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**tukeywilliams** For a real number $\displaystyle x $ suppose that $\displaystyle x + \frac{1}{x} $ is an integer. Prove that $\displaystyle x^{n} + \frac{1}{x^{n}} $ is an integer for all positive integers $\displaystyle n $.

Here is what I did:

Proof: We use induction on $\displaystyle n $. Base case: $\displaystyle x^{1} + \frac{1}{x^{1}} $ is an integer. Inductive step: Suppose that $\displaystyle x^{k} + \frac{1}{x^{k}} $ is an integer for some positive $\displaystyle k $. Then $\displaystyle \left(x^{k} + \frac{1}{x^{k}}\right)\left(x+\frac{1}{x}\right) = \frac{x^{k+2}+x^{k}+x^{2-k}+x^{-k}}{x} $ $\displaystyle = x^{k+1} + x^{k-1} + x^{-k+1} + x^{-k-1} $ $\displaystyle = \left(x^{k+1} + \frac{1}{x^{k+1}}\right) + \left(x^{k-1} + \frac{1}{x^{k-1}}\right) $ (which is an integer by induction hypothesis because an integer multiplied by an integer is an integer). From here, how would I show that $\displaystyle x^{k+1} + \frac{1}{x^{k+1}} $ is an integer?

Thanks