# When does $(p-1)! + 1 = p^k$ hold? [duplicate]

We have a prime $p$ and an integer $k > 0$. When does the following equation stand?

$(p-1)! + 1 = p^k$

I have obviously tried for some little numbers, and in some cases, it stands:

For $p=2$ and $k=1, 1 + 1 = 2$.

For $p=3$ and $k=1, 2 +1 = 3.$

For $p=5$ and $k=2, 24 + 1 = 25.$

Any ideas, how to prove, if there are more, and if yes, what are the solutions?

• No, it was actually a bad suggestion, that's why I deleted it. You've already stated the $p$ is prime, and Wilson theorem implies that only prime numbers can be candidates here, so it brings no additional insights. Mar 18, 2015 at 11:45
• Okay, thanks for helping though. :) Mar 18, 2015 at 11:45
• Nice question -- I suspect it's too hard to answer in general. Note that $p$ always divides $(p-1)! + 1$ and that no prime less than $p$ divides it. However, the number of primes less than $(p-1)!+1$ grows quadratically in $\log(p)$ whereas the number of primes less than $p$ grows linearly in $\log(p)$, so heuristically I suspect this becomes rarer and rarer as $p$ grows. Mar 18, 2015 at 11:48
• But you can combine the fact implied by Wilson theorem, and the fact that "factorial plus $1$ being square" is still an open problem (see here), in order to conclude that your question most likely falls under the same category (i.e., an open problem). Mar 18, 2015 at 11:49
• If this is a homework it means there should be some way to determine it, said that, it doesn't hold for any prime between 6 and 20 so it doesn't seems to have any easy prime clasification, I would try to prove that the answer is that it just holds for $p=2,3,5$ but seems hard. Mar 18, 2015 at 12:13

We show there cannot be any solutions for $p\gt 5$ $$(p-1)!+1 = p^k \implies (p-1)! = p^k-1 = (p-1)\sum\limits_{i=0}^{k-1}p^i$$

Cancel $p-1$ both sides and get $$(p-2)! = \sum\limits_{i=0}^{k-1}p^i$$

Notice that left hand side is divisible by $p-1$ for $p\gt 5$ \begin{align}0&\equiv \sum\limits_{i=0}^{k-1}p^i \pmod{p-1}\\0&\equiv \sum\limits_{i=0}^{k-1}1 \pmod{p-1}\\0&\equiv k \pmod{p-1}\\k&=t(p-1)\end{align}

So we need $k$ to be of form $t(p-1)$

$$(p-1)! + 1 = p^{t(p-1)}$$

Clearly this is impossible because $(p-1)! + 1 \lt p\cdot p\cdots (\text{p-1 times}) = p^{p-1}$

That proves there are no solutions for $p\gt 5$.

• Very nice and simple, thanks! Mar 18, 2015 at 15:13
• Can you spell out why the left hand side is divisible by p−1 for p>5 please? Mar 24, 2015 at 22:46
• n is composite by definition, and therefore its factors must be in (n-1)!? Mar 24, 2015 at 22:57
• The only interesting case is if n is the square of a prime p and p > 2, then 2p is divisible by p and 2p < n, so (n-1)! is divisible by p², which is n. Mar 24, 2015 at 23:03
• It should be mentioned that $t>0$ because it's given that $k>0$ or because $(p-1)!>0$. Also, clearly $t\in\mathbb Z$. Aug 29, 2017 at 20:20