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 

You are on the right track. We need
$$(a+k-8)(a-k-8) = 28$$
Now note that $(a+k-8)$ and $(a-k-8)$ are of the same parity (Why?) and the product is even. Hence, both have to be even. Hence, the possible cases are:
$1$. $(a+k-8)(a-k-8) = 2 \cdot 14$
$2$. $(a+k-8)(a-k-8) = (-2) \cdot (-14)$
$3$. $(a+k-8)(a-k-8) = 14 \cdot 2$
$4$. $(a+k-8)(a-k-8) = (-14) \cdot (-2)$
Now solve each to get possible values of $a$.
