5th APMO 1993 Problem 4 Find all positive integers n for which x n + (x+2) n + (2-x) n = 0 has an integral solution. Solution
: n = 1. There are obviously no solutions for even n, because then all terms are non-negative and at least one is positive. x = -4 is a solution for n = 1. So suppose n is odd n and > 3. If x is positive, then x n + (x+2) n > (x+2) n > (x-2) n , so x n + (x+2) n + (2-x) n > 0. Hence any solution x must be negative. Put x = -y. Clearly x = -1 is not a solution for any n, so if x = -y is a solution then (x+2) = -(y-2) ≤ 0 we have (y+2) n = y n + (y-2) n . Now 4 = ( (y+2) - (y-2) ) divides (y+2) n - (y-2) n . Hence 2 divides y. Put y = 2z, then we have (z+1) n = z n + (z-1) n . Now 2 divides (z+1) n - (z-1) n so 2 divides z, so z+1 and z-1 are both odd. But a n - b n = (a - b)(a n-1 n-2b + a n-3 b 2 + ... + b n-1 ). If a and b are both odd, then each term in (a n-1 n-2b + a n-3 b 2 + ... + b n-1 ) is odd and since n is odd there are an odd number of terms, so (a n-1 n-2b + a n-3 b 2 + ... + b n-1 ) is odd. Hence, putting a=z+1, b=z-1, we see that (z+1) n - (z-1) n = 2(a n-1 n-2b + a n-3 b 2 + ... + b n-1 ) is not divisible by 4. But it equals z n with z even. Hence n must be 1. 5th APMO 1993 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002