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 = 2n2(n + 1)2. Let m = n(n + 1). So N = m2 + m2, N + 2 = (m + 1)2 + (m - 1)2.
Slightly less obviously, N + 1 = (n2 - 1)2 + (n2 + 2n)2.
That is clean, but a little unobvious. Fumbling our way to a solution, we note that n2, n2 + 1 are easy. So we are home if we can do n2 - 1. We need to remember that if A and B are the sum of two squares, then so is AB: (a2 + b2)(c2 + d2) = (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) = n2 - 1, n2, n2 + 1. A starting solution is 8 = 22 + 22, 9 = 32, 10 = 32 + 1.
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001