$r=\pm1$ 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.

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.
