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I enter this equation in Wolfram Alpha : $x^2+x-2y^2=0$ and it gave me something like this :

enter image description here

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.

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It looks like the problem was done by completing the square on the $x$ variable and then using "Pell's equation". I'd suggest looking up the Pell equation to see what's going on. –  coffeemath Jan 30 '13 at 10:34
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2 Answers

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?

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$$8y^2=4x^2+4x=(2x+1)^2-1\implies z^2-8y^2=1\text{ where } z=2x+1$$

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 $x_{k+1},y_{k+1}$

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