# Show that $p^3 \mid (r^{p(p-1)} - 1) \Rightarrow p^2 \mid (r^{p-1}-1)$

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 $$\color{blue}{Y}$$ ?

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

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

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 $$Y\equiv p\pmod{p^2}$$.

• $Y = p^2m + p$ 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 Sep 12, 2021 at 6:29