Exercise 2.1.36 (Solutions of $(p-1)!+1 = p^k$.)

Question

If $p$ is a prime, prove that $(p-1)!+1$ is a power of $p$ if and only if $p=2,3$, or $5$.

Hint: If $p>5$, $(p-1)!$ has factors $2, p-1$ and $(p-1)/2$, and so $(p-1)!$ is divisible by $(p-1)^2$. Then recall Problem 32 in Section 1.2.

Richard Ganaye · Accepted Answer

\begin{proof} If $p = 2$, $(p-1)!+1 = 2^1$. If $p = 3$, $(p-1)!+1 = 3^1$, and if $p = 5$, $(p-1)!+1 = 5^2$. Now assume that $p>5$, where $p$ is prime, and assume for contradiction that $$(p-1)! + 1 = p^k, ext{ where } k\geq 1.$$ Then $2 < \frac{p-1}{2} < p-1$, hence $(p-1)^2 = 2\, \frac{p-1}{2} \, (p-1)$ divides $(p-1)! = p^k - 1$. Therefore $$(p-1)^2 \mid p^k - 1.$$ Hence $$p-1\mid \frac{p^k - 1}{p-1} = 1 + p + p^2 + \cdots + p^{k-1}$$ If we reduce modulo $p-1$, then $p \equiv 1 \pmod{p-1}$, therefore \begin{align*} 0 &\equiv 1 + p + p^2 + \cdots + p^{k-1} \pmod{p-1}\ &\equiv 1+ 1 + \cdots + 1 = k \pmod{p-1}. \end{align*} This shows that $p-1 \mid k$, so $k = (p-1)q$ for some integer $q$. Thus $$(p-1)! = p^{(p-1)q} - 1.$$ Since $p^{p-1} -1$ is a divisor of $p^{(p-1)q} - 1$ (because $p^{(p-1)q} - 1 = (p^{p-1} - 1) \sum_{k=0}^{q-1} p^{(p-1)k}$), we obtain $$p^{p-1} -1 \mid (p-1)! ,$$ a fortiori $$p^{p-1} -1 \leq (p-1)! .$$ But, since every $i \in [\![2,p-1]\!]$ is such that $i \leq p$ \begin{align*} (p-1)! &= 2\cdot 3\cdots (p-1)\ &\leq p^{p-2}\ &< p^{p-1} - 1, \end{align*} because $p^{p-1} - p^{p-2} = p^{p-2}(p-1) >1$. Since $p^{p-1} -1 \leq (p-1)! $ and $(p-1)! < p^{p-1} - 1$ gives a contradiction, $(p-1)!+1$ is never a power of $p$ for $p>5$. \end{proof}

Exercise 2.1.36 (Solutions of $(p-1)!+1 = p^k$.)

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Exercise 2.1.36 (Solutions of $(p-1)!+1 = p^k$.)

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