Finding condition for integral roots of a quadratic equation. I need to find the values of k(possible) for which the quadratic equation $$x^2+2kx+k =0$$ will have integral roots.
So I assumed roots to be $a,b$
Then I got the condition $a+b=-2k$and $a\cdot b=k$; so combining these I get $a+b+2ab=0$;
And now I need to find the integral values of $a,b$ for which this equation is satisfied,how should I procced from here??
Also is there any shorter much elegant way to do this question.
(Note-A hint would suffice)

Given $$x^2+2kx+k=0\Rightarrow x^2+2kx+k^2 = k^2-k\Rightarrow (x+k)^2=k^2-k$$
So we get $$\displaystyle (x+k)^2= \left(\sqrt{k^2-k}\right)^2\Rightarrow x+k =\pm \sqrt{k^2-k} $$
So we get $$\displaystyle x= \pm \sqrt{k^2-k}-k\;,$$ Now here $x\in \mathbb{Z}$
So let $$k^2-k=l^2\Rightarrow 4k^2-4k = 4l^2\Rightarrow (4k^2-4k+1) = (2l)^2+1$$
So we get $$(2k-1)^2-(2l)^2=1\Rightarrow (2k+2l-1)\cdot (2k-2l-1) = 1\times 1 = -1\times 1$$
$\bullet\; $ If $2k+2l-1 = 1$ and $2k-2l-1 = 1\;,$ after solving we get $(k,l) = (1,0)$
$\bullet\; $ If $2k+2l-1 = -1$ and $2k-2l-1 = -1\;,$ after solving we get $(k,l) = (0,0)$
