Solve over integers: $x(x+1) = 2y(y+1)$ Solving $x(x+1) = 2y(y+1)$
I've got this :
y = $\frac{1}{2} \sqrt{2x^2+2x+1}$ so $2x^2+2x+1 = n^2$
so $x \equiv 0 \pmod 4$ or $x \equiv -1 \pmod 4$ and I totally stuck...

$$x(x+1) = 2y(y+1) \Leftrightarrow 4x^2+4x=2(4y^2+4y)\Leftrightarrow \\4x^2+4x+1=2(4y^2+4y+1)-1 \Leftrightarrow (2x+1)^2-2(2y+1)^2=1$$
This is the Pell equation, you want the odd solutions to it.
