The zeroes of the function $f(x)=x^2-ax+2a$ are integers. Find all possible values of a.

Question

So far I have only got 9 from just guess and check. I am thinking of using Vieta's Formula, but I am struggling over the algebra. Can someone give me the first few steps?

Answer 1

Let $u$ and $v$ be the zeroes, then $u+v=a$ and $uv=2a$. That is, $uv = 2(u+v)$, or $u(v-2) = 2v$. Which means $v \neq 2$, and further, $u = \dfrac{2v}{v-2} = 2 + \dfrac{4}{v-2}$. Which means $v-2$ must be one of the divisors (positive or negative) of $4$. So you have six candidates, and all of them WILL provide values for $a$, you just have to compute them.

Answer 2

$u+v=a$ and $uv=2a$ lead to $uv-2u-2v=0$, hence to $(u-2)(v-2)=4$.
$4$ cannot be written as the product of two integers in too many ways, just $1\cdot 4$, $2\cdot 2$, $4\cdot 1$, $(-1)\cdot(-4)$, $(-2)\cdot(-2)$, $(-4)\cdot (-1)$. It follows that $$(u,v)\in\{(3,6),(4,4),(6,3),(1,-2),(0,0),(-2,1)\}$$ and $$ a \in \{-1,0,8,9\}. $$