<s>[INST] $x+\frac{1}{x}$ is an integer Is the following Proof Correct? In particular please comment on the correctness of the given formulas.
Theorem. Given that $x$ is a real number, $x\neq 0$, and $x + \frac{1}{x}$ is an integer. For all $n\ge 1$, $x^n+\frac{1}{x^n}$ is an integer.
Proof. We construct the proof by recourse to Strong-Induction.  Assume for an arbitrary $n\in\mathbf{Z^+}$ that $x^k+\frac{1}{x^k}$ is an integer for any positive integer $k$ strictly less than $n$. Now Consider the following cases.
Case-1: If $n$ is even then for some $l\in\mathbf{Z^+}$, $n = 2l$ thus the Binomial-Theorem implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l-1}\binom{n}{2j}\left(x^{2j}+\frac{1}{x^{2j}}\right)-\binom{n}{l} = x^n+\frac{1}{x^n}$$
from the inductive hypothesis we know that $x+\frac{1}{x}\in\mathbf{Z}$ which implies that $(x+\frac{1}{x})^n \in\mathbf{Z}$ in addition it also follows from the inductive hypothesis that $x^{2j}+\frac{1}{x^{2j}}\in\mathbf{Z}$ for $j\in\{1,2,3,...,l-1\}$ moreover we also that $\binom{n}{r}\in\mathbf{Z^+}$ is always a positive integer implying that $x^n+\frac{1}{x^n}$ is an integer.
Case-2: If $n$ is odd then for some $l\in\mathbf{Z^+}$, $n = 2l+1$ thus the Binomial-Theorem implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l}\binom{n}{2j-1}\left(x^{2j-1}+\frac{1}{x^{2j-1}}\right) = x^n+\frac{1}{x^n}$$
and by using the same reasoning as in the previous case we can deduce that $x^n+\frac{1}{x^n}$ is an integer.
 [/INST] I think it is easier to see that

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

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