Prove that there exist infinitely many integers n such that n, n+1, n+2 are each a sum of two squares of integers.

Start with the square of 4 times a triangular number:

(2n(n+1))^2 (+0)
(2n(n+1)^2+1 Add 1, easy.
(2n(n+1))^2+2 More difficult. We need a partition into 2 squares.

Looking at small values, the derivation of the smaller square becomes obvious:

18 3,3,
146 5,11
578 7,23
1602 9,39
3602 11,59
7058 13,83

(2n(n+1))^2+2 = 4n^4+8n^3+4n^2+2.

4n^4+8n^3+4n^2+2-(2n+1)^2 = (2n^2+2n-1)^2, and we are done.

There are other qualifying triplets that don't belong to this set.

 

Edited on February 9, 2016, 5:14 am

Posted by broll on 2016-02-09 01:33:32