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I've come across a proof where they use the fact that for $p$ prime, $r, m \in \mathbb{N}$ such that $r|m$ it holds that $(p^r - 1)|(p^m-1)$. I've tried to understand why this must be true, but haven't managed to prove it. Does anyone know how to prove it? And does the result hold for more general cases than $p$ being prime, too?

It further says that from $(p^r - 1)|(p^m-1)$ it follows that $(T^{p^r}-T)|(T^{p^m}- T)$. Why's that?

Studentu
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Write $m= kr+ n$ where $0\leq n<r$ Then we have $$p^{kr}p^n \equiv 1 \pmod{p^r-1}$$ Since $p^r \equiv 1 \pmod{p^r-1}$ we have now $$p^n \equiv 1 \pmod{p^r-1}\implies p^r-1\mid p^n-1$$

so if $n\ne 0$ then $$p^r-1\leq p^n-1 \leq p^{r-1}-1\implies p\leq 1 $$ A contradiction, so $r\mid m$.

The other way, if $r\mid m$ then $m=kr$ so we have $$p^m-1 = (p^r)^k-1 = (p^r-1)\Big((p^r)^{k-1}+(p^r)^{k-2}+...p^r+1\Big)$$

nonuser
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  • Thanks for your answer. Can you please explain why the implications in the first four lines of it and the fact that $(2^r - 1)|(p^m - 1)$ do hold? – Studentu Nov 18 '21 at 16:27
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    Typo............. – nonuser Nov 18 '21 at 16:41
  • Oh I see, thank you! Could you please explain the second part, too, i.e., why for $(p^r - 1)|(p^m - 1)$ it follows that the polynomial $T^{p^r}-T$ divides the polynomial $T^{p^m}-T$? – Studentu Nov 20 '21 at 02:17