2026/036) Show that a positive integer m is a sum of two triangular numbers if and only if 4m+1 is a sum of two squares.

m is a sum of 2 triangular numbers 

Let the 2 triangular numbers be $t_n$ and $t_p$

We have 

$t_n = \frac{n(n+1)}{2}$

$t_p = \frac{p(p+1)}{2}$

So we have

$m = t_n +t_p =\frac{n(n+1)}{2} + \frac{p(p+1)}{2}$

Or $4m = 2n(n+1) + 2p(p+1)$

Or $4m + 1= 2n(n+1) + 2p(p+1) + 1$

 Or $4m+1 = 2n^2+2p^2 + 2n + 2p + 1$

using the fact that $2(a^2+b^2) = (a+b)^2 + (a-b)^2$ one can expand the RHS and check 

 we get $4m+1 = (p+n)^2 + (p-n)^2  + 2n + 2p + 1$

Now $4m + 1 = 2t(t+1) + 2k(k+1) + 1 = 2t^2 +2t + 2k^2 + 2k + 1$

$= (t-k)^2 + (t+k)^2 + 2(t+k) +1$

$= (t-k)^2 + (t+k+1)^2$

is sum of 2 squares.  As each step is reversible we can start from bottom and go backwards to prove the other part.