5th USAMO 1976 Problem 3 Find all integral solutions to a 2 + b 2 + c 2 = a 2 b 2 . Solution
: 0, 0, 0. Squares must be 0 or 1 mod 4. Since the rhs is a square, each of the squares on the lhs must be 0 mod 4. So a, b, c are even. Put a = 2a 1 , b = 2b 1 , c = 2c 1 . Then a 1 2 + b 1 2 + c 1 2 = square. Repeating, we find that a, b, c must each be divisible by an arbitrarily large power of 2. So they must all be zero. 5th USAMO 1976 © John Scholes jscholes@kalva.demon.co.uk 11 Jul 2002