Finding condition for integral roots of a quadratic equation.

Question

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)

Answer 1

Here is a hint to develop your existing method. Multiply the equation in $a$ and $b$ by $2$ and add a constant which enables you to factorise it.

Answer 2

HINT:

$$x=\dfrac{-2k\pm\sqrt{4k^2-4k}}2=-k\pm\sqrt{k(k-1)}$$

So we need the product of two consecutive integers namely, $k(k-1)$ to be perfect sqaure

Answer 3

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 $(k,l) = (0,0)$