IMO 1988 SL 22

Origin: KOR

Problem

Let p be the product of two consecutive integers greater than

Solution

Set X = p i=1 xi and w.l.o.g. assume that X \geq0 (if (x1, . . . , xp) is a solution, then (−x1, . . . , −xp) is a solution too). Since x2 \geqx for all integers x, it follows that p i=1 x2 i \geqX. If the last inequality is an equality, then all xi’s are 0 or 1; then, taking that there are a 1’s, the equation becomes 4p + 1 = 4(a + 1) + a−1, which forces p = 6 and a = 5. Otherwise, we have X + 1 \leqp i=1 x2 i = 4p+1X2 + 1, so X \geqp + 1. Also, by the Cauchy–Schwarz inequality, X2 \leqp p i=1 x2 i = 4p 4p+1X2 + p, so X2 \leq4p2 + p and X \leq2p. Thus 1 \leqX/p \leq2. However, p  i=1  xi −X p 2

 x2 i −2X p  xi + X2 p

 x2 i −pX2 p2 = 1 − X2 p(4p + 1) < 1, and we deduce that −1 < xi −X/p < 1 for all i. This finally gives xi \in{1, 2}. Suppose there are b 2’s. Then 3b + p = 4(b + p)2/(4p + 1) + 1, so p = b + 1/(4b −3), which leads to p = 2, b = 1. Thus there are no solutions for any p ̸\in{2, 6}. Remark. The condition p = n(n+1), n \geq3, was unnecessary in the official solution, too (its only role was to simplify showing that X ̸= p −1).