18.781 Homework 9 Due: 15th April 2014 Q 1 (3.4(9)). Let f(x; y) = ax2 + bxy + cy2 be a quadratic form with integral coefficients whose discriminant d is a perfect square, possibly zero. Show that there are integers h1; h2; k1; k2 such that f(x; y) = (h1x + k1y)(h2x + k2y). Proof. If a = 0, then f(x; y) = y(bx + cy) is a factorization into linear polynomials with integer coefficients 2 b c (h1 = 0; k1 = 1; h2 = b; k2 = c). So now assume a 6= 0. Consider the polynomial g(z) = z + a z + a 2 Q[z]. By the assumption that d := b2 − 4ac is a perfect square, say d = k2 for some k 2 Z, we see by the quadratic −b+k −b−k formula that this equation factors into two linear factors in Q[z](g(z) = (z − 2a )(z − 2a )). Now, (ax − −b+k y)(ax − −b−k y) f(x; y) = ay2g(x=y) = 2 2 a As b + k = b − k + 2k, we see that b − k ≡ 0(mod 2) if and only if b + k ≡ 0(mod 2). (b − k)(b + k) = b2 − k2 = −b−k (ax−ly)(ax−(l+k)y) 4ac ≡ 0(mod 2), so 2 j b − k and 2 j b + k. Let 2 = l. We then have f(x; y) = a . This tells us that l(l + k) = ac. Let r = (a; l), and let a = rs. s j l(l + k) as a j l(l + k) and (s; l) = 1, so s j l + k. So we see that can write a = rs for some integers r and s such that rt = l and su = l + k for some other integers t and u, so l l+k f(x; y) = (sx − r y)(rx − s ) Q 2 (3.4(10)). Let f(x; y) = ax2 + bxy + cy2 be a quadratic form with integral coefficients. Show that there exist integers x0; y0, not both 0, such that f(x0; y0) = 0, if and only if the discriminant d of f(x; y) is a perfect square, possibly 0. Proof. If f = 0, then we can pick any pair of integers x0 and y0 and the discriminant of f is a square. So henceforth assume f 6= 0. If the discriminant is a perfect square, then by the previous problem, f(x; y) = (h1x + k1y)(h2x + k2y) for some integers h1; h2; k1; k2, so x0 = k1, y0 = −h1 satisfies f(x0; y0) = 0 (If f 6= 0, then h1 and k1 cannot both be zero). Suppose there exist integers x0; y0, not both 0, such that f(x0; y0) = 0. x0 2 x0 Assume y0 6= 0 (a similar argument works when x0 6= 0). Then we have the equality a + b + c = 0 y0 y0 over the rational numbers. Multiplying by 4a and completing squares, we see (2a x0 + b)2 = b2 − 4ac. So the y0 discriminant d equals l2=m2 for some integers l and m. l2=m2 is an integer (i.e. m2 j l2), so we must have m j l. (If m - l, there exists a prime p such that pα j m and pα - l, then p2α j m2 and therefore p2α j l2 but pα - l implies p2α - l2, a contradiction). Q 3 (3.5(1)). Find a reduced form that is equivalent to the form 7x2 + 25xy + 23y2. 1 −2 Proof. The discriminant of f is 625−644 = −19 which is not a perfect square. Using the matrix , 0 1 0 1 we see that the given form is equivalent to 7x2 − 3xy + y2. Using the matrix , this form is −1 0 1 −1 equivalent to x2 + 3xy + 7y2. Using , this form is now equivalent to x2 + xy + 5y2. This is a 0 1 reduced form. 1 Q 4 (3.5(8)). Let f(x; y) = 44x2 −97xy +35y2. Show that f is equivalent to the form g(x; y) = x(47x−57y). Show that n is represented by f if and only if n can be written in the form n = ab where b ≡ 47a(mod 57). Find the least positive integer n represented by f. Proof. We see that f(x; y) is equivalent to the form f(x; x + y) = −18x2 − 27xy + 35y2 = f 0(x; y) which in turn is equivalent to f 0(x + y; y) = −18x2 − 63xy − 10y2 which in turn is equivalent to −10x2 + 63xy − 18y2. This is in turn equivalent to −10(x + 6y)2 + 63(x + 6y)y − 18y2 = −10x2 − 57xy, and this in turn equivalent to −10x2 − 57x(−x + y) = 47x2 − 57xy. n is represented by f if and only if n is represented by g. If n is represented by g, then n = a(47a−57y) for some integers a and y. Let b = 47a−57y, then b ≡ 47a(mod 57). Conversely, if n = ab and b ≡ 47a(mod 57), then let y 2 Z be such that b−47a = −57y, then n = a(47a−57y) which shows n is represented by g and hence by f. Q 5 (3.7(3)). Show that any positive definite binary quadratic form of discriminant −3 is equivalent to f(x; y) = x2 + xy + y2. Show that a positive integer n is properly represented by f if and only if n is of the form n = 3α Q pβ, where α = 0 or 1 and all the primes p are of the form 3k + 1. Proof. By Theorem 3:19, if f = ax2 + bxy + cy2 is a reduced positive definite binary quadratic form of discriminant −3, then a = 1. As f is reduced, either b = 0 or b = 1. But if b = 0, then the discriminant is −4c which cannot be −3, so b = 1. b2 − 4ac = −3 now tells us that 1 − 4c = −3, so c = 1. The set of integers properly represented by a form is the same for equivalent forms. As f is the unique positive definite reduced quadratic form of discriminant −3, if at all a binary quadratic form of discriminant −3 (which has to be either positive definite or negative definite) represents a positive integer n, it is represented by f. So by Theorem 3:13, we see that this is possible if and only if the congruence x2 ≡ −3(mod 4n) is solvable. x2 ≡ −3(mod 8) is not solvable (as we can see by explcitly writing out square classes modulo 8, or applying Corollary 2:44), so 2 - n. So the congruence x2 ≡ −3(mod 4n) is solvable if and only if x2 ≡ −3(mod n) is solvable (x2 ≡ −3(mod 4) is solvable and (4; n) = 1). Also, x2 ≡ −3(mod 9) is not solvable (again we can α Q αi see this explcitly by looking at residue classes modulo 9). Let n = 3 pi for some primes pi ≥ 5. By 2 αi what we have just remarked α 2 f0; 1g. By Hensel's lemma, we now see that x ≡ −3(mod pi ) is solvable 2 αi if and only if x ≡ −3(mod pi) is solvable for pi ≥ 5 (any solution to this lifts uniquely modulo pi as any solution has to be a unit (as −3 is a unit), and 2 is also a unit, so the derivative 2x evaluated at this solution is non-vanishing). −3 −1 3 p (pi−1)=2 i (pi−1)=2 = = (−1) (−1) = pi(mod 3) pi pi pi 3 2 So we see that x ≡ −3(mod pi) is solvable if and only if pi ≡ 1(mod 3) for pi ≥ 5. Putting all of this together, we see that x2 ≡ −3(mod 4n) is sovable if and only if n = 3α Q pβ, where α = 0 or 1 and all the primes p are of the form 3k + 1, and therefore n is properly represented by f if and only if n is of this particular form. Q 6 (3.7(6)). Show that any positive definite quadratic form of discriminant −23 is equivalent to exactly 2 2 2 2 2 2 one of the form f0(x; y) = x + xy + 6y , f1(x; y) = 2x + xy + 3y or f2(x; y) = 2x − xy + 3y . Show that −23 −23 if p = −1 then p is not represented by any of these forms. Show that if p = 1 then p has a total of 4 representations by these forms. Show that in the latter case either p has four representations by f0 or 2 representations apiece by f1 and f2. Determine which of these cases applies when p = 139. Proof. Every positive definite quadratic form is equivalent to a unique reduced positive definite form, so it suffices to show that f0, f1 and f2 are all the distinct reduced positive definite binary quadratic forms. If f(x; y) = ax2 + bxy + cy2 is a reduced positive definite binary quadratic form of discriminant −23, then 0 < a < p23=3 by Theorem 3:19. This implies a 2 f1; 2g. If a = 1, then b 2 f0; 1g, but b = 0 is not possible as in this case, the discriminant equals −4ac which cannot equal −23 6≡ 0(mod 4). So if a = 1, then b = 1, 2 and as b − 4ac = −23, c = 6, so we get the reduced form f0. If a = 2, then b 2 {−1; 0; 1; 2g. b 6= 0(mod 2) as b ≡ 0(mod 2) implies the discriminant b2 − 4ac is zero modulo 2 but −23 is not. So b 2 {−1; 1g. In this case b2 − 4ac = −23 forces c = 3.