Prove that rational numbers $a,b$ are integers if $a+b$ and $ab$ are integers

Question

I have been trying to prove this via divisibility, assuming that $a=\frac{n}{m}$ and $b=\frac{r}{q}$ for some $n,m,r,q$ in

Ints($m$,$q$ not $0$), but I'm completely stuck here. Any help?

Answer 1

Hint: Apply the rational root theorem to

$$ x^2 - ( a + b) x + ab $$

Answer 2

Let $n:=a+b$ and $m:=ab.$ Then $a^2-an+m=0$ and hence $$a=\dfrac{n\pm\sqrt{n^2-4m}}{2}.$$ note that since $a\in\Bbb Q,$ then $n^2-4m$ must be a perfect square.

Then necessarily $2\mid n\pm\sqrt{n^2-4m}$ (if $n$ is odd, then $n^2-4m$ is odd $\Longrightarrow$ $n\pm\sqrt{n^2-4m}$ is even; if $n$ is even, then $n^2-4m$ is even $\Longrightarrow$ $n\pm\sqrt{n^2-4m}$ is even).