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Let $p\in \Bbb{P},\alpha\in \Bbb{N}$. Find all of the solutions of the following equation: $$(p-1)!+1=p^{\alpha} \ \ \ ,p>6$$

My attempt

We can rewrite the equation as follows:

$$(p-1)!=p^{\alpha}-1$$ $$(p-1)!=(p-1)(1+p+p^2+...+p^{\alpha-1})$$ $$(p-2)!=1+p+p^2+...+p^{\alpha-1}$$

Then the best I could find out is that:

$$(p-2)!\equiv \alpha \pmod{p-1}$$

And in general if $ 2\leq k \leq p-1$:

$$k^{\alpha}=1 \pmod{p-k}$$

But then I don't know how to continue. Thank you for your time

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  • $\begingroup$ I think taking $v_p$ may help, you get the $\alpha$ in $\endgroup$ – user665856 May 1 at 7:56
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    $\begingroup$ Hint: $1+p+\dots+p^{\alpha-1} \equiv ? \pmod{p-1}$. And $(p-2)! \equiv ? \pmod{p-1}$. $\endgroup$ – Marco Vergamini May 1 at 7:57
  • $\begingroup$ @MarcoVergamini what is $(p-2)! \mod p-1$ ? $\endgroup$ – Eureka May 1 at 8:00
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    $\begingroup$ Well, if $p$ is a prime greater than $6$, I bet $2$ and $(p-1)/2$ are two different numbers in $1, 2, \dots, p-2$, so... $\endgroup$ – Marco Vergamini May 1 at 8:02
  • $\begingroup$ @MarcoVergamini $0$ ? $\endgroup$ – Eureka May 1 at 8:03
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Observe if $p>6$ then $(p-1)^2 | p^\alpha -1$ and hence $p-1 |\frac{p^{\alpha}-1}{p-1}=\sum_{i=0}^{\alpha-1} p^i\implies p-1 | \alpha \implies \alpha \geq p-1 $ but then $p^{\alpha}\ge p^{p-1}> (p-1)!+1$

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