$r+r^{-1}$ integral implies $r^n+r^{-n}$ integral

Question

Suppose r + 1/r is an integer. r is real and positve. How to prove r^n + 1/r^n is an integer by induction for all natural numbers n.

Answer 1

Consider $(r+\frac{1}{r})^n$ which you know is an integer. If you expand this out and group together terms of the form $r^i + \frac{1}{r^i}$ you see that $$(r+\frac{1}{r})^n = \sum_{i=0}^{\left \lfloor{\frac{n}{2}}\right \rfloor } \binom{n}{i} \left(r^{n-2i} + \frac{1}{r^{n-2i}} \right)$$

You know that the left hand side is an integer and by induction you have that all of the terms on the right hand side except for the leading term $r^{n} + \frac{1}{r^{n}}$ are integers. Hence $r^{n} + \frac{1}{r^{n}}$ must be an integer as well.

Answer 2

Try the following. By hypothesis if $n=1$ then the relation holds. Suposse that for every $m$ such that $m\leq n$ the relation holds. Therefore consider the product $$(x+x^{-1})(x^n+x^{-n}) $$ and deduce the result from the induction hypothesis.

Answer 3

Apply mathematical induction using the identity: $$x^{n+2}+x^{-n-2} = (x+x^{-1})(x^{n+1}+x^{-n-1})-(x^{n}+x^{-n})$$