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...
 A: Solving it as quadratic equation in terms of $x$ we have:
$$x^2 + x -2y^2 - 2y = 0$$
$$x= \frac{-1 \pm \sqrt{1 + 8y^2 + 8y}}{2}$$
Now we have that $8y^2 + 8y + 1 = m^2$, as $x$ is an integer. Again solving for $y$ we have:
$$y = \frac{-8 \pm \sqrt{64 + 32m^2 - 32}}{16}$$
So we must have that $2(m^2 + 1)$ is a square. So we get $n^2 - 2m^2 = 2$. This Diophantene Equation has infinitely many solutions. For example:
$$n_k + m_k\sqrt{2} = (10 + 7\sqrt{2})(3 + 2\sqrt{2})^k$$
generates infinitely many. Now note that the solutions for $n$ are even, but never divisible by $4$, so we have $m^2 + 1 = 2s^2$, where $s$ is odd. So:
$$y = \frac{-8 \pm 8s}{16} \in \mathbb{Z}$$
$$x = \frac{-1 \pm \sqrt{2s^2 - 1}}{2} \in \mathbb{Z}$$
So we have that each solution of $n^2 - 2m^2 = 2$ generates a solution of $x(x+1)=2y(y+1)$, so there are infinitely many of them. Some examples are: $(x,y) = (3,2),(20,14),(119,84),(696,492) \dots$
A: $$x(x+1) = 2y(y+1) \Leftrightarrow 4x^2+4x=2(4y^2+4y)\Leftrightarrow \\4x^2+4x+1=2(4y^2+4y+1)-1 \Leftrightarrow (2x+1)^2-2(2y+1)^2=1$$
This is the Pell equation, you want the odd solutions to it.
