Exercise 2.8.7 (Number of solutions of $x^{p-1} \equiv 1 \pmod p$; of $x^{p-1} \equiv 2 \pmod p$)

Question

If $p$ is an odd prime, how many solutions are there to $x^{p-1} \equiv 1 \pmod p$, to $x^{p-1} \equiv 2 \pmod p$?

Richard Ganaye · Accepted Answer

\begin{proof} 
By Fermat's theorem, every integer $x$ such that $x
ot \equiv 0 \pmod p$ satisfies $x^{p-1} \equiv 1 \pmod p$. So there are $p-1$ solutions to  the congruence $x^{p-1} \equiv 1 \pmod p$.

Therefore $x^{p-1} 
ot \equiv 2 \pmod p$ for every integer $x$. The congruence $x^{p-1} \equiv 2 \pmod p$ has no solution.
\end{proof}

Exercise 2.8.7 (Number of solutions of $x^{p-1} \equiv 1 \pmod p$; of $x^{p-1} \equiv 2 \pmod p$)

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Exercise 2.8.7 (Number of solutions of $x^{p-1} \equiv 1 \pmod p$; of $x^{p-1} \equiv 2 \pmod p$)

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