Exercise 2.1.52* (Solutions of $(p-2)! -1 = p^k$)

Question

Show that if $p$ is prime then $p \mid ((p-2)! - 1)$, but that if $p>5$ then $(p-2)! - 1$ is not a power of $p$.

Hint: Consider congruences modulo $p-1$.

Richard Ganaye · Accepted Answer

\begin{proof} Using Wilson's theorem, \begin{align*} (p-2)! &\equiv - (p-2)! (p-1)\ &\equiv -(p-1)!\ &\equiv 1 \pmod p. \end{align*} So $p \mid (p-2)! - 1.$ Now consider the case $p>5$, and assume for contradiction that $(p-2)! - 1 = p^k$ for some integer $k\geq 0$. Since $p>5$, $2 < \frac{p-1}{2} \leq p-2$. Indeed, \begin{align*} 2 < \frac{p-1}{2} &\iff 5 < p\ \frac{p-1}{2} \leq p-2 &\iff p-1 \leq 2p- 4 \iff 3 \leq p. \end{align*} Therefore we obtain $p-1 = 2 \cdot \frac{p-1}{2} \mid (p-2)!$, so \begin{align} (p-2)! \equiv 0 \pmod {p-1}. \end{align} But $p\equiv 1 \pmod{p-1}$, thus \begin{align} (p-2)! = p^k + 1 \equiv 2 \pmod {p-1}. \end{align} Then (1) and (2) imply $2 \equiv 0 \pmod {p-1}$, so $p-1 \mid 2$, and $p=2$ or $p=3$. This is in contradiction with the hypothesis $p>5$. If $p>5$ then $(p-2)! - 1$ is not a power of $p$. \end{proof}

Exercise 2.1.52* (Solutions of $(p-2)! -1 = p^k$)

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Exercise 2.1.52* (Solutions of $(p-2)! -1 = p^k$)

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