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$x^n + \frac{1}{x^n} \in \mathbb{Z}$ (is an integer), for all positive integers $n$, where $x$ is rational.

I've surmised that the only rational numbers that satisfy $x$ are 1 and -1.

Thus, as you grow in size with $n$, the answer will always either be 1 or -1.

If $x = 1$, the result will always be 1. if $x = - 1$, the result will be positive 1 if $n$ is even and -1 if $n$ is odd.

But what is the inductive step?

$x^{n+1} + \frac{1}{x ^ {n+1}} = (x^n)(x) + \frac{1}{(x^n)(x)} \in \mathbb{Z}$ (is an integer).

rubberchicken
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compguy24
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2 Answers2

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Hint $\,\ x = \pm1\,\Rightarrow\, x^{-1} = x.\,$ That times $\,x^{-n} = -x^{n} + k\,$ yields $\,x^{-(n+1)} = -x^{n+1}\! + kx$

Bill Dubuque
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HINT:

$$\left(x^n+\frac1{x^n}\right)\left(x+\frac1{x}\right)=\left(x^{n+1}+\frac1{x^{n+1}}\right)+\left(x^{n-1}+\frac1{x^{n-1}}\right)$$

$$\iff x^{n+1}+\frac1{x^{n+1}}=\left(x^n+\frac1{x^n}\right)\left(x+\frac1{x}\right)- \left(x^{n-1}+\frac1{x^{n-1}}\right)$$

lab bhattacharjee
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