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$x\in \mathbb R\;\;$ such that $\;\;x+\frac{1}{x} \in \mathbb Z.$

Prove by induction: $$x+\frac{1}{x} \in \mathbb Z\implies x^n+\frac{1}{x^n} \in \mathbb Z, \;\;\;\forall n\in \mathbb N$$

I used the $complete/strong$ induction.

The base $n=1$ is already in the statement.

Assumption: Let $P(n)\ldots,\;x^n+\frac{1}{x^n} \in \mathbb Z$ be true, as well as $$P(1), P(2),...P(n-2),P(n-1).$$ Step:$\;P(n+1)$ is true.

$x+\frac{1}{x} \in \mathbb Z\;\wedge x^n+\frac{1}{x^n} \in \mathbb Z\implies \Big(x+\frac{1}{x}\Big)\Big(x^n+\frac{1}{x^n}\Big)\in \mathbb Z$

$x^{n+1}+\frac{1}{x^{n-1}}+x^{n-1}+\frac{1}{x^{n+1}}=\underbrace{\Big(x^{n-1}+\frac{1}{x^{n-1}}\Big)}_{\in\mathbb Z}+\Big(x^{n+1}+\frac{1}{x^{n+1}}\Big)$

$x^{n+1}+\frac{1}{x^{n+1}}=\Big(x+\frac{1}{x}\Big)\Big(x^n+\frac{1}{x^n}\Big)-{\Big(x^{n-1}+\frac{1}{x^{n-1}}\Big)}\in\mathbb Z$

PinkyWay
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1 Answers1

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Here is a slightly different take.

Let $a_n=x^n+\frac{1}{x^n}$. Then $a_{n+2}=b a_{n+1}-a_n$, because $x$ and $\frac{1}{x}$ are roots of $t^2=bt-1$, where $b=x+\frac{1}{x}$.

Since $a_0=1 $ and $a_1=b$ are integers, it follows by induction that $a_n$ is an integer for all $n$.

This is exactly your proof!

lhf
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