It seems to overlap with other content.

Question

It seems to overlap with other content. Sorry for the confusion.

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

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.

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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, $x=\frac pq\in\mathbb{Z}$.

Answer 5

$\underbrace{\dfrac{a}b\!+\!\dfrac{c}d}_{\large (a,b)\,=\,1}\! =n\in\Bbb Z,\ \dfrac{a}b\not\in\Bbb Z\overset{\large \exists\, p\ {\rm prime}}\Rightarrow$ $\begin{array}{}p\nmid a\\ p\mid b\end{array}$ $\Rightarrow$ $\,\dfrac{a}b\dfrac{c}d = \dfrac{a}b\!\!\!\dfrac{\,(bn\!-\!a)}{b}\!\not\in\Bbb Z\,$ by $\,\begin{array}{} p\nmid a,\, bn\!-\!a\\ p\mid b\end{array}\ $ QED