Let $n,d$ be positive integers and $p$ a positive prime number.
How can I show that: $$ \gcd(x^{p^n} - x , x^{p^d} - x ) = x^{p^{\gcd(n,d)}} - x $$
Can someone at least give me some hint? I'm kinda lost at the moment.
Here's where I've got:
Suppose w.l.o.g that $n > d$, because if $n=d$ than the result is trivial. Therefore $n = qd + r$. Hence: $$ x^{p^n} = x^{p^{qd + r}} = x^{p^{qd}} x^{p^{r}} \implies x^{p^n} - x = x^{p^{qd}} x^{p^{r}} - x $$ (✰) Now if there's a way to show that: $$ x^{p^n} - x = q(x)\big(x^{p^d} - x \big) + x^{p^r} - x $$ Then it would follow that $$ \gcd(x^{p^n} - x , x^{p^d} - x) = \gcd(x^{p^d} - x , x^{p^r} - x) $$ Then we would have a relation between eculidean algorithm with the integers $n$ and $d$ and the polynomials, so the result would be analogous, i.e, $x^{p^{\gcd(n,d)}} - x$.
I'm actually stuck at (✰).
I've been trying to show that for hours but still can't figure it out... Thanks.
