Recently I tried to solve a diophantine equation
$n(n+1)=4m(m+1)$ with $n,m\in\mathbb{Z}$
which resulted from an other equation. But how can one show, that there are no non-trivial solutions.
Obvsiously there are four solutions. But is there an elemantary way to show, that there are no more?
Thanks in advance.
Multiplying by $4$ and rewriting, you get:
$$(2n+1)^2-1 = 4\left[(2m+1)^2-1\right]$$
Rewriting, you get: $$(4m+2)^2-(2n+1)^2 = 3$$
This means $$(4m+2n+3)(4m-2n+1)=3.$$