
96.2. Determine all real numbers $x$, such that

$$
x^{n}+x^{-n}
$$

is an integer for all integers $n$.


Solution. Set $f_{n}(x)=x^{n}+x^{-n}$. $f_{n}(0)$ is not defined for any $n$, so we must have $x \neq 0$. Since $f_{0}(x)=2$ for all $x \neq 0$, we have to find out those $x \neq 0$ for which $f_{n}(x)$ is an integer foe every $n>0$. We note that

$$
x^{n}+x^{-n}=\left(x+x^{-1}\right)\left(x^{n-1}+x^{1-n}\right)-\left(x^{n-2}+x^{2-n}\right)
$$

From this we obtain by induction that $x^{n}+x^{-n}$ is an integer for all $n>1$ as soon as $x+x^{-1}$ is an integer. So $x$ has to satisfy

$$
x+x^{-1}=m
$$

where $m$ is an integer. The roots of this quadratic equation are

$$
x=\frac{m}{2} \pm \sqrt{\frac{m^{2}}{4}-1}
$$

and they are real, if $m \neq-1,0,1$.
