Pythagorean Triples, Fermat Descent

Lecture 23 Pythagorean Triples, Fermat Descent Diophantine Equations - We start with Pythagorean Triples (x, y, z) where x2 + y 2 = z 2 . Problem is to find all Pythagorean triples. Reductions - can scale triples, so can assume gcd(x, y, z) = 1. We are looking for primitive solutions. Enough to classify solutions. Suppose (x, y, z) is such a triple. Then (x, y) = (y, z) = (x, z) = 1, for if a|x and a|y then a2 x02 + a2 y 02 = z 2 ⇒ a|z, contradicts primitivity. So, x, y, z ∈ N, coprime in pairs. x, y can’t both be even, and can’t both be odd (else x2 + y 2 ≡ 2 mod 4, which can’t be a square), so are opposite parity. wlog assume x odd and y even. x2 + y 2 = z 2 ⇒ y 2 = z 2 − x2 = (z − x)(z + x)  y 2  z − x   z + x  y2 ⇒ = = 4 2 2 2

Still all integers. z−x and z+2 x are coprime, since if a|both, then a|sum ⇒ a|z, 2 and a|difference ⇒ a|x. So we have two coprime integers, whose product is a square ⇒ both are squares. z+x z−x = r2 , = s2 for some r, s ∈ N 2 2 ⇒ z = r2 + s2 , x = r2 − s2  y 2 ⇒ = r2 s2 ⇒ y = 2rs z ⇒ (x, y, z) = (r2 − s2 , 2rs, r2 + s2 ) eg., r = 1, s = 2 ⇒ (3, 4, 5) Since x > 0, r > s and r and s must have opposite parity (since r2 ± s2 odd), and gcd(r, s) = 1 (else if a|rs, then a|x = r2 − s2 and a|z = r2 + s2 ). 2 A more geometrical way of seeing the result, if x2 + y 2 = z 2 ⇒ xz 2 + yz = 1. Set α = xz , β = yz ⇒ α2 + β 2 = 1. In other words, finding solutions to x2 + y 2 = z 2 in integers and solutions to α2 + β 2 = 1 in rationals is equivalent. Find all rational points on curve x2 + y 2 = 1 (unit circle) - obvious solution (1, 0). Suppose (u, v) is some other rational solution. Consider the line joining (1, 0) 1 and (u, v). Slope m = m= v u−1 is rational since u, v are rational. Line is y = m(x − 1). v ⇒ v = m(u − 1) u−1 ⇒ u2 + v 2 = 1 ⇒ u2 + m2 (u − 1)2 = 1 m2 (u − 1)2 = 1 − u2 = (1 − u)(1 + u) 1+u 6 0) ⇒ m2 = (can divide since 1 − u = 1−u If we solve for u we get u = m2 −1 m2 +1 ,  (u, v) = which means that v = m(u − 1) = m2 − 1 −2m , m2 + 1 m2 + 1 −2m m2 +1 .  ,m ∈ Q to get integer solution, set m = − rs , then you get    r2 − s2 2rs  2   − s2 , |{z} 2rs , r2 + s2 )  r2 + s2 , r2 + s2  ⇒ (r| {z } | {z } | {z } | {z } y x z u v We can apply this general method to any (nice) conic curve (plane curve cut out by equation of total degree in x, y of 2 - circles, parabola, hyperbolas, ellipses). Given a conic curve, and we know at least one rational point on it, then this slope method will give us all the rational points on the curve. (NOTE - not always true that conic curves have rational points - eg., x2 + y 2 + 1 = 0 has no real (or rational) points) Remember Fermat’s Last Theorem - xn +y n = z n has no non-trivial (ie., xyz 6= 0) solutions if n ≥ 3. We’ll show this for n = 4. Proof uses method called Fermat’s infinite descent: Given any integer solution, can produce a smaller integer solution. Since only finitely many positive integers smaller than initial solution, there cannot be any solutions. We’ll actually show that there does not exist x, y, z ∈ N such that x4 + y 4 = z 2 , which is stronger. Theorem 82. x4 + y 4 = z 2 has no solutions in Z>0 Proof. Suppose there is a solution. Let z be the size of the solution. Then there’s a possibly smaller solution with x, y coprime, for if a|xy then a4 |z 2 ⇒ a2 |z ⇒

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2 x 4 + ay = az2 . So we may as well assume x, y coprime. Then x, y, z are a coprime in pairs. x4 + y 4 = z 2 can be rewritten as (x2 )2 + (y 2 )2 = z 2 , with 2 x2 , y 2 , z coprime in pairs. wlog, we can assume x odd and y even, and by what we know of Pythagorean triplets, (x2 , y 2 , z) = (r2 − s2 , 2rs, r2 + s2 ) r>s>0 (r, s) = 1 and are opposite parity First equations gives that x2 + s2 = r2 . We have x odd and r + s odd, which forces s to be even, r odd, and so (x, s, r) = (t2 − u2 , 2tu, t2 + u2 ) t>u>0 (t, u) = 1 and are opposite parity Now y 2 = 2rs, y even, s even.  y 2 2 =r s 2 s 2 and r and are coprime since r, s are coprime, which means that r and 2s must both be squares. Define 2s = m2 and r = n2 , and so 2s = tu = m2 , and with t, u coprime, t and u are squares and we can write t = k 2 , u = l2 . Plug into r = t2 + u2 ⇒ n2 = k 4 + l4 , which gives another solution (k, l, n) to equation, with k, l, n > 0. n ≤ n2 = r ≤ r2 < r2 + s2 = z and so this is a smaller solution.  Techniques for Diophantine Equations and interesting examples Say we need to show some diophantine equation (some polynomial equation with integer coefficients) has no solutions. If you can show that there are no solutions mod m, then there are no solutions (known as checking for local solutions) Eg. x2 = 3 + 4y 3 has no solutions because mod 4 this says x2 ≡ 3 mod 4 Eg. x3 + y 3 − 7z 3 = 3 has no solutions because this says x3 + y 3 ≡ 3 mod 7, and a cube mod 7 can only be 0, ±1 (±1 since x6 = (x3 )2 ≡ 1 mod 7 by FlT) Eg. Only solution to x3 + 2y 3 + 4z 3 = 0 is (0, 0, 0). 3 Proof. By infinite descent. Assume non-trivial, produce smaller. Assume gcd(x, y, z) = 1 (otherwise can get smaller solution). mod 2 ⇒ x3 ≡ 0 mod 2 ⇒ x ≡ 0 mod 2. Set x = 2x0 8x03 + 2y 3 + 4z 3 = 0 4x03 + y 3 + 2z 3 = 0 ⇒ y = 2y 0 ⇒ z = 2z 0 ⇒ 2|x, y, z and so (x0 , y 0 , z 0 ) is smaller solution  Eg. y 2 = x3 + D is a classical equation Theorem 83. y 2 = x3 + 7 has no integer solutions. Proof. y 2 ≥ 0 so x ≥ −1, so RHS is nonzero. Now if x were even, we’d have y 2 ≡ 7 ≡ 3 mod 4, which is impossible, so x is odd, which means that x3 + 7 is even, so y is even. Rewrite as y 2 + 1 = x3 + 8 = (x + 2)(x2 − 2x + 4) So if some prime p divides y 2 + 1, then p has to be odd and y 2 ≡ −1 mod p, which means that p ≡ 1 mod 4. But also, x3 ≡ y 2 − 7 ≡ 0 − 7 mod 4 ≡ 1 mod 4, and so x ≡ 1 mod 4, which means that x + 2 ≡ 3 mod 4 ≥ 1. So there exists a prime dividing x + 2 which is 3 mod 4, and so p|y 2 + 1, which is a contradiction.  4