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.
On
$$8y^2=4x^2+4x=(2x+1)^2-1\implies z^2-8y^2=1\text{ where } z=2x+1$$
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 $x_{k+1},y_{k+1}$
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?