10th Balkan 1993

p is prime and k is a positive integer > 1. Show that we can find positive integers (m, n) ≠ (1, 1) such that (m^p + n^p)/2 = ( (m + n)/2 )^k iff k = p.

Solution

x = 2, y = 2 is a solution if k = p. Let z = (x+y)/2. If x+y is odd, then 2z^m is not integral. Contradiction. Hence x+y is even. x=y=1 is not allowed, so z is at least 2. Hence (x^p+y^p)/2 < (x+y)^p/2 ≤ (2z)^p/2 < z^2p, so m < 2p. But (x^p + y^p)/2 ≥ z^p, so m ≥ p.

Put x = z+k, y = z-k. If k = 0, then lhs = z^p, rhs = z^m, so m = p and then any z is a solution. If k is non-zero, then ((z+k)^p+(z-k)^p)/2= z^p + pC2 z^p-2k² + ... + pCp-1 z k^p-1 = z^m. So p divides z^m - z^p

10th Balkan 1993

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002