## An integral expression

A real number $$x$$ is given such that $$x+\frac{1}{x}$$ is an integer. Prove that for every natural number $$n$$ the value $$x^n+\frac{1}{x^n}$$ is also an integer.

• #### Solution

For $$n=1$$ the assumption applies directly.

For $$n=2$$ we proceed from the formula: $$\displaystyle\left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}$$.

The left side and the number 2 are integers, so $$x^2+\frac{1}{x^2}$$ must also be an integer.

For $$n+1$$ we proceed from the formula

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

which we may transform to

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

By the inductive hypothesis for $$n$$ and $$n-1$$, the expression on the right-hand side is a product of integers minus an integer, and is hence an integer.  