Solution by the Organizers (Based on Euler’s Original Proof 1738)

Problem of the Week 9, Spring 2006 Solution by the organizers (Based on Euler’s original proof 1738) . We will prove that the only rational solutions to the equation m2 − n3 = 1 are (m, n) = (−1, 0), (1, 0), (0, −1), (3, 2), (−3, 2). If n = 0 then m = ±1 are all possible solutions, similarly if m = 0 then n = −1 is the only solution. Suppose n, m 6= 0 and n = a/b with a, b ∈ Z, b > 0, and gcd(a, b) = 1. Then m2 b4 = a3 b + b4 , (1) that is b (a3 + b3 ) = b (a + b) (a2 − ab + b2 ) is a non-zero square. Let c = a + b, then equation (1) becomes ¡ ¢ m2 b4 = bc 3b2 − 3bc + c2 . (2) If c = 3b then a = 2b and n = 2 which gives the solutions m = ±3. From now on assume c 6= 3b. Notice that gcd(b, c) = gcd(b, a + b) = gcd(b, a) = 1 and gcd(3b2 − 3bc + c2 , b) = gcd(c2 , b) = 1. Assume that the pair (b, c) satisfies that |b| is as small as possible such that the right-hand side of (2) is a square. We will show that, once c = 3b is excluded, there are no other solutions by the method of infinite descent. That is we will find a new pair (b0 , c0 ) with |b0 | < b. To do this we first divide in two cases. Case 1 3 does not divide c. Then gcd(3b2 − 3bc + c2 , c) = gcd(3b2 , c) = gcd(b2 , c) = 1. Thus in equation (2) we have the product of three integers pairwise coprime which equals a square. Additionally b > 0 and 3b2 − 3bc + c2 ≥ 3(b − c/2)2 ≥ 0. Thus each of b, c, and 3b2 −3bc+c2 are squares. Then there are positive integers p and q such that gcd(p, q) = 1 and µ ¶2 p 2 2 3b − 3bc + c = b −c q 2 2 2bpq bp + c2 . = − 2 q q Thus 3q 2 − 2pq b = . c 3q 2 − p2 (3) Now we divide in two subcases. Case1.1. 3 does not divide p. Suppose P is a prime common divisor of 3q2 − 2pq and 3q 2 − p2 . Then either P |q or P |(3q − 2p). If P |q then P |p2 which is a contradiction since gcd(p, q) = 1. Thus P |(3q − 2p) on the other hand P | (3q 2 − 2pq) − (3q 2 − p2 ) = p2 − 2pq. Thus either P |p or P |(p − 2q). If P |p then P |3q 2 but P 6= 3 (since 3 does not divide p), so P |q2 which is a contradiction. Therefore we must have that P |(p − 2q). Thus P | (3q − 2p) + 2 (p − 2q) = −q which is a case we considered before. Therefore our conclusion is that gcd(3q2 − 2pq, 3q2 − p2 ) = 1. 1 This implies that b = 3q 2 − 2pq and c = 3q2 − p2 or b = 2pq − 3q2 and c = p2 − 3q 2 . However since c is a square we must have c ≡ 0, 1 (mod 4) and 3q 2 − p2 ≡ 3, 2 (mod 4) since we cannot have both p and q even. Therefore the first option is discarded and we have that b = 2pq − 3q 2 and c = p2 − 3q 2 . Now recall that c is a square, so there are positive integers r, s with gcd(r, s) = 1 such that ³ r ´2 p2 − 3q 2 = p − q s 2pqr r2 q 2 + 2 . = p2 − s s This implies that p r2 + 3s2 = q 2rs and which means that b 3s2 − 3rs + r2 2p − 3 = = , q2 q rs ¡ 2 ¢ r2 s2 b 2 = rs 3s − 3rs + r q2 and the left-hand side of the equation is a square since b is a square, thus we obtained an equation of the same form as (2) but 0 < s < b which we will check later. Case1.2. 3 divides p. Let p = 3P then equation (3) becomes b q 2 − 2P q = 2 . c q − 3P 2 By an argument similar to the previous case we can check that gcd(q 2 − 2P q, q 2 − 3P 2 ) = 1. And again similarly to last case 3P 2 − q 2 cannot be a square because it fails modulo 4. Therefore we must have q 2 − 3P 2 = c and q 2 − 2P q = b. But c is a square, so there are positive integers r, s such that ³ r ´2 2 2 q − 3P = q − P s 2qrP r2 P 2 + 2 . = q2 − s s This implies that 2P 4sr = 2 q 3s + r2 2 and 2P 3s2 − 4sr + r2 (3s − r) (s − r) b = = 1 − = . 2 2 2 q q 3s + r 3s2 + r2 Then 2 ¢ ¡ 2 b (3s2 + r2 ) 2 (3s − r) (s − r) . = 3s + r q2 Note that the left hand side is a square since b is a square, and if we let t = s − r and u = 3s − r then the right hand side becomes ¡ ¢ tu 3t2 − 3ut + u2 . Similarly to last case we also have that 0 < t < b (To be checked later). Case 2 3 divides c. Let c = 3C. Then gcd(3C, b) = 1 and equation (2) becomes ¡ ¢ m2 b4 = bC b2 − 3bC + 3C 2 . 9 Moreover gcd(b2 − 3bC + 3C 2 , b) = gcd(3C 2 , b) = gcd(C 2 , b) = 1, and gcd(b2 − 3bC + 3C 2 , C) = gcd(b2 , C) = 1. Thus b, C, and b2 − 3bC + 3C 2 are all squares and then we can apply Case 1 from here. In each of the three cases we obtain an expression like the right hand side of equation (2) which is equal to a square, however in each case we contradict the minimality of the solution. Therefore there are no other solutions. Now we go over the technical aspects of proving that 0 < s < b in Case 1.1 and 0 < t < b in Case 1.2. Claim 3 In Case 1.1 we have that 0 < s < b. Proof. First note that b = q (2p − 3q) so we have that 0 < q ≤ b. Also we know that q/p = 2rs/ (r2 + 3s2 ). Now note that gcd (s, r2 + 3s2 ) = gcd (s, r2 ) = 1 and gcd (r, r2 + 3s2 ) = gcd (r, 3s2 ) = gcd (r, 3). Therefore 2rs and r2 + 3s2 have a greatest common factor of 1, 2, 3 or 6. So if 3 - r then the greatest common factor is 2, thus q = rs or q = 2rs and consequently s ≤ q. On the other hand if 3|r then the greatest common factor is 3 or 6. Thus q = rs/3 or 2rs/3 and consequently rs/3 ≤ q. However 3 ≤ r, thus s ≤ q. In any case we conclude that 0 < s ≤ q ≤ b. Equality occurs if b = q which implies that 2p − 3q = 1 and then ¡ ¢ 4c = 4 p2 − 3q2 = (2p)2 − 12q2 = (1 + 3q)2 − 12q 2 = 1 + 6q − 3q 2 = 1 + 6b − 3b2 . But c is a square, thus 6b − 3b2 + 1 = 3b (2 − b) + 1 > 0. But if b > 2 then 3b (2 − b) < −6. Also if b = 2 then 1 = 2p − 3b = 2p − 6 which is impossible by parity; and finally if b = 1 then c = 1 which gives a = 0 that was excluded to begin with. 3 Claim 4 In Case 1.2 we have that 0 < |t| < b. Proof. Similarly to last case b = q (q − 2P ), so 0 < q ≤ b. Also P/q = 2sr/ (3s2 + r2 ) and again gcd (2sr, 3s2 + r2 ) is either 1, 2, 3, or 6. In all cases we can deduce that q ≥ (3s2 + r2 )/6. We have that s = (u − t)/2 and r = (u − 3t) /2 thus à µ ¶2 µ ¶2 ! 1 u−t u − 3t q ≥ 3 + (4) 6 2 2 ¢ 1¡ 2 u − 4ut + 5t2 ≥ 4 ¢ 1¡ (u − 2t)2 + t2 . ≥ 4 If |t| ≥ 4 then t2 ≥ 4 |t| and consequently q ≥ |t| > 0. If |t| = 3 and (u − 2t) 6= 0 then (u − 2t)2 + t2 ≥ 10 and then q ≥ 3 = |t|. Otherwise u − 2t = 0 and s = 3t/2 which is not an integer. If |t| = 2 and |u − 2t| ≥ 2 then (u − 2t)2 + t2 ≥ 8 and thus q ≥ |t|. The other possibilities are as follows, if u − 2t = 0 then r = −t/2 and s = 3t/2 which is impossible since both must be positive. If u − 2t = ±1 then s = (t ± 1) /2 which is not an integer. If |t| = 1 then since q is an integer we must have q ≥ 1 = |t|. Finally, if t = 0 then s = r = u/2 which implies that P = 1, q = 2, and b = 0 which is impossible. In all possibilities we obtain q ≥ |t| > 0, with equality if b = q which implies that b − 2P = q − 2P = 1 and then ¶2 µ b−1 2 2 2 c = q − 3P = b − 3 2 2 (b + 3) − 12 = . 4 But c is a square, say c = C 2 , then the last equation becomes 12 = (b + 3)2 − (2C)2 = (b + 3 + 2C) (b + 3 − 2C), and since both factors must be even in order for b to be integer we must have b + 3 + 2C = 6 and b + 3 − 2C = 2 which gives b = C = 1 and then a = c − b = C 2 − b = 0 which was excluded. Thus equality never happens. 4