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 
 A: 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 $a$.
A: 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?
A: Another way:
We can set  $a^2-16a+36$ to $(a-n)^2$ where $n$ is some integer
$\displaystyle \implies n^2=36-16a+2an$ which is even $\implies 2|n^2\iff 2|n$ as $2$ is prime.
So, we can set $n=2m$ where $m$ is some integer
$\displaystyle \implies 4m^2=36-16a+4am\implies a=\frac{m^2-9}{m-4}=m+4+\frac7{m-4}$
$\displaystyle\implies m-4$ must divide  $7$ and $m-4\ne0$ as $a$ is finite
A: Hint: The question tells you the value of $a$ is unique: you just need to find it. You don't have to go through the effort of obtaining a "complete solution". Have you tried just trying examples?
Hint 2: They ask for a nonzero value of $a$. Maybe $a=0$ is special, and studying that particular case will be informative.
