0
$\begingroup$

This question already has an answer here:

Suppose that $r$ is a real number such that $r + \frac{1}{r}$ is an integer. Prove that $r^{2017} + \frac{1}{r^{2017}}$ is an integer.

I thought of using prove by induction, but I don't think it really suits in this case. I've also thought of using modulos in some kind of way but not sure how. Also, can we assume that both $r^{2017}$ and $\frac{1}{r^{2017}}$ are both integers, since adding them make up an integer?

$\endgroup$

marked as duplicate by Ross Millikan number-theory Mar 29 '18 at 2:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ "can we assume that two numbers are both integers, since their sum is an integer?" no, 1/2 + 1/2 is an integer, but neither summand is. $\endgroup$ – ziggurism Mar 29 '18 at 2:13
  • $\begingroup$ Another duplicate is here $\endgroup$ – Ross Millikan Mar 29 '18 at 2:17
  • $\begingroup$ sorry about that. didn't realize this question have been asked. $\endgroup$ – Stupid Wannabe Mathematician Mar 29 '18 at 2:26
2
$\begingroup$

$(r+1/r)^2=r^2+1/r^2+2$ implies that $r^2+1/r^2$ is an integer.

Assume $r^k+1/r^k$ is integer for all $k<n$.

Then $(r+1/r)^n=r^n+1/r^n+\binom{n}{1}(r^{n-2}+1/r^{n-2})+\binom{n}{2}(r^{n-4}+1/r^{n-4})+...$. Therefore, $r^n+1/r^n$ is an integer.

$\endgroup$
1
$\begingroup$

If $r + 1/r = m$, then $r$ and $1/r$ are both roots of the quadratic $x^2 - m x + 1$. Then $a_n = c r^n + d/r^n$ satisfies the recursion $a_{n+2} - m a_{n+1} + a_n = 0$ for any constants $c$ and $d$, in particular for $c=d=1$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.