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If $p \nmid r$ is an odd prime, show $$p^3 \mid (r^{p(p-1)} - 1) \Rightarrow p^2 \mid (r^{p-1}-1)$$


My attempt:

$$r^{p(p-1)}-1 = (\color{red}{r^{p-1}-1})(\color{blue}{r^{(p-1)(p-1)} +r^{(p-1)(p-2)} + \cdots + r^{(p-1)2 }+r^{p-1}+1}) = \color{red}{X}\color{blue}{Y}$$

$p \mid \color{red}{X}$ by Euler,
and $\color{blue}{Y} \equiv (\underbrace{1+1+\cdots}_{p-1\text{ times}})+1 =p\equiv 0 \pmod p$.
So the right hand side $\color{red}{X}\color{blue}{Y}$ is divisible by $p^2$.
How to show the other $p$ doesn't divide $\color{blue}{Y}$ ?

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2 Answers 2

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I don't offhand see any way to use what you did to prove the result. Instead, Fermat's little theorem shows that

$$r^{p-1} = ap + 1 \tag{1}\label{eq1A}$$

for some integer $a$. This then gives

$$\begin{equation}\begin{aligned} \left(r^{p-1}\right)^{p} - 1 & = (ap + 1)^{p} - 1 \\ & = (ap)^p + p(ap)^{p-1} + \ldots + \frac{p(p-1)}{2}(ap)^2 + p(ap) + 1 - 1 \\ & = p^2\left(a^{p}p^{p-2} + a^{p-1}p^{p-2} + \ldots + \frac{p-1}{2}(a^2)(p) + a \right) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Note every term inside the brackets has a factor of $p$ except for the last one, i.e., $a$. Since $p^3 \mid (r^{p(p-1)} - 1)$, this means $p \mid a$. Thus, \eqref{eq1A} gives that

$$p^2 \mid r^{p-1} - 1 \tag{3}\label{eq3A}$$

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You can note that $$ Y=(r^{p-1}-1)[(r^{p-1})^{p-2}+2(r^{p-1})^{p-3}+\dots+(p-2)r^{p-1}+(p-1)]+p. $$ Clearly $p\mid (r^{p-1}-1)$, and the term inside square bracket is, modulo $p$, the same as $1+2+\dots+(p-1)=p\frac{p-1}2=0$ so you get $Y\equiv p\pmod{p^2}$.

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  • $\begingroup$ $Y = p^2m + p$ so $Y \equiv p \not \equiv 0 \pmod{p^2}$ You're awesome! I still have to make sense of that factorization... thank you so much again XD $\endgroup$
    – across
    Commented Sep 12, 2021 at 6:29

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