29th IMO shortlist 1988/22 solution

29th IMO 1988 shortlist

Problem 22

Show that there are only two values of N for which (4N+1)(x₁² + x₂² + ... + x_N²)= 4(x₁ + x₂ + ... + x_N)² + 4N + 1 has an integer solution x_i.

Answer

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 x_i 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 x_i is a solution, then -x_i is also a solution. So let us assume that ∑ x_i ≥ 0. If some x_i is not 0 or 1, then either some x_i is negative, or some x_i > 1. In either case we have x_i² > x_i, so ∑ x_i < ∑ x_i². Put X = ∑ x_i. Then X + 1 ≤ ∑ x_i² = (4/(4N+1) X² + 1, so X ≥ (4N+1)/4 and hence X ≥ N+1.

By Cauchy, we have X² ≤ N ∑ x_i² = (4N/(4N+1) X² + N, so X² ≤ N(4N+1), so X ≤ 2N. Hence 1 < X/N ≤ 2.

We have ∑ (x_i - X/N)² = (∑ x_i²) - X²/N = 1 - X²/(N(4N+1)) < 1. Hence each |x_i - X/N| < 1. Hence 0 < x_i < 3. But x_i 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.

29th IMO shortlist 1988