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A friend asked me this question:

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

Clearly one direction is obvious, namely that $p=2,3,5$ implies $(p - 1)! + 1$ is a power of $p$.

The other direction is not clear to me. Since by Wilson's theorem $p$ divides $(p - 1)! + 1$ so we need to show that if there are no other prime factors then $p=2,3,5$. Can someone give me a hint for establishing this?

Thanks

  • See https://math.stackexchange.com/questions/805068/to-find-all-n-such-that-n-11-is-a-perfect-power-of-n?noredirect=1&lq=1 for the classification of $n\in\Bbb{Z}_{>0}$ such that $(n-1)!+1$ is a power of $n$. – 19021605 Sep 17 '25 at 07:19

2 Answers2

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Let $p^k - 1 = (p-1)!$. Cancel the factor of $(p-1)$, we get $$p^{k-1} + \cdots + 1 = (p-2)!$$ Now consider mod $p-1$ to get information about $k$. When can the equality ever hold?

  • Can you be a little more explicit? What do you mean by considering mod $p-1$ to get information about $k$? –  Jan 27 '13 at 13:36
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    @shahab, show that when $p > 5$, left hand side = k mod $p-1$ while RHS = 0 mod $p-1$. Show that this implies $k \ge p-1$. –  Jan 27 '13 at 19:54
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Not sure if only one hint would help, you could try using more hints as needed. First, the statement holds for all natural numbers $n \notin \{2, 3, 5\} $, not just for primes. One way to proceed is to show that for $n > 5$:
1) $n$ cannot be even if it satisfies the equation
2) for $n$ odd, $(n-1)^2 \space | \space (n-1)!$
3) from (2) above and the given equation, show $(n-1) \space | \space k$ which implies $n-1 \le k$
4) show that $\forall n > 2, (n-1)! < n^{n-1} - 1$

Hope that helps!

Macavity
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