$r=\pm1$ are the only rationals with $\,r+1/r\in \Bbb Z$ (sum with its reciprocal is an integer) 
Can sum of a rational number and its reciprocal be an integer?

My brother asked me this question and I was unable to answer it. 
The only trivial solutions which I could think of are $1$ and $-1$.
As to what I tried, I am afraid not much. I have never tried to solve such a question, and if someone could point me in the right direction, maybe I could complete it on my own.
Please don't misunderstand my question. 
I am looking for a rational number $r$ where $r + \frac{1}{r}$ is an integer.
 A: It seems like you are asking for a rational number $n$ with the property that
$$n+\frac{1}{n}$$
is an integer. Let $z$ be an integer. Then we have
$$n+\frac{1}{n}=z$$
and
$$n^2+1=zn$$
$$n^2-zn+1=0$$
and by the quadratic formula,
$$n=\frac{z\pm\sqrt{z^2-4}}{2}$$
And so $z$ must be an integer, and $z^2-4$ must be a perfect square. This can only happen when $z=\pm2$, so we have
$$n=\frac{\pm2\pm\sqrt{2^2-4}}{2}$$
$$n=\frac{\pm2}{2}$$
$$n=\pm 1$$
Looks like you've found the only solutions!
A: Let $r = \frac mn$ 
so $r + \frac 1r = \frac mn + \frac nm = \frac {m^2 + n^2}{mn}$
Let $p$ be prime so that  $p|m$ but $p\not \mid n$.  Then $p\not \mid m^2 + n^2$ and $r + \frac 1r$ is not an integer.  The same would apply for any $q$ prime that divides $n$ but not $m$.  
So for $r + \frac 1r$ to be an integer $m$ and $n$ must have the same prime factors.
But we express $r = \frac mn$ "in lowest terms", then $m$ and $n$ have no prime factors in common.  So $m$ and $n$ can not have any prime factors!  There are only two numbers that do not have any prime factors.  Those are $\pm 1$.
So $r = \frac {\pm 1}{\pm 1} = 1, -1$.  The two trivial answers.  Those are the only answers. 
A: Lemma(1): Let $a$ & $b$ be integers such that $ab \mid a^2+b^2$. 
If $\gcd(a,b)=1$, then prove that $a=\pm b$. 
Proof: We claim that $ab=\pm 1$.


*

*Proof of the claim: Suppose on contrary; that $1 < |ab|$. 
So there exist a prime number $p$, which divides $ab$; i.e. $p \mid ab$. 
Without loss of generality we can assume that $p \mid a$. 
So $p$ must divides $b^2=(a^2+b^2)-a^2$. 
[Because $p$ divides both of the $(a^2+b^2)$ & $a^2$.] 
So we can conclude that $p$ must divides $b$; 
which is an obvious contradiction with the assumption that $\gcd(a,b)=1$.


So we can conclude that $a=\pm 1$ & $b=\pm 1$; which implies that $a=\pm b$.


Lemma(2): Let $a$ & $b$ be integer such that $ab \mid a^2+b^2$. 
Prove that $a=\pm b$. 
Proof: Let $d:=\gcd(a,b)$, 
so there exist integers $a^{\prime}$ & $b^{\prime}$ such that: 
$$ 
a=da^{\prime} \ , \ \ \ \ \ \ \ 
b=db^{\prime} \ , \ \ \ \ \ \ \ 
\gcd(a^{\prime},b^{\prime})=1 . $$
The relation $ab \mid a^2+b^2$, implies that there is an integer $k$, 
such that: 
$$
k(ab) = a^2+b^2  
\Longrightarrow 
k\big( (da^{\prime})(db^{\prime}) \big) = (da^{\prime})^2+(db^{\prime})^2 
\Longrightarrow 
k\big( a^{\prime}b^{\prime} \big) = (a^{\prime})^2+(b^{\prime})^2 , 
$$ 
so we obtain a pair $(a^{\prime},b^{\prime})$ such that: 
$$a^{\prime}b^{\prime} \mid (a^{\prime})^2+(b^{\prime})^2 \ , 
\ \ \ \ \ \ \ \ \ \ \ \ 
\gcd(a^{\prime},b^{\prime})=1 .$$
So by Lemma(1) we have: 
$$a=d(a^{\prime})=d(\pm b^{\prime})=\pm d(b^{\prime})=\pm b .$$




Let $\dfrac{r}{s}$ 
be an arbitrary non-zero rational number, i.e. 
$r,s \neq 0$. 
Suppose that $\dfrac{r}{s}+\dfrac{s}{r}=n$ for some integer $n$.
Then we have: $\dfrac{r^2+s^2}{rs}=n$;
which implies $rs \mid r^2+s^2$;
so we can conclude that $r=\pm s$.
A: Let $\frac{m}{n}+\frac{n}{m}=k$, where $\gcd(m,n)=1$ and $\{m,n,k\}\subset \mathbb N$.
Thus, $m^2+n^2=kmn$, which gives that $m^2$ divisible by $n$ and $n^2$ divisible by $m$.
Try to end it now.
A: Key Idea $\ r\ \&\ 1/r\,$ have integer sum & product so by RRT both are integers, so $\,r =\pm1.\,$
For convenience we reproduce the proof below, slightly generalized to $\,r\ \&\ c/r,\,$ for $\,c\in\Bbb Z$.
Lemma $ $ If  $\ r\in \Bbb Q,\,c\in\Bbb Z\ $ then $\ r + c/r = b\in\Bbb Z \iff r,\, c/r \in \Bbb Z\,\ $ [OP is  $\,c \!=\! 1\Rightarrow r=\pm1 ]$
Proof $\ (\overset{\times\ r}\Longrightarrow)\,\ \  r^2 +c = b\, r \,\overset{\rm\small RRT}\Rightarrow\,r\in \Bbb Z\,$ $\,\Rightarrow\,r\mid c\,$ by $ $ RRT = Rational Root Test. $\,\ (\Leftarrow)\ $ Clear.

Remark $ $ More generally if $\ a\, r + c/r = b\ $ for $\,a,b,c\in\Bbb Z\,$ then scaling by $\,r\,$ we deduce as above $\ a\,r^2 - b\,r + c = 0\,$ so if $\, r = e/d,\ \gcd(e,d)=1\,$ RRT $\Rightarrow  e\mid c,\  d\mid a.\,$ If $\,a,c\,$ have $\rm\color{#c00}{few}$ factors then only a $\rm\color{#c00}{few}$ possibilities exist for $\,r,\,$ e.g. if $\,a,c\,$ are primes then $\,\pm r = 1,\, c,\,1/a,\,$ or $\,c/a\,$.
[Or $\ ar\,\ \&\,\ c/r\,$ have integer sum & product so RRT $\Rightarrow$ both $\in\Bbb Z\,$ so $\,ar = ae/d\in\Bbb Z\Rightarrow d\mid a,\,$ and $\,c/r = cd/e\in\Bbb Z\Rightarrow e\mid c,\,$ by $\,d,e\,$ coprime and Euclid's Lemma].
These are special cases of ideas going back to Kronecker, Schubert and others which relate the possible factorizations of a polynomial to the factorizations of its values. In fact we can devise a simple (but inefficient) polynomial factorization algorithm using these ideas. For more on this viewpoint see this answer and its links.
A: Suppose $\frac pq+\frac qp =n$ then $p^2+q^2=pqn$ for integers $p,q,n$. As a quadratic in $p$ this is $p^2-qnp+q^2=0$ so that $$p=\frac {qn\pm \sqrt {q^2n^2-4q^2}}{2}$$ so that for the square root to yield an integer we require $n^2-4=m^2$ for some integer  $m$. The only two integer squares which differ by $4$ are $0$ and $4$, so $n=\pm 2$ and the only solutions are $p=\pm q$.
A: This is equivalent to the quadratic formula solutions, but I like it a little better.
Suppose that $r=\frac{a}{b}$, and $r+\frac{1}{r}=\frac{a}{b} + \frac{b}{a} = k$ is an integer.  We can rewrite this equation as $a^2 + b^2 = kab$, and multiplying by $4$ completing the square gives us: $$(2a-kb)^2 = (k^2 - 4)b^2$$
For this equation to hold, $k^2 - 4$ must be a square.  The squares are $0,1,4,9,\ldots$ with growing consecutive differences, so this is only possible if $k^2=4$, or $k=\pm 2$.
Finally, this gives us $(2a-kb)^2 = 0$, or $a=\pm b$.  In other words, $r=\pm 1$.
A: Suppose there does exist number $r$ with that property
$$r + \frac{1}{r}=n$$
$$ \frac{1}{r}=n-r$$
$n-r$ must be an integer since both are integers.
So we are looking for $r$ that has a property that $1/r$ is an integer. It is not hard to see that it is true only for $1,-1$.
I think this is the easiest and best solutions ( as my classmate said everyone thinks his soultion is the best)
A: $$\frac{m}{n} + \frac{n}{m} = \frac{m^{2} + n^{2}}{mn}$$
is equal to some integer $z$, provided that $m^{2} + n^{2} = kmn$ for some integer $k$; which implies that 
$$ m^{2} - kmn + n^{2} = 0;$$
which furthermore implies that $k = 2$; and so
$$(m - n)^{2} = 0;$$
Therefore, $m = n$; giving the solutions $\frac{m}{n} = \pm1.$ 
