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This question already has an answer here:

Suppose that there is a real number $r$ such that $r+\frac{1}{r}$ is an odd integer.Then $r$ is irrational.

Let $r\in \Bbb Q$ then $r=\dfrac{p}{q}\implies \dfrac{p}{q}+\dfrac{q}{p}=$odd

$\implies \dfrac{p^2+q^2}{pq}$ is odd integer.

Since $r+\frac{1}{r}$ is an odd integer$\implies pq\mid p^2+q^2$

How to derive a contradiction from here?

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marked as duplicate by Bill Dubuque abstract-algebra Sep 29 '17 at 14:42

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    $\begingroup$ $p$ and $q$ have no common factor (co-primes). $\endgroup$ – Ahmad Sep 29 '17 at 14:07
  • $\begingroup$ People answering this question should note that it does not specify that the real numbers or rationals or integers involved are positive. $\endgroup$ – Mark Bennet Sep 29 '17 at 14:23
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You should have stated that $\frac pq$ is in lowest terms. $pq|p^2+q^2$ implies that $p$ divides $q$ and that $q$ divides $p$.

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  • $\begingroup$ Does this use the oddness of $r+\frac{1}{r}$ or just that it is not $2$ with $r=1$? $\endgroup$ – Henry Sep 29 '17 at 14:09
  • $\begingroup$ @Henry It implies $p=q$, hence $r=1$ and $r+1/r=2$, which contradicts oddness. $\endgroup$ – Aaron Sep 29 '17 at 14:10
  • $\begingroup$ @Henry It is not using oddness at all, but divisibility properties of the integers. You should note that $r=-1$ is another possibility. The conclusion is that no odd integers are possible for rational $r$. Neither are most of the even integers. $\endgroup$ – Mark Bennet Sep 29 '17 at 14:21
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We may assume $\gcd(p,q)=1$. Then $p|pq$ and $pq|p^2+q^2$ implies $p|p^2+q^2$ and subsequently, since $p|p^2$, $p|q^2$. Since $p|q^2$ and $p|p^2$, we know $p|\gcd(p^2,q^2)=\gcd(p,q)^2=1$. Thus $p=1$. Then we have $\frac{p^2+q^2}{pq}=\frac{1+q^2}{q}$. Since $q|q^2$ and $q|1+q^2$, we arrive at $q|1\rightarrow q=1$. Then $r=\frac{p}{q}=1$, but $1+\frac{1}{1}$ is not an odd integer.

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We have $$ \frac{{p^{\,2} + q^{\,2} }}{{pq}} = \frac{{\left( {p + q} \right)^{\,2} - 2pq}}{{pq}} = \frac{{\left( {p + q} \right)^{\,2} }}{{pq}} - 2 $$

so $$ \frac{{\left( {p + q} \right)^{\,2} }}{{pq}} = n\quad \Rightarrow \quad \left( {p + q} \right)^{\,2} - n\,pq = 0 $$ and the parity table tells you that n cannot be odd $$ \begin{array}{c|lcr} {p\backslash q} & & {\rm o} & {\rm e} \\ \hline {\rm o} & & {n\,{\rm even}} & \emptyset \\ {\rm e} & & \emptyset & {n\,{\rm even}} \\ \end{array} $$

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