30th IMO 1989 shortlist Problem 12 a, b, c, d, m, n are positive integers such that a 2 + b 2 + c 2 + d 2 = 1989, a + b + c + d = m 2 and the largest of a, b, c, d is n 2 . Find m and n. Solution
: m = 9, n = 6. 7 4 = 2401 > 1989, so n  4. Hence n = 5 or 6. Assume a ≥ b ≥ c ≥ d, so a = n 2 . k has the same parity as k 2 . But a 2 + b 2 + c 2 + d 2 = 1989 is odd, so a + b + c + d = m 2 is odd. If n = 36, then b 2 + c 2 + d 2 = 693, so by Cauchy, (b + c + d) 2 ≤ 3.693. Hence b + c + d ≤ 45. Hence 36  b 2 + c 2 + d 2 = 1989 - a 2 , so a 2 - 49a + 206 > 0. Put f(x) = x 2 - 49x + 206. We can easily check that f(4) = f(45) = 26, f(5) = f(44) = -14, so f(x)  (B + C + D) 2 . Contradiction. So there are no solutions with m = 9, n = 5. 30th IMO shortlist 1989 © John Scholes jscholes@kalva.demon.co.uk 28 Dec 2002 Last corrected/updated 28 Dec 02