If the sum and product of two rational numbers are both integers, then the two rational numbers must be integers. I would appreciate if somebody could help me with the following problem.
Prove that if the sum and product of two rational numbers are both integers, then the two rational numbers must be integers.
Thanks!
 A: Let $p,q$ be rational. Then $p+q=n\in\mathbb{Z}\implies p=n-q$. So let $\displaystyle q=\frac{a}{b}$ be in lowest terms. We then have $(n-\frac{a}{b})\frac{a}{b}=m\in\mathbb{Z}\implies na-\frac{a^2}{b}=mb\implies\frac{a^2}{b}\in\mathbb{Z}\implies \frac{a}{b}\in\mathbb{Z}$ since $\frac{a}{b}$ are coprime. So $q$, hence also $p$ are integers.
A: By plugging in $x=a$ and $x=b$, we see that
$$
x^2-(a+b)x+ab=0
$$
As shown in this answer, a rational root of a monic polynomial with integer coefficients must be an integer.

Importing the Referenced Answer
It has been suggested that specializing the proof in the above referenced answer to quadratic polynomials might be useful.
Suppose $x=\frac pq$, where $ps+qr=1$, is a root of $x^2+mx+n=0$, where $m,n\in\mathbb{Z}$.
Subsitute $x=\frac{1-qr}{qs}$
$$
\frac{(1-qr)^2}{q^2s^2}+m\frac{1-qr}{qs}+n=0
$$
Multiply by $pqs^2$
$$
\left(\frac pq-2pr+pqr^2\right)+pms(1-qr)+npqs^2=0
$$
cancelling yields
$$
\frac pq=2pr-pqr^2+pms(qr-1)-npqs^2
$$
In particular, $x=\frac pq\in\mathbb{Z}$.
A: This may not be the simplest proof, but I think it's pretty.  Let your two rational numbers be $r,s$.  Then the polynomial $f(x)=(x-r)(x-s)=x^2-(r+s)x+rs$ has integer coefficients by your hypotheses.  By Gauss' Lemma, this polynomial must be reducible over the integers, and hence $r,s$ are both integers.
A: Starting with the equation provided by Vadim above, $x^2-(r+s)x + rs=0$, take any rational root $a/b$, with $\gcd(a,b)=1$.  We get, after clearing the denominator:
$a^2 -b(r+s)x +rsb^2=0$,which can be rewritten to show  $a^2$ is a multiple of $b$ which contradicts the fact that $a$ and $b$ are co-prime. (This is essentially the proof that a rational number that is an algebraic integer is an integer, for the quadratic case). 
A: $\underbrace{\dfrac{a}b\!+\!\dfrac{c}d}_{\large (a,b)\,=\,1}\! =n\in\Bbb Z,\ \dfrac{a}b\not\in\Bbb Z\overset{\large \exists\, p\ {\rm prime}}\Rightarrow$  $\begin{array}{}p\nmid a\\ p\mid b\end{array}$ $\Rightarrow$ $\,\dfrac{a}b\dfrac{c}d = \dfrac{a}b\!\!\!\dfrac{\,(bn\!-\!a)}{b}\!\not\in\Bbb Z\,$ by $\,\begin{array}{} p\nmid a,\, bn\!-\!a\\ p\mid b\end{array}\ $ QED 
