Find all integral solutions to a2 + b2 + c2 = a2b2.
Solution
Answer: 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 = 2a1, b = 2b1, c = 2c1. Then a12 + b12 + c12 = 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.
© John Scholes
jscholes@kalva.demon.co.uk
11 Jul 2002