💡Okay, so I need to prove that if m and n are relatively prime positive integers, then the polynomial x^m - y^n is irreducible over the complex numbers. Hmm, let me start by recalling what it means for a polynomial to be irreducible. A polynomial is irreducible if it cannot be factored into the product of two nonconstant polynomials. So, in this case, I need to show that x^m - y^n cannot be written as f(x,y) * g(x,y) where both f and g are nonconstant polynomials.Alright, let's assume the opposite for contradiction. Suppose that x^m - y^n is reducible. Then, there exist nonconstant polynomials f(x,y) and g(x,y) such that x^m - y^n = f(x,y) * g(x,y). Now, I need to find some properties or constraints on f and g that would lead to a contradiction.One approach I remember is evaluating the polynomial at specific points. Maybe if I substitute certain values for x and y, I can get some information about f and g. Let's try plugging in x = z^n and y = z^m for some complex number z. Then, the left-hand side becomes (z^n)^m - (z^m)^n = z^{mn} - z^{mn} = 0. So, 0 = f(z^n, z^m) * g(z^n, z^m). This means that either f(z^n, z^m) = 0 or g(z^n, z^m) = 0 for all z.But f and g are polynomials, and if a polynomial is zero for infinitely many points, it must be the zero polynomial. So, either f(z^n, z^m) is identically zero or g(z^n, z^m) is identically zero. Without loss of generality, let's assume f(z^n, z^m) = 0 for all z.Now, let's fix y and consider the equation as a polynomial in x. For a fixed y ≠ 0, there are m distinct complex numbers z such that z^m = y. So, for each of these z, f(z^n, y) = 0. This suggests that f(x, y) has m distinct roots in x for each fixed y. Therefore, the degree of f in x must be at least m.Similarly, if we fix x and consider the polynomial in y, for each fixed x ≠ 0, there are n distinct complex numbers w such that w^n = x. So, for each of these w, g(x, w^m) = 0. This implies that g(x, y) has n distinct roots in y for each fixed x, so the degree of g in y must be at least n.But wait, the original polynomial x^m - y^n has degree m in x and degree n in y. If f has degree at least m in x, and g has degree at least n in y, then their product would have degree at least m + n, which is higher than the degree of x^m - y^n. That seems like a problem because the degrees should add up when multiplying polynomials.Hmm, maybe I made a mistake there. Let me think again. The degree of the product f * g should be the sum of the degrees of f and g. But in this case, the degree of x^m - y^n is max(m, n), not m + n. So, perhaps I need to consider the degrees more carefully.Wait, actually, when considering the total degree, which is the sum of the degrees in each variable, the total degree of x^m - y^n is m + n. If f has degree m in x and g has degree n in y, then their product would have total degree m + n, which matches. But if f has higher degree in x or g has higher degree in y, that would cause the total degree to exceed m + n, which is not possible. So, f must have degree exactly m in x, and g must have degree exactly n in y.But then, if f has degree m in x, and g has degree n in y, and m and n are coprime, does that lead to a contradiction? Maybe I need to use the fact that m and n are coprime somewhere here.Another idea: perhaps consider the polynomial x^m - y^n as a polynomial in x with coefficients in the field of rational functions in y. Then, if it factors, it would factor as a product of polynomials in x with coefficients in that field. But since x^m - y^n is a binomial, maybe it's irreducible over that field.Wait, but I'm not sure about that approach. Maybe I should think about the roots of the polynomial. If x^m - y^n factors, then each factor would correspond to some roots of the equation x^m = y^n. But since m and n are coprime, the equation x^m = y^n has only trivial solutions where x and y are both zero, right? No, that's not true. For example, if m = 2 and n = 3, then x^2 = y^3 has solutions like x = t^3 and y = t^2 for some t.But in that case, the polynomial x^2 - y^3 can be factored as (x - y^{3/2})(x + y^{3/2}), but since we are working over the complex numbers, these factors are not polynomials unless the exponents are integers. Wait, actually, in the complex numbers, we can take roots, but the factors would involve fractional exponents, which are not polynomials. So, maybe that's why x^m - y^n is irreducible.But I need to formalize this. Let me think about the degrees again. Suppose f and g are polynomials such that f * g = x^m - y^n. Then, f and g must each have degrees dividing the total degree. But since m and n are coprime, the only way for the degrees to divide is if one of them is 1, but that would make the other one equal to x^m - y^n, which is not possible because both f and g are nonconstant.Wait, that might not be the right way to think about it. The total degree is m + n, but the degrees in x and y are m and n respectively. So, maybe I need to consider the degrees in each variable separately.Suppose f has degree a in x and b in y, and g has degree c in x and d in y. Then, we have a + c = m and b + d = n. Since m and n are coprime, the only way to write m as a sum of two nonnegative integers is if one of them is 0 and the other is m, or if they are both positive but their greatest common divisor divides m. But since m and n are coprime, I think this might lead to a contradiction.Wait, actually, if a and c are such that a + c = m, and b + d = n, and m and n are coprime, then perhaps one of a or c must be 0, but that would make f or g a constant polynomial, which contradicts the assumption that both are nonconstant. Hmm, that might be a way to go.Let me try to formalize this. Suppose f and g are nonconstant polynomials such that f * g = x^m - y^n. Then, looking at the degrees in x, we have deg_x(f) + deg_x(g) = m. Similarly, deg_y(f) + deg_y(g) = n. Since m and n are coprime, the only way to write m as a sum of two nonnegative integers is if one of them is 0 and the other is m, or if they are both positive but their greatest common divisor divides m. But since m and n are coprime, the only way for deg_x(f) and deg_x(g) to add up to m is if one of them is 0 and the other is m. Similarly for the degrees in y.But if deg_x(f) = 0, then f is a polynomial in y only, and similarly, if deg_x(g) = 0, then g is a polynomial in y only. But then, f * g would be a polynomial in y only, which cannot equal x^m - y^n unless f or g is a constant. But we assumed both are nonconstant, so this leads to a contradiction.Wait, that seems promising. Let me check again. If f and g are both nonconstant, then their degrees in x must both be positive, right? Because if one of them had degree 0 in x, it would be a polynomial in y only, and then the other would have degree m in x, making their product have degree m in x, which is fine. But then, looking at the degrees in y, we have deg_y(f) + deg_y(g) = n. Similarly, since n is coprime with m, the only way to write n as a sum of two nonnegative integers is if one is 0 and the other is n. But if f is a polynomial in y only, then deg_y(f) = something, and deg_y(g) would have to be n - deg_y(f). But if f is a polynomial in y only, then g would have to have degree 0 in y, meaning g is a polynomial in x only. But then, f is a polynomial in y only and g is a polynomial in x only, so their product f * g would be a polynomial in x and y with terms of the form x^a y^b, but x^m - y^n has only two terms, so this would require that f and g are monomials. But then, f * g would be a single term, which cannot equal x^m - y^n, which has two terms. Therefore, this is a contradiction.So, this suggests that our initial assumption that x^m - y^n is reducible is false, hence x^m - y^n is irreducible over the complex numbers when m and n are coprime.Wait, let me make sure I didn't miss anything. I assumed that if f and g are nonconstant, then one of them must have degree 0 in x or y, leading to a contradiction. But what if both f and g have positive degrees in both x and y? Is that possible?Suppose f has degree a in x and b in y, and g has degree c in x and d in y, with a + c = m and b + d = n. Since m and n are coprime, the only way for a and c to add up to m is if one of them is 0, but we just considered that case. Alternatively, if a and c are both positive, then their greatest common divisor must divide m, but since m and n are coprime, this might not necessarily lead to a contradiction. Hmm, maybe I need a different approach.Another idea: use the fact that x^m - y^n is a binomial, and binomials are often irreducible under certain conditions. Maybe I can use the Eisenstein's criterion, but that usually applies to polynomials in one variable. Alternatively, consider the polynomial ring in two variables and use some irreducibility criteria there.Wait, in the ring C[x,y], a polynomial is irreducible if it cannot be factored into the product of two nonconstant polynomials. So, if x^m - y^n factors, then it must factor into two polynomials, each of which is nonconstant in at least one variable.But earlier, I considered the case where one polynomial is in x only and the other is in y only, leading to a contradiction. What if both f and g have mixed terms, i.e., depend on both x and y? Is that possible?Suppose f and g are both polynomials in both x and y. Then, when we set x = z^n and y = z^m, we get f(z^n, z^m) * g(z^n, z^m) = 0. As before, this implies that either f(z^n, z^m) = 0 or g(z^n, z^m) = 0 for all z. Without loss of generality, assume f(z^n, z^m) = 0 for all z. Then, f(x,y) must vanish on the curve x = z^n, y = z^m, which is the parametrization of the curve x^m = y^n.But since m and n are coprime, this curve is smooth and irreducible. Therefore, f must be a power of the defining polynomial of this curve, which is x^m - y^n. But then, f would be a multiple of x^m - y^n, which would make g a constant polynomial, contradicting the assumption that both f and g are nonconstant.Therefore, this leads to a contradiction, so our initial assumption that x^m - y^n is reducible must be false. Hence, x^m - y^n is irreducible over the complex numbers when m and n are coprime.I think that covers it. By assuming reducibility, we derived that one of the factors must be a multiple of x^m - y^n, which would make the other factor constant, contradicting the nonconstant assumption. Therefore, x^m - y^n is irreducible.

💡First, identify the given number, which is 1.0.Next, set up the equation based on the problem statement:3 times the number minus the unknown value equals twice the number.Substitute the known number into the equation:3 * 1.0 - v = 2 * 1.0.Simplify the equation:3 - v = 2.Solve for the unknown value (v):v = 3 - 2.Finally, determine that the value to be subtracted is 1.

💡First, I need to determine the total number of ways to select 4 people out of the 7 available (4 boys and 3 girls). This can be calculated using the combination formula ( binom{7}{4} ), which equals 35.Next, I must exclude the scenarios where the selection consists of only boys or only girls to ensure that both genders are represented. For selections with only boys, there is only one way to choose all 4 boys, which is ( binom{4}{4} = 1 ).For selections with only girls, since there are only 3 girls, it's impossible to choose 4 girls. Therefore, the number of ways to choose only girls is 0.Finally, subtract the invalid selections from the total number of selections: ( 35 - 1 - 0 = 34 ).Thus, the total number of valid selections that include both boys and girls is 34.

💡Alright, so I have this problem here where I'm given a complex number z that satisfies the equation zi = 2 - i. And I need to figure out which of the options A, B, C, or D is correct. Let me try to break this down step by step.First, I need to find what z is. The equation is zi = 2 - i. Hmm, okay, so z multiplied by i equals 2 minus i. To solve for z, I should probably divide both sides by i. But wait, dividing by i can be tricky because it's a complex number. Maybe I can multiply both the numerator and the denominator by the complex conjugate of i to make it easier. The complex conjugate of i is -i, right?So, if I take the equation zi = 2 - i and divide both sides by i, it becomes z = (2 - i)/i. To simplify this, I'll multiply the numerator and the denominator by -i:z = (2 - i)/i * (-i)/(-i) = (2*(-i) - i*(-i)) / (i*(-i)).Let me compute the numerator first: 2*(-i) is -2i, and -i*(-i) is +i². But i² is equal to -1, so that becomes -2i + (-1) = -2i - 1.Now, the denominator: i*(-i) is -i², which is -(-1) = 1.So, putting it all together, z = (-2i - 1)/1 = -1 - 2i.Okay, so z is equal to -1 - 2i. Now, let's check each of the options one by one.**Option A: z² + 2z - 5 = 0**Let me compute z² first. z is -1 - 2i, so z squared is (-1 - 2i)². Expanding this, I get:(-1)² + 2*(-1)*(-2i) + (-2i)² = 1 + 4i + 4i².Since i² is -1, this becomes 1 + 4i - 4 = -3 + 4i.Now, let's compute 2z: 2*(-1 - 2i) = -2 - 4i.Adding z² and 2z: (-3 + 4i) + (-2 - 4i) = -5 + 0i = -5.Subtracting 5: -5 - 5 = -10.So, z² + 2z - 5 equals -10, not 0. Therefore, option A is incorrect.**Option B: The complex conjugate of z is 1 + 2i**The complex conjugate of z, which is -1 - 2i, is obtained by changing the sign of the imaginary part. So, the conjugate should be -1 + 2i. But option B says it's 1 + 2i. That's not correct because the real part is also changed from -1 to 1, which isn't how conjugation works. So, option B is incorrect.**Option C: The point corresponding to z is in the fourth quadrant**In the complex plane, the real part is the x-coordinate, and the imaginary part is the y-coordinate. So, z = -1 - 2i corresponds to the point (-1, -2). Quadrants are labeled as follows: first quadrant is (+, +), second is (-, +), third is (-, -), and fourth is (+, -). Since both the real and imaginary parts are negative, this point is in the third quadrant, not the fourth. Therefore, option C is incorrect.**Option D: |z| = sqrt(5)**The modulus of z, |z|, is calculated as the square root of (real part squared plus imaginary part squared). So, for z = -1 - 2i:|z| = sqrt((-1)^2 + (-2)^2) = sqrt(1 + 4) = sqrt(5).That's exactly what option D states. So, option D is correct.Let me just recap to make sure I didn't make any mistakes. I found z by solving zi = 2 - i, which gave me z = -1 - 2i. Then, I checked each option:- A: Plugging z into the equation didn't give 0, so A is wrong.- B: The conjugate was incorrectly stated, so B is wrong.- C: The point is in the third quadrant, not the fourth, so C is wrong.- D: The modulus is indeed sqrt(5), so D is correct.I think that's solid. I don't see any errors in my calculations, so I'm confident that D is the correct answer.**Final Answer**boxed{D}

💡Okay, so I need to find the volume of a regular hexagonal pyramid. The pyramid has a height ( h ) and the radius ( R ) of the circumscribed sphere. Hmm, let me think about how to approach this.First, I remember that the volume of a pyramid is given by the formula:[V = frac{1}{3} times text{Base Area} times text{Height}]So, I need to find the area of the base, which is a regular hexagon, and then multiply it by the height ( h ) and divide by 3.But wait, the problem gives me the radius ( R ) of the circumscribed sphere, not the side length of the hexagon. So, I need to relate ( R ) to the side length of the hexagon.I recall that in a regular hexagon, the radius of the circumscribed circle (which is the same as the distance from the center to any vertex) is equal to the side length of the hexagon. So, if the radius ( R ) is given, then the side length ( a ) of the hexagon is also ( R ). Is that correct?Wait, no, that might not be right. Let me think again. In a regular hexagon, the radius of the circumscribed circle is equal to the side length. So, if the radius ( R ) is given, then the side length ( a = R ). Okay, that seems correct.So, if the side length ( a = R ), then the area of the regular hexagon can be calculated using the formula:[text{Area} = frac{3sqrt{3}}{2} a^2]Substituting ( a = R ) into the formula, we get:[text{Area} = frac{3sqrt{3}}{2} R^2]Now, plugging this into the volume formula:[V = frac{1}{3} times frac{3sqrt{3}}{2} R^2 times h]Simplifying this, the 3 in the numerator and denominator cancels out:[V = frac{sqrt{3}}{2} R^2 h]Wait, but hold on. The problem mentions a regular hexagonal pyramid with a height ( h ) and radius ( R ) of the circumscribed sphere. Is the radius ( R ) referring to the radius of the base's circumscribed circle or the radius of the sphere that circumscribes the entire pyramid?I think I might have made a mistake here. The radius ( R ) is of the circumscribed sphere, not just the base. So, this sphere encompasses the entire pyramid, including the apex. Therefore, ( R ) is not just the radius of the base's circumscribed circle but the radius of the sphere that passes through all the vertices of the pyramid.Hmm, that complicates things a bit. So, I need to relate the radius ( R ) of the circumscribed sphere to the side length of the hexagon and the height ( h ) of the pyramid.Let me visualize the pyramid. It's a regular hexagonal pyramid, so the base is a regular hexagon, and the apex is directly above the center of the base. The circumscribed sphere would pass through all the vertices of the base and the apex.So, the sphere has its center somewhere along the axis of the pyramid, which is the line from the apex perpendicular to the base. Let me denote the center of the sphere as ( O ). The distance from ( O ) to any vertex of the base and to the apex is ( R ).Let me denote the center of the base as ( M ). The distance from ( M ) to any vertex of the base is the radius of the base's circumscribed circle, which is equal to the side length ( a ) of the hexagon. So, ( MA = a ).The height of the pyramid is ( h ), which is the distance from the apex ( P ) to the base. So, ( PM = h ).Now, considering the sphere, the center ( O ) lies somewhere along the line ( PM ). Let me denote the distance from ( O ) to ( M ) as ( x ). Then, the distance from ( O ) to ( P ) would be ( h - x ).Since ( O ) is the center of the sphere, the distance from ( O ) to any vertex of the base (like ( A )) is ( R ). So, in triangle ( OMA ), we have:[OM^2 + MA^2 = OA^2]Substituting the known values:[x^2 + a^2 = R^2]Similarly, the distance from ( O ) to the apex ( P ) is also ( R ). So, in triangle ( OMP ):[OM^2 + PM^2 = OP^2]But wait, ( OP ) is the distance from ( O ) to ( P ), which is ( h - x ). So,[x^2 + h^2 = (h - x)^2]Wait, that can't be right because expanding ( (h - x)^2 ) gives ( h^2 - 2hx + x^2 ). So,[x^2 + h^2 = h^2 - 2hx + x^2]Subtracting ( x^2 + h^2 ) from both sides:[0 = -2hx]Which implies ( x = 0 ). That would mean the center of the sphere is at the base's center ( M ). But if that's the case, then the distance from ( O ) to the apex ( P ) is ( h ), which should be equal to ( R ). So, ( h = R ). But that seems contradictory because the sphere would then have radius ( R = h ), but the base's radius is ( a ), so unless ( a = 0 ), which isn't possible, this doesn't make sense.I must have made a mistake in setting up the equations. Let me try again.The distance from ( O ) to ( A ) is ( R ), so in triangle ( OMA ):[OM^2 + MA^2 = OA^2 implies x^2 + a^2 = R^2]The distance from ( O ) to ( P ) is also ( R ), so in triangle ( OMP ):[OM^2 + PM^2 = OP^2 implies x^2 + h^2 = R^2]Wait, so both equations give:[x^2 + a^2 = R^2]and[x^2 + h^2 = R^2]Therefore, setting them equal:[x^2 + a^2 = x^2 + h^2 implies a^2 = h^2 implies a = h]So, the side length ( a ) of the hexagon is equal to the height ( h ) of the pyramid. That's interesting.So, if ( a = h ), then the area of the base is:[text{Area} = frac{3sqrt{3}}{2} a^2 = frac{3sqrt{3}}{2} h^2]Then, the volume of the pyramid is:[V = frac{1}{3} times frac{3sqrt{3}}{2} h^2 times h = frac{sqrt{3}}{2} h^3]But wait, the problem mentions the radius ( R ) of the circumscribed sphere. If ( a = h ), then from the earlier equation ( x^2 + a^2 = R^2 ), and since ( a = h ), we have:[x^2 + h^2 = R^2 implies x = sqrt{R^2 - h^2}]But earlier, we found that ( x = 0 ), which led to a contradiction unless ( R = h ). So, perhaps my assumption that ( O ) lies on the line ( PM ) is incorrect? Or maybe I need to reconsider the position of ( O ).Wait, no, in a regular pyramid, the center of the circumscribed sphere must lie along the axis of the pyramid, which is ( PM ). So, ( O ) must lie somewhere on ( PM ).But if ( a = h ), then ( R ) must satisfy ( R^2 = x^2 + a^2 ) and ( R^2 = x^2 + h^2 ). Since ( a = h ), both equations are consistent, but they don't help me find ( R ) in terms of ( h ).Wait, maybe I need to express ( R ) in terms of ( h ) and ( a ). Let me try that.From ( x^2 + a^2 = R^2 ) and ( x^2 + h^2 = R^2 ), subtracting the two equations:[(x^2 + a^2) - (x^2 + h^2) = R^2 - R^2 implies a^2 - h^2 = 0 implies a = h]So, again, ( a = h ). Therefore, ( R^2 = x^2 + a^2 = x^2 + h^2 ). But without another equation, I can't solve for ( x ) or ( R ).Wait, perhaps I need to consider another point on the sphere. For example, the apex ( P ) is at distance ( R ) from ( O ), so ( OP = R ). Similarly, any base vertex ( A ) is at distance ( R ) from ( O ). So, considering both, we have:1. ( OA = R ): ( sqrt{x^2 + a^2} = R )2. ( OP = R ): ( sqrt{(h - x)^2 + 0} = R ) (since ( P ) is along the axis, its distance from ( O ) is just ( |h - x| ))So, from equation 2:[|h - x| = R implies h - x = R quad text{or} quad x - h = R]But since ( x ) is the distance from ( O ) to ( M ), and ( M ) is below ( P ), ( x ) must be less than ( h ). So, ( h - x = R implies x = h - R )Substituting into equation 1:[(h - R)^2 + a^2 = R^2]Expanding:[h^2 - 2hR + R^2 + a^2 = R^2]Simplifying:[h^2 - 2hR + a^2 = 0]But earlier, we found that ( a = h ). So, substituting ( a = h ):[h^2 - 2hR + h^2 = 0 implies 2h^2 - 2hR = 0 implies 2h(h - R) = 0]Which gives ( h = 0 ) or ( h = R ). ( h = 0 ) is not possible, so ( h = R ).Wait, so if ( h = R ), then the height of the pyramid is equal to the radius of the circumscribed sphere. Then, the side length ( a = h = R ).So, the area of the base is:[text{Area} = frac{3sqrt{3}}{2} R^2]And the volume is:[V = frac{1}{3} times frac{3sqrt{3}}{2} R^2 times R = frac{sqrt{3}}{2} R^3]But the problem states the pyramid has height ( h ) and radius ( R ). If ( h = R ), then the volume is ( frac{sqrt{3}}{2} R^3 ). But the problem doesn't specify that ( h = R ); it just gives ( h ) and ( R ).So, perhaps my earlier assumption that ( a = h ) is incorrect. Maybe I need to find a relationship between ( a ), ( h ), and ( R ) without assuming ( a = h ).Let me go back. From the two equations:1. ( x^2 + a^2 = R^2 )2. ( (h - x)^2 = R^2 )From equation 2:[(h - x)^2 = R^2 implies h - x = R implies x = h - R]Substituting into equation 1:[(h - R)^2 + a^2 = R^2]Expanding:[h^2 - 2hR + R^2 + a^2 = R^2]Simplifying:[h^2 - 2hR + a^2 = 0 implies a^2 = 2hR - h^2]So, the side length ( a ) is:[a = sqrt{2hR - h^2}]Therefore, the area of the base is:[text{Area} = frac{3sqrt{3}}{2} a^2 = frac{3sqrt{3}}{2} (2hR - h^2)]Simplifying:[text{Area} = frac{3sqrt{3}}{2} times h(2R - h) = frac{3sqrt{3}}{2} h(2R - h)]Now, the volume of the pyramid is:[V = frac{1}{3} times text{Area} times h = frac{1}{3} times frac{3sqrt{3}}{2} h(2R - h) times h]Simplifying:[V = frac{sqrt{3}}{2} h^2 (2R - h)]So, the volume is:[V = frac{sqrt{3}}{2} h^2 (2R - h)]Wait, let me double-check this. If ( h = R ), then the volume becomes:[V = frac{sqrt{3}}{2} R^2 (2R - R) = frac{sqrt{3}}{2} R^3]Which matches what I got earlier when assuming ( h = R ). So, that seems consistent.Another check: if ( h = 0 ), the volume should be 0, which it is. If ( h = 2R ), the term ( (2R - h) ) becomes 0, so the volume is 0, which makes sense because the apex would be at the same height as the sphere's radius, making the pyramid flat.Therefore, I think this formula is correct.**Final Answer**The volume of the regular hexagonal pyramid is boxed{frac{1}{2} h^2 sqrt{3} (2R - h)}.

💡Okay, so I have this problem where I need to show that the sequence ( x_n ) defined by ( x_1 = a ), ( x_2 = b ), and for ( n geq 3 ), ( x_n = frac{x_{n-1}^2 + x_{n-2}^2}{x_{n-1} + x_{n-2}} ) is not an integer for any ( n geq 3 ). Here, ( a ) and ( b ) are relatively prime integers greater than 1. Hmm, interesting.Let me start by understanding the sequence better. The first two terms are given as ( a ) and ( b ). Then each subsequent term is the average of the squares of the two previous terms, divided by their sum. So, it's like a weighted average, but with squares. I wonder if this sequence has any particular properties or if it converges to something.But the problem isn't about convergence; it's about whether the terms can be integers. So, I need to show that starting from two relatively prime integers greater than 1, the sequence never hits an integer again. That seems non-trivial. Maybe I can use induction or some number theory here.Let me try induction. The base cases are ( n = 1 ) and ( n = 2 ), which are given as integers ( a ) and ( b ). So, the base cases are fine. Now, suppose that for some ( k geq 2 ), ( x_k ) and ( x_{k-1} ) are integers. Then, I need to show that ( x_{k+1} ) is not an integer.Wait, but the problem says ( a ) and ( b ) are relatively prime. Maybe that plays a role in the divisibility of the terms. Let me think about the expression for ( x_n ):( x_n = frac{x_{n-1}^2 + x_{n-2}^2}{x_{n-1} + x_{n-2}} )I can rewrite this as:( x_n = frac{x_{n-1}^2 + x_{n-2}^2}{x_{n-1} + x_{n-2}} = frac{(x_{n-1} + x_{n-2})^2 - 2x_{n-1}x_{n-2}}}{x_{n-1} + x_{n-2}}} )Simplifying that, I get:( x_n = x_{n-1} + x_{n-2} - frac{2x_{n-1}x_{n-2}}{x_{n-1} + x_{n-2}}} )Hmm, so ( x_n ) is equal to the sum of the two previous terms minus twice their product divided by their sum. That seems a bit complicated, but maybe I can analyze the fractional part here.If ( x_n ) is an integer, then ( frac{2x_{n-1}x_{n-2}}{x_{n-1} + x_{n-2}}} ) must also be an integer because the sum and difference of integers are integers. So, ( frac{2x_{n-1}x_{n-2}}{x_{n-1} + x_{n-2}}} ) must be an integer.Let me denote ( d = gcd(x_{n-1}, x_{n-2}) ). Since ( a ) and ( b ) are relatively prime, I wonder if this gcd remains 1 throughout the sequence. If that's the case, then ( x_{n-1} + x_{n-2} ) divides ( 2x_{n-1}x_{n-2} ). But since ( gcd(x_{n-1}, x_{n-2}) = 1 ), ( x_{n-1} + x_{n-2} ) must divide 2. Because ( x_{n-1} ) and ( x_{n-2} ) are both greater than 1, their sum is greater than 2, which would mean that ( x_{n-1} + x_{n-2} ) cannot divide 2. Therefore, this leads to a contradiction, implying that ( x_n ) cannot be an integer.Wait, let me check that reasoning again. If ( gcd(x_{n-1}, x_{n-2}) = 1 ), then ( x_{n-1} + x_{n-2} ) divides ( 2x_{n-1}x_{n-2} ). But since ( gcd(x_{n-1} + x_{n-2}, x_{n-1}) = gcd(x_{n-2}, x_{n-1}) = 1 ), similarly ( gcd(x_{n-1} + x_{n-2}, x_{n-2}) = 1 ). Therefore, ( x_{n-1} + x_{n-2} ) must divide 2. However, since ( x_{n-1} ) and ( x_{n-2} ) are both greater than 1, their sum is at least ( a + b ), which is greater than 2 because ( a, b > 1 ). Therefore, ( x_{n-1} + x_{n-2} ) cannot divide 2, which means our assumption that ( x_n ) is an integer must be false.So, if I can show that ( gcd(x_{n}, x_{n-1}) = 1 ) for all ( n geq 1 ), then this argument holds, and ( x_n ) cannot be an integer for ( n geq 3 ). Let me try to prove that ( gcd(x_{n}, x_{n-1}) = 1 ) by induction.Base case: ( gcd(x_1, x_2) = gcd(a, b) = 1 ) by the problem statement.Inductive step: Suppose ( gcd(x_{k}, x_{k-1}) = 1 ) for some ( k geq 2 ). We need to show that ( gcd(x_{k+1}, x_k) = 1 ).From the recurrence relation:( x_{k+1} = frac{x_k^2 + x_{k-1}^2}{x_k + x_{k-1}} )Let me denote ( d = gcd(x_{k+1}, x_k) ). Then ( d ) divides ( x_{k+1} ) and ( x_k ). Therefore, ( d ) divides ( x_{k+1} times (x_k + x_{k-1}) = x_k^2 + x_{k-1}^2 ). Since ( d ) divides ( x_k ), it also divides ( x_k^2 ). Therefore, ( d ) must divide ( x_{k-1}^2 ).But ( gcd(x_k, x_{k-1}) = 1 ) by the inductive hypothesis, so ( gcd(d, x_{k-1}) = 1 ). Therefore, ( d ) must divide 1, which implies ( d = 1 ). Thus, ( gcd(x_{k+1}, x_k) = 1 ).Therefore, by induction, ( gcd(x_n, x_{n-1}) = 1 ) for all ( n geq 1 ).Going back to the earlier argument, since ( gcd(x_{n-1}, x_{n-2}) = 1 ), ( x_{n-1} + x_{n-2} ) divides 2, but ( x_{n-1} + x_{n-2} geq a + b > 2 ), which is a contradiction. Therefore, ( x_n ) cannot be an integer for any ( n geq 3 ).I think this makes sense. The key was to show that the gcd remains 1 throughout the sequence, which allows us to conclude that the denominator in the expression for ( x_n ) cannot divide the numerator, hence ( x_n ) is not an integer.**Final Answer**boxed{x_n} text{ is not an integer for } n geq 3.