Russian 1997 Problem 5 Find all integral solutions to (m 2 - n 2 ) 2 = 1 + 16n.
(m,n) = (±1,0), (±4,3), (±4,5) Solution Clearly n cannot be negative. On the other hand, if (m,n) is a solution, then so is (-m,n). If n = 0, then m = ±1. If m = 0, then n 4 = 1 + 16n, which has no solutions (because n must divide 1, but 1 is not a solution). So let us assume m and n are positive. We have m 2 = n 2 + √(1+16n) or m 2 = n 2 - √(1+16n). In the first case, obviously m > n. But (n+1) 2 = n 2 + 2n + 1 which is greater than n 2 + √(1+16n) for (2n+1) 2 > 1 + 16n or n > 3. So if n > 3, then m 2 lies strictly between n 2 and (n+1) 2 , which is impossible. It is easy to check that n = 1, 2 do not give solutions, but n = 3 gives the solution m = 4. Similarly, in the second case we must have n ≤ 5, and it is easy to check that n = 1, 2, 3, 4 do not give solutions, but n = 5 gives the solution m = 4. Thanks to Suat Namli Russian 1997 © John Scholes jscholes@kalva.demon.co.uk 7 February 2004 Last corrected/updatd 7 Feb 04