using product-to-sum again. Let's see:So, 4 sinα sinβ sinγ = 4 * [ (cos(α - β) - cos(α + β)) / 2 ] * sinγ = 2 [cos(α - β) sinγ - cos(α + β) sinγ]. Then, apply the identity sinA cosB = [sin(A + B) + sin(A - B)] / 2 to each term.First term: 2 * [ cos(α - β) sinγ ] becomes 2 * [ (sin(γ + α - β) + sin(γ - α + β)) / 2 ] = sin(γ + α - β) + sin(γ - α + β).Second term: -2 * [ cos(α + β) sinγ ] becomes -2 * [ (sin(γ + α + β) + sin(γ - α - β)) / 2 ] = - sin(γ + α + β) - sin(γ - α - β).So combining these, RHS becomes sin(γ + α - β) + sin(γ - α + β) - sin(γ + α + β) - sin(γ - α - β). Hmm, this seems complicated. Maybe I should check if there's a condition on α, β, γ? The original problem didn't specify any, but sometimes these identities hold under certain conditions. For example, in a triangle, α + β + γ = π. Wait, if that's the case, then maybe the identity holds when α + β + γ = π. Let me check if that's necessary.Suppose α + β + γ = π. Let's see. Then, γ = π - α - β. Substitute into the RHS: 4 sinα sinβ sin(π - α - β) = 4 sinα sinβ sin(α + β), since sin(π - x) = sinx. So RHS becomes 4 sinα sinβ sin(α + β). Let's compute the LHS under this condition.LHS: sin2α + sin2β + sin2γ. Substitute γ = π - α - β. Then sin2γ = sin[2(π - α - β)] = sin(2π - 2α - 2β) = -sin(2α + 2β), since sin(2π - x) = -sinx. So LHS becomes sin2α + sin2β - sin(2α + 2β).Hmm, can I simplify this? Let's compute sin2α + sin2β - sin(2α + 2β). Let me apply the sum-to-product identity on sin2α + sin2β first: 2 sin(α + β) cos(α - β). Then subtract sin(2α + 2β) which is 2 sin(α + β) cos(α + β). So the expression becomes 2 sin(α + β) cos(α - β) - 2 sin(α + β) cos(α + β) = 2 sin(α + β) [cos(α - β) - cos(α + β)].Now, using the identity cosA - cosB = -2 sin[(A + B)/2] sin[(A - B)/2]. Let's apply this: cos(α - β) - cos(α + β) = -2 sin[ ( (α - β) + (α + β) ) / 2 ] sin[ ( (α - β) - (α + β) ) / 2 ].Simplify the arguments: first term inside sin is (2α)/2 = α. Second term inside sin is (-2β)/2 = -β. So this becomes -2 sinα sin(-β) = -2 sinα (-sinβ) = 2 sinα sinβ. Therefore, cos(α - β) - cos(α + β) = 2 sinα sinβ.Therefore, the expression becomes 2 sin(α + β) * 2 sinα sinβ = 4 sinα sinβ sin(α + β), which is exactly equal to the RHS when γ = π - α - β. So, under the condition that α + β + γ = π (i.e., the angles are those of a triangle), the identity holds.But wait, the original problem didn't specify that α, β, γ are angles of a triangle. So is the identity only valid under that condition? Let me test with some values. Suppose α = β = γ = π/3, so they add up to π. Then LHS: sin(2π/3) + sin(2π/3) + sin(2π/3) = 3*(√3/2) = 3√3/2 ≈ 2.598. RHS: 4 sin(π/3)^3 = 4*(√3/2)^3 = 4*(3√3/8) = 3√3/2 ≈ 2.598. So it works here.Now, let's try angles that don't add up to π. Let's take α = β = γ = π/4. Then α + β + γ = 3π/4 ≠ π. LHS: sin(π/2) + sin(π/2) + sin(π/2) = 3*1 = 3. RHS: 4*(sinπ/4)^3 = 4*(√2/2)^3 = 4*(2√2/8) = √2 ≈ 1.414. These are not equal. So the identity doesn't hold here, which means the identity is only valid when α + β + γ = π. Therefore, the original problem probably assumes that α, β, γ are angles of a triangle, even though it wasn't stated. Maybe I should mention that in the solution.So, to summarize, the identity holds if α + β + γ = π. Then, using that condition, we can substitute γ = π - α - β into the equation and show both sides are equal through trigonometric identities. Alternatively, perhaps there's a more general proof without assuming the angle sum, but given the counterexample when the sum isn't π, it's necessary.Alternatively, maybe there's another approach using complex exponentials or other identities. Let me think. For example, using the identity for sin2α + sin2β + sin2γ. Alternatively, express everything in terms of products. Let me try another method.Using the identity: sinA + sinB + sinC - 4 sin(A/2) sin(B/2) sin(C/2) = 0 when A + B + C = 2π. Wait, not sure. Maybe not directly applicable here.Wait, another idea: If α + β + γ = π, then 2α + 2β + 2γ = 2π. So maybe setting A = 2α, B = 2β, C = 2γ, then A + B + C = 2π. Then, perhaps using the identity for the sum of sines when their sum is 2π. Let me check a known identity.There's an identity that states that if A + B + C = 2π, then sinA + sinB + sinC = 4 sin(A/2) sin(B/2) sin(C/2). Let's verify that. Let A = 2α, B = 2β, C = 2γ, so A + B + C = 2π, so 2α + 2β + 2γ = 2π ⇒ α + β + γ = π. Then, sinA + sinB + sinC = 4 sin(A/2) sin(B/2) sin(C/2) ⇒ sin2α + sin2β + sin2γ = 4 sinα sinβ sinγ. Which is exactly the identity we need! Therefore, this identity is a specific case of the more general identity when A + B + C = 2π.Therefore, the key here is recognizing that the given identity is a form of the known identity for three angles summing to 2π, with each angle being double of α, β, γ. Hence, the condition α + β + γ = π is necessary.So, the conclusion is that the identity holds if and only if α + β + γ = π, and the proof follows from the general identity for angles summing to 2π. Alternatively, we can prove it directly using sum-to-product formulas as I did earlier when substituting γ = π - α - β.Therefore, the step-by-step proof would involve stating the condition α + β + γ = π, then expressing one angle in terms of the others, substituting into the equation, and simplifying using trigonometric identities to show both sides are equal.Alternatively, using the identity for A + B + C = 2π: sinA + sinB + sinC = 4 sin(A/2) sin(B/2) sin(C/2), substituting A = 2α, B = 2β, C = 2γ, which gives the desired identity.Hence, the identity is proven under the condition that α + β + γ = π. The original problem might have omitted this condition, but it's necessary for the identity to hold.

✅:The given trigonometric identity (sin 2alpha + sin 2beta + sin 2gamma = 4 sin alpha sin beta sin gamma) holds under the condition that (alpha + beta + gamma = pi) (i.e., the angles are those of a triangle). Here's the step-by-step proof:Step 1: Assume (alpha + beta + gamma = pi). Express (gamma) in terms of (alpha) and (beta): [gamma = pi - alpha - beta.]Step 2: Substitute (gamma) into the left-hand side (LHS): [sin 2alpha + sin 2beta + sin 2gamma = sin 2alpha + sin 2beta + sin[2(pi - alpha - beta)].]Simplify (sin 2gamma): [sin[2pi - 2alpha - 2beta] = -sin(2alpha + 2beta) quad (text{since } sin(2pi - x) = -sin x).]Thus, the LHS becomes: [sin 2alpha + sin 2beta - sin(2alpha + 2beta).]Step 3: Simplify using sum-to-product identities: First, combine (sin 2alpha + sin 2beta): [sin 2alpha + sin 2beta = 2sin(alpha + beta)cos(alpha - beta).]Subtract (sin(2alpha + 2beta)): [2sin(alpha + beta)cos(alpha - beta) - sin(2alpha + 2beta).]Note that (sin(2alpha + 2beta) = 2sin(alpha + beta)cos(alpha + beta)). Substitute this: [2sin(alpha + beta)cos(alpha - beta) - 2sin(alpha + beta)cos(alpha + beta).]Factor out (2sin(alpha + beta)): [2sin(alpha + beta)left[cos(alpha - beta) - cos(alpha + beta)right].]Step 4: Simplify the cosine difference: Use the identity (cos A - cos B = -2sinleft(frac{A+B}{2}right)sinleft(frac{A-B}{2}right)): [cos(alpha - beta) - cos(alpha + beta) = -2sinalphasin(-beta) = 2sinalphasinbeta.]Substitute back: [2sin(alpha + beta) cdot 2sinalphasinbeta = 4sinalphasinbetasin(alpha + beta).]Step 5: Relate (sin(alpha + beta)) to (singamma): Since (alpha + beta = pi - gamma), we have: [sin(alpha + beta) = sin(pi - gamma) = singamma.]Thus, the LHS simplifies to: [4sinalphasinbetasingamma,]which matches the right-hand side (RHS).Conclusion: The identity (sin 2alpha + sin 2beta + sin 2gamma = 4 sin alpha sin beta sin gamma) is valid when (alpha + beta + gamma = pi). This condition is essential for the equality to hold, as demonstrated through substitution and trigonometric identities.[boxed{sin 2alpha + sin 2beta + sin 2gamma = 4 sin alpha sin beta sin gamma quad text{when} quad alpha + beta + gamma = pi}]

🔑:To prove the given trigonometric identity:[sin 2alpha + sin 2beta + sin 2gamma = 4 sin alpha sin beta sin gamma]we follow these steps:1. Expand the Left-hand Side using Sum-to-Product Identities: We start by using the sum formulas for sine functions. The sum of two sine functions can be written as a product: [ sin 2alpha + sin 2beta = 2 sinleft(frac{2alpha + 2beta}{2}right) cosleft(frac{2alpha - 2beta}{2}right) ] Simplifying the arguments inside the trigonometric functions: [ sin 2alpha + sin 2beta = 2 sin(alpha + beta) cos(alpha - beta) ]2. Relate to (sin 2gamma): We are given (gamma) as part of the equation. Using the identity for the double angle of sine: [ sin 2gamma = 2 sin gamma cos gamma ] We need to express (sin 2 gamma) in a way that can be combined with the other terms. It can also be written to show the relationship with (alpha) and (beta). Notice, [ sin 2gamma = 2 sin gamma cos gamma = -2 sin gamma cos (alpha + beta) ] since (alpha + beta + gamma = pi).3. Combine the Expressions: Next, we see the identities and try to combine them. Adding up the terms we have: [ sin 2alpha + sin 2beta + sin 2gamma = 2 sin(alpha + beta) cos(alpha - beta) + 2 sin gamma cos gamma ]4. Use a Cosine Difference Identity: Recall the trigonometric identity for the difference of cosines: [ cos (alpha - beta) - cos (alpha + beta) = -2 sin alpha sin beta ] Therefore, [ 2 sin(alpha + beta) [cos (alpha - beta)] + 2 sin gamma cos [gamma] ]5. Relate the sum of Sines with a product of sines: Observe how the terms combine to give the right-hand side. We get: [ 2 sin alpha sin beta (cos (alpha - beta) + cos (alpha + beta)) = 4 sin alpha sin beta sin gamma ] Since (cos (alpha - beta) + cos (alpha + beta) = 2 sin alpha sin beta).Therefore, combining these together, we have:[sin 2alpha + sin 2beta + sin 2gamma = 4 sin alpha sin beta sin gamma](boxed{sin 2alpha + sin 2beta + sin 2gamma = 4 sin alpha sin beta sin gamma})

🧠:Okay, let me try to figure out this problem step by step. So we have a 2x2 square with numbers in each cell, all distinct. The conditions given are that the sum of the first row equals the sum of the second row, and the product of the first column equals the product of the second column. We need to prove that the total sum of all four numbers is zero. Hmm, interesting.First, let me visualize the grid. Let's denote the numbers as follows:First row: a, bSecond row: c, dSo the grid looks like:a bc dGiven that all numbers are pairwise distinct. So a ≠ b, a ≠ c, a ≠ d, b ≠ c, b ≠ d, c ≠ d.Now, the sum of the first row equals the sum of the second row. That gives us the equation:a + b = c + d ...(1)Then, the product of the first column equals the product of the second column. So:a * c = b * d ...(2)We need to prove that a + b + c + d = 0.Okay, let's start with equation (1): a + b = c + d. If I rearrange this, it becomes a + b - c - d = 0. So that's the same as (a + b + c + d) - 2(c + d) = 0, but not sure if that helps yet. Alternatively, maybe express one variable in terms of others. For example, from equation (1), we can write c = a + b - d. Then substitute this into equation (2). Let's try that.Substituting c = a + b - d into equation (2):a * (a + b - d) = b * dExpanding the left side:a² + a b - a d = b dLet's bring all terms to one side:a² + a b - a d - b d = 0Factor terms where possible. Let's see:a² + a b - d(a + b) = 0Hmm, maybe factor a from the first two terms:a(a + b) - d(a + b) = 0Oh, nice! Then factor out (a + b):(a + b)(a - d) = 0So this product equals zero, which means either a + b = 0 or a - d = 0.But wait, the problem states that all numbers are pairwise distinct. So a ≠ d, so a - d ≠ 0. Therefore, the only possibility is that a + b = 0.But from equation (1), a + b = c + d. So if a + b = 0, then c + d = 0 as well. Therefore, the total sum a + b + c + d = 0 + 0 = 0. Which is exactly what we needed to prove. Wait, so is that it? That seems straightforward. But let me check again to make sure I didn't skip any steps or make a mistake in the algebra.Starting again:From equation (1): a + b = c + d.From equation (2): a c = b d.Substituted c = a + b - d into equation (2):a(a + b - d) = b dExpanding: a² + a b - a d = b dThen, moving all terms to left:a² + a b - a d - b d = 0Factoring: a(a + b) - d(a + b) = 0Which is (a + b)(a - d) = 0.Since a ≠ d, a - d ≠ 0, so a + b = 0. Then c + d = a + b = 0. Therefore, total sum is 0.Wait, but the problem states all numbers are pairwise distinct. So if a + b = 0, then b = -a. But does that introduce any contradictions with the distinctness? Let's see. Suppose a and b are negatives of each other. Then c + d = 0 as well, so d = -c. So the grid would look like:a -ac -cBut then, all numbers must be distinct. So a ≠ -a implies a ≠ 0. Similarly, c ≠ -c implies c ≠ 0. Also, a ≠ c, a ≠ -c, -a ≠ c, -a ≠ -c. Since all numbers are distinct, this holds as long as a ≠ c and a ≠ -c, and c ≠ 0, a ≠ 0.But the problem doesn't state that the numbers can't be negatives of each other, just that they are pairwise distinct. So this seems okay. For example, a = 1, b = -1, c = 2, d = -2. Then all are distinct, and the sums of rows are 0, and products of columns are 1*2 = 2 and (-1)*(-2) = 2. So that works. The total sum is 1 + (-1) + 2 + (-2) = 0. So that's a valid example.But is this the only case? Let me check if there are other possibilities where the total sum isn't zero, but the conditions are still met. Wait, according to the algebra above, if the conditions hold, then a + b must be zero, so total sum must be zero. So regardless of the values, as long as they meet the given conditions, the total sum is zero. Therefore, the proof is valid.But let me test another example to be sure. Suppose a = 2, b = -2, c = 3, d = -3. Then rows sum to 0, columns products: 2*3 = 6 and (-2)*(-3) = 6. Total sum is 0. All numbers are distinct. So that works.Another example with different numbers. Let's say a = 3, b = -1, then sum of the first row is 2. Then c + d must be 2. Let's pick c = 4, d = -2. Now check column products: 3*4 = 12 and (-1)*(-2) = 2. But 12 ≠ 2. So this doesn't satisfy the column product condition. Therefore, such a grid wouldn't work. Therefore, the only way to satisfy both conditions is when a + b = 0 and c + d = 0, leading to total sum zero.Therefore, the conclusion is that the sum of all four numbers must be zero. Hence proved.Wait, but in my first substitution, I assumed c = a + b - d. Is that valid? Because from equation (1), c = (a + b) - d, yes. So that substitution is correct.Therefore, the key step is factoring the equation to (a + b)(a - d) = 0, which gives a + b = 0, since a ≠ d. Then, c + d = a + b = 0, so total sum is zero. Therefore, the proof is complete.I think that's solid. Let me check once more for any hidden assumptions or possible mistakes. For example, could there be a case where even if a + b = 0, the products of the columns don't equal? But in the example above, when a + b = 0 and c + d = 0, then the column products are a*c and b*d. Since b = -a and d = -c, then b*d = (-a)*(-c) = a*c, so they are equal. Therefore, as long as a + b = 0 and c + d = 0, the column products are equal. Therefore, the conditions are satisfied.Therefore, the only solutions are those where a + b = 0 and c + d = 0, hence the total sum is zero, and the numbers are distinct as long as a, c are non-zero and a ≠ c, a ≠ -c, etc.Therefore, the proof is correct.

✅:To prove that the sum of all four numbers in the 2 times 2 grid is zero under the given conditions, let the grid entries be a, b (first row) and c, d (second row). The conditions are:1. Row sums are equal: a + b = c + d tag{1}2. Column products are equal: ac = bd tag{2}Step 1: From equation (1), express c in terms of a, b, d: c = a + b - d.Step 2: Substitute c into equation (2): a(a + b - d) = bd. Expanding and rearranging: a^2 + ab - ad - bd = 0. Factor: a(a + b) - d(a + b) = 0 implies (a + b)(a - d) = 0.Step 3: Since all numbers are distinct, a neq d, so a - d neq 0. Thus: a + b = 0. From equation (1), this implies: c + d = 0.Step 4: Summing all four numbers: a + b + c + d = (a + b) + (c + d) = 0 + 0 = 0.Conclusion: The total sum of the four numbers is zero. This satisfies the problem’s requirements, as distinctness is maintained (e.g., a neq -a, c neq -c, etc.), and the column product equality holds because ac = (-a)(-c) = bd.boxed{0}

🔑:Let's denote the four numbers in the (2 times 2) square as follows:[begin{array}{|c|c|}hlinea & b hlinec & d hlineend{array}]We are given two conditions:1. The sum of the numbers in the first row is equal to the sum of the numbers in the second row:[a + b = c + d]2. The product of the numbers in the first column is equal to the product of the numbers in the second column:[a c = b d] Step-by-Step Solution:1. Start with the equations given: [ begin{cases} a + b = c + d a c = b d end{cases} ]2. From the first equation, solve for one of the variables. Let's solve for (c): [ c = a + b - d ]3. Substitute (c = a + b - d) into the second equation (a c = b d): [ a (a + b - d) = b d ]4. Distribute (a) on the left-hand side: [ a^2 + a b - a d = b d ]5. Rearrange the equation to gather all terms involving (d) on one side: [ a^2 + a b = a d + b d ]6. Factor (d) out of the right-hand side: [ a^2 + a b = d(a + b) ]7. Since (a) and (d) are distinct (all numbers are distinct), divide both sides by (a + b): [ a = d ]8. Given that (a) and (d) are distinct, the only solution to this equation when (a) and (d) are equal and non-zero is if: [ a + b = 0 ]9. Substitute (a + b = 0) into (c + d = 0) from the first equation, yielding: [ c + d = 0 ]10. Therefore, the sum of all four numbers is: [ a + b + c + d = 0 + 0 = 0 ] ConclusionThus, we have proven that the sum of all four numbers in the (2 times 2) square is zero:[boxed{0}]

🧠:Okay, so I have this problem here where I need to prove that there's a constant a > 1, which doesn't depend on m or n, such that if m/n is less than sqrt(7), then 7 - (m²/n²) is at least a/n². And then I also need to find the maximum possible value of a. Alright, let's try to unpack this step by step.First, let me restate the problem in my own words. Given two positive integers m and n, if their ratio m/n is less than the square root of 7, then the difference between 7 and the square of their ratio (m²/n²) should be greater than or equal to some constant a divided by n squared. The goal is to show that such a constant a exists and find the largest possible a that works for all m and n.Hmm. So, essentially, we need to bound how close m/n can get to sqrt(7) from below, and ensure that the difference 7 - (m/n)² is always at least a/n². The key here is that a has to be independent of m and n, so it's a universal constant for all positive integers m and n satisfying m/n < sqrt(7).I remember that when dealing with approximations of irrational numbers by rationals, there's something called Dirichlet's approximation theorem or maybe results from continued fractions that give bounds on how close rationals can get to irrationals. But I'm not sure if that's directly applicable here. Maybe this problem is related to the theory of Diophantine approximations?Alternatively, maybe I can approach this algebraically. Let's let x = m/n. Then the condition is x < sqrt(7), and we need to show that 7 - x² >= a/n². Rearranging, this would mean that (sqrt(7) - x)(sqrt(7) + x) >= a/n². Since x < sqrt(7), both (sqrt(7) - x) and (sqrt(7) + x) are positive. So, the product of these two terms is 7 - x².But we need to bound 7 - x² from below. Since x is close to sqrt(7), (sqrt(7) - x) will be small, but (sqrt(7) + x) will be approximately 2*sqrt(7). So, maybe we can approximate 7 - x² ≈ 2*sqrt(7)*(sqrt(7) - x). Therefore, if we can bound (sqrt(7) - x) from below, then multiplying by 2*sqrt(7) would give us a lower bound for 7 - x².But x is m/n, so (sqrt(7) - m/n) needs to be bounded below by something. If we can find a lower bound for sqrt(7) - m/n, then multiplying by 2*sqrt(7) would give us the required a/n².Wait, but how do we find such a lower bound? For any irrational number, like sqrt(7), the difference between the irrational and a rational number m/n can be made arbitrarily small by choosing m/n close enough. However, the trade-off is that n has to be large. There's a theorem by Liouville that says algebraic numbers can't be approximated too closely by rationals, but sqrt(7) is algebraic of degree 2, so maybe Liouville's theorem gives a bound here.Liouville's theorem states that for an irrational algebraic number α of degree d, there exists a constant c such that |α - m/n| > c/n^d for all integers m, n with n > 0. In our case, sqrt(7) is degree 2, so d=2. Therefore, there exists a constant c such that |sqrt(7) - m/n| > c/n² for all m, n. Then, using that, we can write sqrt(7) - m/n > c/n² (since m/n < sqrt(7)), so multiplying by (sqrt(7) + m/n) gives 7 - (m/n)^2 > c/n²*(sqrt(7) + m/n). But (sqrt(7) + m/n) is at least sqrt(7) (since m/n is positive), so we get 7 - (m/n)^2 > c*sqrt(7)/n². Therefore, setting a = c*sqrt(7), we would have the required inequality.But Liouville's theorem gives a specific constant c. For Liouville's theorem, the constant c is typically 1/(d+1), but maybe for quadratic irrationals, there is a better constant. Wait, actually, Liouville's theorem's constant depends on the specific algebraic number. Let me check. The general form of Liouville's theorem is |α - m/n| > c/n^{d}, where c = 1/(H(α) + 1), but I'm not sure. Maybe for quadratic irrationals, we can get a better bound. Wait, maybe I need to compute the exact constant here.Alternatively, perhaps I can approach this problem using continued fractions. Since sqrt(7) is a quadratic irrational, its continued fraction is periodic, so maybe the convergents of the continued fraction can be used to find the minimal distance between sqrt(7) and rationals m/n, thereby giving the maximal a.Alternatively, maybe this is more straightforward. Let's suppose that we need to find the minimal value of n²(7 - (m²/n²)) over all positive integers m, n with m/n < sqrt(7). Then, a would be the infimum of these values. Since m and n are integers, n²(7 - (m²/n²)) = 7n² - m². So, a is the minimal value of 7n² - m² over all m, n with m < n*sqrt(7). Therefore, the maximum a is the minimal such 7n² - m².Therefore, we need to find the minimal positive value of 7n² - m² where m, n are positive integers with m < n*sqrt(7). Then, a would be that minimal value. So, the problem reduces to finding the minimal positive integer value of 7n² - m². Since 7n² - m² must be positive, m must be the closest integer less than n*sqrt(7).Therefore, for each n, let m_n be the integer part of n*sqrt(7), i.e., m_n = floor(n*sqrt(7)). Then, compute 7n² - m_n² for each n, and find the minimal such value. The minimal value across all n would be the maximal a.But since sqrt(7) is irrational, the fractional parts of n*sqrt(7) are dense in [0,1), so the differences (sqrt(7) - m_n/n) can be made arbitrarily small, but the corresponding 7n² - m² might not necessarily get arbitrarily small. Wait, but 7n² - m² = (sqrt(7)n - m)(sqrt(7)n + m). If sqrt(7)n - m is approximately 1/(sqrt(7)n + m), then the product would be approximately (1/(sqrt(7)n + m))*(2 sqrt(7) n) ) ≈ 2 sqrt(7) n / (2 sqrt(7) n) ) = 1. But this is too hand-wavy.Alternatively, let's compute 7n² - m² for small n and see if we can spot a pattern or find the minimal value.Let's start with n=1: m must be less than sqrt(7) ≈ 2.645, so m=2. Then 7(1)^2 - 2^2 = 7 -4=3.n=2: m < 2*sqrt(7) ≈5.291, so m=5. 7(4) -25=28-25=3.n=3: m <3*sqrt(7)≈7.937, so m=7. 7*9 -49=63-49=14.n=4: m <4*sqrt(7)≈10.583, so m=10. 7*16 -100=112-100=12.n=5: m <5*sqrt(7)≈13.228, so m=13. 7*25 -169=175-169=6.n=6: m <6*sqrt(7)≈15.874, so m=15. 7*36 -225=252-225=27.n=7: m <7*sqrt(7)≈18.520, so m=18. 7*49 -324=343-324=19.n=8: m <8*sqrt(7)≈21.166, so m=21. 7*64 -441=448-441=7.n=9: m <9*sqrt(7)≈23.811, so m=23. 7*81 -529=567-529=38.n=10: m <10*sqrt(7)≈26.458, so m=26. 7*100 -676=700-676=24.Hmm, so for n=1,2 we get 3, n=3 gives 14, n=4 gives 12, n=5 gives 6, n=6 gives 27, n=7 gives 19, n=8 gives 7, n=9 gives 38, n=10 gives 24. The minimal value so far is 3 for n=1,2 and 6 for n=5, 7 for n=8. Wait, but 3 is the smallest. Wait, but maybe for larger n, this value can get even smaller?Wait, but maybe there's a pattern here. Let's check n=11: m <11*sqrt(7)≈29.103, so m=29. 7*121 -841=847-841=6. So here, 6 again. Hmm, n=11 gives 6.n=12: m <12*sqrt(7)≈31.749, m=31. 7*144 -961=1008-961=47.n=13: m <13*sqrt(7)≈34.395, m=34. 7*169 -1156=1183-1156=27.n=14: m <14*sqrt(7)≈37.040, m=37. 7*196 -1369=1372-1369=3.Wait, n=14 gives 3 again. So at n=14, m=37. Let's check 37/14≈2.642, which is less than sqrt(7)≈2.645. So 37²=1369, 7*(14)^2=1372. So 1372-1369=3. So that's the same as n=1,2,14. So 3 is recurring.Wait, but n=5 and n=11 gave 6, n=8 gave 7, n=14 gave 3. So perhaps 3 is the minimal possible value? But then, when n=1, m=2, 7 -4=3, so 3/1²=3. So a=3. But wait, the problem states that a>1. So 3 is greater than 1. So maybe the maximum possible a is 3? But we need to confirm that 7 - m²/n² is always at least 3/n² for all m,n with m/n <sqrt(7). But wait, in n=5, we had 6/n², which is 6/25=0.24, and 3/n² would be 3/25=0.12. So 6/n² is larger than 3/n², so that's okay. But the minimal value across all n is 3, so 3 is the minimal such that 7 - m²/n² >=3/n². Therefore, the maximum a is 3.But wait, hold on. When n=14, 7n² -m²=3, so 7 - (m/n)^2=3/n², so that equality holds. Similarly for n=1, 7 -4=3=3/1². So 3 is achieved for n=1 and n=14. So if we take a=3, then 7 -m²/n² >=3/n² for all m,n with m/n <sqrt(7). But wait, for other n, like n=5, the difference is 6/n² which is greater than 3/n², so 3 is indeed the minimal possible a. So a=3 is the maximum value such that the inequality holds. Therefore, the answer is 3.But let me confirm this. Suppose there exists some m,n where 7n² -m² <3. But according to our calculations, the minimal value is 3. Let me check n=4: 12, which is 12/16=0.75. If we set a=3, then 3/16=0.1875, so 0.75 >=0.1875, which holds. For n=3, 14/9≈1.555, which is larger than 3/9≈0.333. So it seems that 3 is indeed the minimal such value.But perhaps there's a case where 7n² -m² is between 1 and 3? For example, if n= something larger. Let's test n=15: m <15*sqrt(7)≈39.686, so m=39. 7*225 -39²=1575-1521=54. 54/225=0.24.n=16: m <16*sqrt(7)≈42.332, m=42. 7*256 -1764=1792-1764=28. 28/256≈0.109, which is larger than 3/256≈0.0117. So 0.109>0.0117.Wait, but if we have n=1, 7n² -m²=3. For n=2, same. For n=14, same. So maybe 3 is indeed the minimal possible value. Hence, the maximum a is 3.But to be thorough, let's check n=7. For n=7, m=18. 7*49 - 324=343-324=19. 19/49≈0.387>3/49≈0.061.Another way to think: suppose that 7n² -m² >=3 for all m,n with m/n <sqrt(7). If this is true, then a=3 is the maximal possible. If there exists some m,n where 7n² -m² <3, then a must be smaller. But according to our earlier trials, the minimal value achieved is 3. So unless there is some m,n with 7n² -m² <3, which we haven't found yet, 3 is the answer.But perhaps we can look for solutions to 7n² -m² =1 or 2. Let's see. Suppose 7n² -m² =k, where k is 1 or 2. For k=1, 7n² -m²=1. Then, m²=7n² -1. Let's see if this is possible.Check small n:n=1: m²=7-1=6, not a square.n=2: m²=28-1=27, not a square.n=3: m²=63-1=62, not a square.n=4: 112-1=111, not a square.n=5: 175-1=174, no.n=6: 252-1=251, nope.n=7: 343-1=342, no.n=8: 448-1=447, no.n=9: 567-1=566, no.n=10: 700-1=699, nope.Similarly, k=2: 7n² -m²=2.So m²=7n² -2.n=1: 7-2=5, not a square.n=2: 28-2=26, no.n=3:63-2=61, no.n=4:112-2=110, nope.n=5:175-2=173, nope.n=6:252-2=250, no.n=7:343-2=341, no.n=8:448-2=446, no.n=9:567-2=565, no.n=10:700-2=698, nope.So it seems that there are no solutions for k=1 or 2. Therefore, the minimal k is 3. Thus, the maximum a is 3.Alternatively, maybe there's a theoretical reason why 7n² -m² cannot be 1 or 2. Let's consider modulo 7. For 7n² -m² =k, rearranged as m² ≡ -k mod7. The quadratic residues modulo 7 are 0,1,2,4. So -k mod7 must be a quadratic residue.For k=1: m² ≡-1 mod7. -1 mod7 is 6. 6 is not a quadratic residue modulo 7 (since the residues are 0,1,2,4). Therefore, no solution.For k=2: m²≡-2 mod7. -2 mod7=5. 5 is not a quadratic residue modulo7. Therefore, no solution.For k=3: m²≡-3 mod7. -3 mod7=4. 4 is a quadratic residue (2²=4). Therefore, possible. For example, when n=1, m=2: m²=4≡4 mod7, and 7(1)^2 -4=3, so k=3. Similarly, when n=14, m=37: 37²=1369≡1369 mod7. 1369/7=195*7 +4, so 1369≡4 mod7. Therefore, 7n² -m²=3≡3 mod7.Therefore, only k≡3 mod7 can be possible when m²≡4 mod7. But since 3 is achievable, the minimal k is 3, hence the maximal a is 3.Therefore, the constant a=3 is the maximum value satisfying the given condition.Final AnswerThe maximum value of ( a ) is boxed{3}.

✅:Given that ( m ) and ( n ) are positive integers, we need to prove that there exists a constant ( a > 1 ) independent of ( m ) and ( n ) such that if ( frac{m}{n} < sqrt{7} ), then ( 7 - frac{m^2}{n^2} geq frac{a}{n^2} ), and determine the maximum value of ( a ).First, we recognize that the problem requires bounding the difference ( 7 - left( frac{m}{n} right)^2 ) from below. Let ( x = frac{m}{n} ). The condition ( x < sqrt{7} ) implies we need to show ( 7 - x^2 geq frac{a}{n^2} ). This can be rewritten as ( ( sqrt{7} - x )( sqrt{7} + x ) geq frac{a}{n^2} ). Since ( x ) is close to ( sqrt{7} ), ( sqrt{7} - x ) is small, and ( sqrt{7} + x ) is approximately ( 2sqrt{7} ).Using Liouville's theorem for quadratic irrationals, we know there exists a constant ( c ) such that ( | sqrt{7} - frac{m}{n} | > frac{c}{n^2} ). This leads to ( 7 - left( frac{m}{n} right)^2 geq frac{2sqrt{7}c}{n^2} ). However, we need to find the exact minimal value of ( 7n^2 - m^2 ) for positive integers ( m ) and ( n ).By examining small values of ( n ) and corresponding ( m ), we find:- For ( n = 1 ), ( m = 2 ): ( 7 cdot 1^2 - 2^2 = 3 )- For ( n = 2 ), ( m = 5 ): ( 7 cdot 2^2 - 5^2 = 3 )- For ( n = 14 ), ( m = 37 ): ( 7 cdot 14^2 - 37^2 = 3 )We observe that the minimal value of ( 7n^2 - m^2 ) is 3. Checking modulo 7, we confirm that ( 7n^2 - m^2 = 1 ) or ( 2 ) is impossible due to quadratic residues. Thus, the minimal value is indeed 3.Therefore, the maximum value of ( a ) is (boxed{3}).

🔑:1. Understanding the problem statement We need to prove the existence of a constant (a > 1) such that for any positive integers (m) and (n), when [ frac{m}{n} < sqrt{7} ] we have the inequality [ 7 - frac{m^2}{n^2} geq frac{a}{n^2} ]2. Starting the analysis We know that if ( frac{m}{n} < sqrt{7} ), then ( m^2 < 7n^2 ). Rearranging, we get: [ 7n^2 - m^2 > 0 ]3. Considering the values of ( m^2 pmod{7} ) Since ( m ) is an integer, ( m^2 ) can only be congruent to ( 0, 1, 2, ) or ( 4 mod 7 ). This can be written as: [ m^2 equiv 0, 1, 2, 4 pmod{7} ]4. Substituting in the main inequality Using the above congruences, we have: [ 7n^2 - m^2 equiv 0, 6, 5, 3 pmod{7} ] Here, (7n^2 - m^2) can be 0 (removed since it leads to a contradiction because then ( frac{m}{n} = sqrt{7}) which isn't allowed), or it can be 6, 5, or 3 modulo 7. Since ( 7n^2 - m^2 > 0 ), given the non-negativity and integer properties, we must have: [ 7n^2 - m^2 geq 3 ]5. Determining the constant ( a ) Thus, the inequality becomes: [ 7n^2 - m^2 geq 3 ] Dividing both sides by (n^2) results in: [ 7 - frac{m^2}{n^2} geq frac{3}{n^2} ] This aligns with our requirement to find an (a) such that: [ 7 - frac{m^2}{n^2} geq frac{a}{n^2} ] By comparison, the best value for (a) is: [ a = 3 ]6. Verification by specific example Let's verify it by choosing specific values for (n) and (m). Take ( n = 2 ) and ( m = 5 ), then: [ frac{m}{n} = frac{5}{2} = 2.5 ] which is less than ( sqrt{7} approx 2.64575 ). We check: [ 7 - frac{m^2}{n^2} = 7 - frac{25}{4} = frac{28}{4} - frac{25}{4} = frac{3}{4} ] and: [ frac{3}{n^2} = frac{3}{4} ] Therefore, ( 7 - frac{m^2}{n^2} = frac{3}{n^2} ). Conclusion: The maximum value of (a) such that ( 7 - frac{m^2}{n^2} geq frac{a}{n^2} ) for ( frac{m}{n} < sqrt{7} )(boxed{3})