IMO 1989 SL 15

Origin: IRE

Problem

Let a, b, c, d, m, n be positive integers such that a2+b2+c2+d2 = 1989, a+b+c+d = m2, and the largest of a, b, c, d is n2. Determine, with proof, the values of m and n.

Solution

By Cauchy’s inequality, 44 < \sqrt 1989 < a + b + c + d \leq \sqrt 2 \cdot 1989 < 90. Since m2 = a + b + c + d is of the same parity as a2 + b2 + c2 + d2 = 1989, m2 is either 49 or 81. Let d = max{a, b, c, d}. Suppose that m2 = 49. Then (49 −d)2 = (a + b + c)2 > a2 + b2 + c2 = 1989 −d2, and so d2 −49d + 206 > 0. This inequality does not hold for 5 \leqd \leq44. Since d \geq

1989/4 > 22, d must be at least 45, which is impossible because 452 > 1989. Thus we must have m2 = 81 and m = 9. Now, 4d > 81 implies d \geq21. On the other hand, d < \sqrt 1989, and hence

d = 25 or d = 36. Suppose that d = 25 and put a = 25 −p, b = 25 −q, c = 25−r with p, q, r \geq0. From a+b+c = 56 it follows that p+q+r = 19, which, together with (25 −p)2 + (25 −q)2 + (25 −r)2 = 1364, gives us p2 + q2 + r2 = 439 > 361 = (p + q + r)2, a contradiction. Therefore d = 36 and n = 6. Remark. A little more calculation yields the unique solution a = 12, b = 15, c = 18, d = 36.