<s>[INST] Integer $a$ , If $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots If the equation $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots. Then the integer value of $a$ is
$\bf{My\; Try}::$ Let $\alpha,\beta\in \mathbb{Z}$ be the roots of the equation . Then $\alpha+\beta = (6-a)$ and $\alpha\cdot \beta = a$
Now $D = (a-6)^2-4a = $ perfect square. So $a^2+36-12a-4a = k^2$. where $k\in \mathbb{Z}$
$a^2-16a+36=k^2\Rightarrow (a-8)^2-28=k^2\Rightarrow (a-8)^2-k^2 = 28$
Now How can i solve after that
Help Required
Thanks 
 [/INST] From the first line of your solution, $\alpha.\beta+\alpha+\beta=6-a+a=6$, so $(\alpha+1)(\beta+1)=7$
What are the factors of 7?
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