32nd IMO 1991 shortlist

Find all positive integer solutions to 3^m + 4ⁿ = 5^k.

Answer

3² + 4² = 5²

Solution

5^k = (-1)^k mod 3, and 4ⁿ = 1 mod 3, so k must be even. Put k = 2K. Similarly 3^m = (-1)^m mod 4, and 5^k = 1 mod 4, so m must be even. Put m = 2M.

Now 2²ⁿ = 4ⁿ = 5^2K - 3^2M = (5^k + 3^M)(5^K - 3^M). Hence for some non-negative integers a, b with (a + b) + b = 2n, we have 5^K + 3^M = 2^a+b, 5^K - 3^M = 2^b. Subtracting, we get 2·3^M = 2^b(2^a - 1), so b must be 1 and 3^M = 2^a - 1. But 3^M = 1 or 3 mod 8, so a = 1 or 2. But a + 2b is even, so a = 2. Hence M = 1 and n = 2. So we have 3² + 4² = 5^k. Hence k = 2.

32nd IMO shortlist 1991

© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2003
Last corrected/updated 1 Jan 03