Square and triangular numbers are expressed as
Further on this can be expressed as
Taking $a=2m+1$ and $b=2n$ the expression becomes
$2b^2=a^2-1$ or $1=a^2-2b^2$
After a bit of factoring previous equation becomes
One of the solutions is $(a,b)=(3,2)$ and $(m,n)=(1,1)$. From here additional solutions can be found recursively.
Once there is a solution say $(m,n)$ there is another $(1+im+jn, 1+km+ln)$ for some integers $i,j,k,l$. I need help proving this.