If $x^2 + ax + b =0$ has an integer solution then show that it divides $b$

Question

If $x^2 + ax + b =0$ has integer solution then show that it divides $b$

Now $$x = \frac {-a + \sqrt{a^2 - 4b}}{2}$$ Since $x$ has integer solution so $2$ must divide numerator. So $a =2k$ and $\sqrt{a^2 - 4b}=2m$. How do i proceed?

Answer 1

Suppose it has integer solution $\alpha$ and $\beta$, then we have

$$(x-\alpha)(x-\beta)=0$$

That is $$x^2-(\alpha+\beta)x + \alpha\beta = 0$$

That is $\alpha \beta = b$

Answer 2

$$b=-x(x+a)$$ so that both $x$ and $x+a$ divide $b$.

Answer 3

The question (when $a,b\in\mathbb{Z}$) and its proof are a special case of a more general result.

Let $p=x^n+a_{n-1}x^{n-1}+\ldots+a_1 x+a_0$ be a polynomial of degree $n\geq 1$ with integer coefficients. Then $p(u)=0$ and $u\in\mathbb{Z}$ implies that $u$ is a divisor of $a_0$.

The proof immediately follows from the fact that $u$ divides $a_l u^l$ for all $l\geq 1$.