Let \(b_1 < b_2 < b_3 < \dots\) be the sequence of all natural numbers which are sum of squares of two natural numbers. Prove that there exists infinite natural numbers like \(m\) which \(b_{m+1}-b_m=2015\).
It's easy to force something to not be a sum of squares through CRT - the real challenge is forcing \(x\) and \(x + 2015\) to both be sums of squares. Therefore, we force this algebraically, and hope that this is enough to win.