Re: [math-fun] sum of 2 triangular numbers is a triangular number

I mentioned this on Feb 19 2020 and several people sent very helpful replies. For squares, we have the theorem that all solutions to x^2+y^2=z^2 in integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. Expanding on the replies from Michael Collins and Rich, it looks like the following is the analog for triangular numbers T_n = n(n+1)/2. ALL solutions to T_n+T_k=T_m are given by the list of what one might call Trithagorean triples: these are the triples [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q is any odd divisor of T less than n, plus these triples with the first two coordinates swapped. I haven't found this in the literature, but it can hardly be new.