Prove that there are infinitely many triples of positive integers (m, n, p) satisfying 4mn - m - n = p2 - 1, but none satisfying 4mn - m - n = p2.
Solution
We can take (m,n,p) = (3k2,1,3k) (thanks to Suat Namli), or (k(k-1)/2, k(k+1)/2,k2-1) as solutions of 4mn-m-n = p2-1.
Suppose m, n, p satisfy 4mn-m-n = p2, so (4m-1)(4n-1) = 4p2+1 = (2p)2 + 1. Now suppose q is any prime dividing the lhs. Then (2p)2 = -1 mod q. But (2p)2 = 1 mod q (Fermat), so q = 1 mod 4. But 4m-1 must have at least one prime factor = 3 mod 4. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
12 February 2004
Last corrected/updated 12 Feb 04