Proof that $a+\frac{1}{a}\in\mathbb{Z}$ iff $a=\pm1$ I showed someone how to prove by induction that if $a+\frac{1}{a}\in\mathbb{Z}$ then also $a^n+\frac{1}{a^n}\in\mathbb{Z}$. He noted that there was no need for induction since obviously $a\in\{1,-1\}$. (Irrelevant, as the purpose of the assignment was to use induction.)
I've had trouble trying to prove this seemingly trivial fact. (For example, one sub-trouble was trying to show that if $a$ is irrational then so is $a+\frac{1}{a}$.)
I thought of assuming $a\in\mathbb{R}$, but if you have any other interesting assumption please share it. Note that this fails for $\mathbb{C}$ (since $i+\frac{1}{i}=0$).
 A: As already mentioned, this is false. However, if you assume $a$ is an integer, you can write
$$ a+ \frac{1}{a} = \frac{a^2+1}{a} =n$$
so $a|a^2+1$ and $a|a^2$ as well, hence $a | (a^2+1-a^2) = 1$. The only integers that satisfy this are $a=\pm 1$.
Similarly for $a = p/q \in Q$, $p,q$ coprime, we can write
$$\frac{p^2+q^2}{pq} = n$$
so that $p | p^2 + q^2$, hence $p|q^2$. But $\gcd(q,p) =1$, so $p | 1 \cdot q^2$ implies $p | 1$ or $p = \pm 1$. Similarly, we can show $q | p^2 + q^2$ and hence $q | p^2$ to conclude $q = \pm 1$ as well. So in this case, if $a$ is rational, we must have $a = \pm 1$ as well.
A: Take $a=\dfrac{3-\sqrt{5}}{2}$, then $a+\dfrac{1}{a}=3$.
A: This is false as you can see by solving the quadratic equation $a^2 -na+1 = 0$ which has real solutions for all values of $n$ except $n=0$.
A: This is false.  For $f(x)=x+1/x$ then $f$ is continuous on $(0,\infty)$.  Furthermore $\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow\infty} f(x)=\infty$ and $f(1)=2$, whence for every integer $n\geq 2$ there exists a real number $a$ such that $f(a)=n$ by the intermediate value theorem.  Similarly on $(-\infty,0)$.
A: It is a theorem that if you have two reduced fractions (that is, the greatest common divisor of the numerator and denominator is 1), then they have equal denominators if their sum is an integer. Of course, from your expression, it is pretty clear that they are both reduced fractions. So, this means that their denominators are equal. This implies that $a=1$. Of course, this is only considering positive integers, but this same thing can be extended to negative integers by considering the absolute value. Thus, we have that $$|a|=1 \Rightarrow a = \pm 1$$ 
This is assuming that a is an integer, though. If it is not, this does not hold.
A: A simple counterexample is $\ \ 4\, =\, 2+\sqrt{3}\,+\, \overbrace{\dfrac{1}{2+\sqrt{3}}}^{\Large 2\ -\ \sqrt 3}$
There are infinitely many more like that
$$ 2n\, =\, n+\sqrt{n^2\!-\!1}\,+\, \overbrace{\dfrac{1}{n+\sqrt{n^2\!-\!1}}}^{\Large n\ -\ \sqrt{n^2-1}}\quad$$
