An integral expression
Task number: 3322
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.
