Show that we can find infinitely many triples N, N + 1, N + 2 such that each member of the triple is a sum of one or two squares.

**Solution**

Take N = 2n^{2}(n + 1)^{2}. Let m = n(n + 1). So N = m^{2} + m^{2}, N + 2 = (m + 1)^{2} + (m - 1)^{2}.

Slightly less obviously, N + 1 = (n^{2} - 1)^{2} + (n^{2} + 2n)^{2}.

That is clean, but a little unobvious. Fumbling our way to a solution, we note that n^{2}, n^{2} + 1 are easy. So we are home if we can do n^{2} - 1. We need to remember that if A and B are the sum of two squares, then so is AB: (a^{2} + b^{2})(c^{2} + d^{2}) = (ac + bd)^{2} + (ad - bc)^{2}. Thus if we can find a solution for n - 1, n, n + 1, then we can also find one for (n + 1)(n - 1) = n^{2} - 1, n^{2}, n^{2} + 1. A starting solution is 8 = 2^{2} + 2^{2}, 9 = 3^{2}, 10 = 3^{2} + 1.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001