### 29th IMO 1988 shortlist

Problem 22

Show that there are only two values of N for which (4N+1)(x12 + x22 + ... + xN2)= 4(x1 + x2 + ... + xN)2 + 4N + 1 has an integer solution xi.

N = 2 has the solution 9(12 + 22) = 4(1 + 2)2 + 9.
N = 6 has the solution 25(12 + 12 + 12 + 12 + 12 + 02) = 4(1 + 1 + 1 + 1 + 1 + 0)2 + 25.

Solution

Suppose every xi is 0 or 1. Suppose there are M 1s. Then we have (4N+1)M = 4M2 + 4N+1, so (4N+1)(M-1) = 4M2. Since 4N+1 is odd, 4 must divide M-1. So M is odd. So 8 does not divide M-1. Since M-1 and M are coprime, no prime p > 2 can divide M-1. Hence M = 5. So N = 6. That is a possible solution (as shown explicitly above).

If xi is a solution, then -xi is also a solution. So let us assume that ∑ xi ≥ 0. If some xi is not 0 or 1, then either some xi is negative, or some xi > 1. In either case we have xi2 > xi, so ∑ xi < ∑ xi2. Put X = ∑ xi. Then X + 1 ≤ ∑ xi2 = (4/(4N+1) X2 + 1, so X ≥ (4N+1)/4 and hence X ≥ N+1.

By Cauchy, we have X2 ≤ N ∑ xi2 = (4N/(4N+1) X2 + N, so X2 ≤ N(4N+1), so X ≤ 2N. Hence 1 < X/N ≤ 2.

We have ∑ (xi - X/N)2 = (∑ xi2) - X2/N = 1 - X2/(N(4N+1)) < 1. Hence each |xi - X/N| < 1. Hence 0 < xi < 3. But xi is an integer, so it must be 1 or 2. Suppose there are M 1s. Then there are N-M 2s and we have: 4MN - 3M - 4M2 - 1 = 0. Hence M divides 1, so M = 1. Hence N = 2, and we have the other solution shown above.