How can I find the integer solutions to $x^2+x-2y^2=0$? I enter this equation in Wolfram Alpha :   $x^2+x-2y^2=0$
and it gave me something like this :

and I am wondering how this solution is found and how to know if a given equation would guarantee to have integer solution.
Can anyone teach me what's behind the hood?
Thanks.
 A: Hint:
Rewrite the equation: $x^2+x=2y^2$, that is
$$x(x+1)=2y^2$$
Notice that $(x,x+1)=1$, so we can write $y^2$ as $2a^2b^2$, here $(a,b)=1$ and $2\nmid b$.
Then we get
$$\left\{\begin{array}\\x=2a^2\\x+1=b^2\end{array}\right.\text{ or }\left\{\begin{array}\\x+1=2a^2\\x=b^2\end{array}\right.$$
Hence,
$$b^2-2a^2=\pm 1$$
So we get two Pell-like equations. And you know what's going on then?
A: $$8y^2=4x^2+4x=(2x+1)^2-1\implies z^2-8y^2=1\text{ where } z=2x+1$$
Using this,
$$z_k\pm2\sqrt2y_k=(z_1\pm2\sqrt2\cdot y_1)^k$$
By observation, $z_1=\pm3,y_1=\pm1$ as $z^2-8y^2=1=3^2-8\cdot1$
If we take $z_1=3,y_1=1$
  $$z_k+2\sqrt2y_k=(3+2\sqrt2)^k \text{ and } z_k-2\sqrt2y_k=(3-2\sqrt2)^k$$
Now solve for $z_k,y_k$
Similarly for the other combinations 
$z_1=-3,y_1=-1$
$z_1=3,y_1=-1$
and $z_1=-3,y_1=1$
From the recurrence relation in Wikipedia, if $x_1,y_1,x_k,y_k$ are integers, so will be $x_{k+1},y_{k+1}$
A: Expanding on answerr by @lab bhattacharjee:
z^2 = 8.y^2 + 1 gives us equation of form x^2 = N.y^2 + 1 
which can be done simply https://en.wikipedia.org/wiki/Chakravala_method. 
started in 6th century CE India
