Simple proof that $a$ is coprime Prove that if $a$ divides $x^n-1$ and $x^m-1$, then $a$ is coprime with $x$.
I think this should be easy but I can't think of a way to do it.
 A: Assume that $n\ge 1$. Let $d$ be a common divisor of  $a$ and $x$. 
(i) Since $a$ divides $x^n-1$, it follows that $d$ divides $x^n-1$. 
(ii) Since $d$ divides $x$, it follows that $d$ divides $x^n$. 
From (i) and (ii), we conclude that $d$ divides $x^n-(x^n-1)$, and therefore $d$ divides $1$.   
We have shown that any common divisor $d$ of $a$ and $x$ divides $1$. So $a$ and $x$ cannot have a common divisor greater than $1$, and therefore $a$ and $x$ are relatively prime. 
A: As a matter fact, I fiddled around with the problem and got this: 
Let $m>n$  then if $a$ divides $x^m-1$ and $x^n-1$, it divides $(x^m-1)-(x^n-1) = x^m-x^n = x^n(x^{m-n} - 1)$ Now if $a$ divides $x^n$ then clearly they're not coprime. Let's assume it divides $x^{m-n}-1$ instead, then it must divide $(x^{m-n}-1)-(x^n-1)=x^{m-n}-x^n = x^n(x^{m-2n}-1)$ Now again, if it doesn't divide $x^n$ then it must divide the other factor, and we can keep going on like this till $m-kn<0$ and at that point $a$ must divide $x^n$ which means it is not coprime with $x$.
Combining this with André Nicolas's proof, I guess it means that $x^m-1$ and $x^n-1$ are relatively prime so the hypothesis isn't even true . No such $a$ exists.
