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As the title says,

Find all positive integers $n$ for which $(n-1)!+1$ is a power of $n$.

The solutions I've found are $\{2,3,5\}$ (thanks Brandon!), but I'm having difficulties proving that these are the only ones. What I've got so far is that $n$ must be prime since $(n-1)!+1$ would not be congruent to $n$ if $n$ were composite.

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    $\begingroup$ $n=2,3$ are solutions as well. $\endgroup$ – Brandon Carter Dec 18 '11 at 7:12
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Well, for primes $p < 1290,$ those for which $$ 1 + (p-1)! \equiv 0 \pmod{p^2}$$ are the three $$\{ 5, \; \; 13, \; \; 563 \}$$

As soon as $p \geq 7,$ we have $(p-1)! > p^3,$ so we need $ 1 + (p-1)! \equiv 0 \pmod{p^3},$ and, in fact, $ 1 + (p-1)! \equiv 0 \pmod{p^4}.$ Quite rare.

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  • $\begingroup$ On the other hand, $(13-1)!+1=13^2 \times 2834329$ $\endgroup$ – J. M. is a poor mathematician Dec 18 '11 at 9:20
  • $\begingroup$ @J.M. yes, I factored these for $p \leq 61$ in gp-Pari, I don't see any chance for more solutions, which is what the large unpredictable prime factor with $p=13$ tells me. I know $v_{563} (1 + 562!) \geq 2$ but not the exact value. $\endgroup$ – Will Jagy Dec 18 '11 at 9:29
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This is really an old problem in ML. Note that, the equation doesn't hold for primes $>5$ which is in fact the Liouvilles'theorem

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    $\begingroup$ What does "ML" mean? If it is not a standard or common abbreviation (and it appears so to me), then it may be better to write it out in full. $\endgroup$ – Srivatsan Dec 25 '11 at 9:03
  • $\begingroup$ It's pretty famous though! MathLinks. $\endgroup$ – Ehsan M. Kermani Dec 25 '11 at 10:44

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