# Find all positive integers $n$ for which $(n-1)!+1$ is a power of $n$ [duplicate]

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.

• $n=2,3$ are solutions as well. Dec 18, 2011 at 7:12

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.
• On the other hand, $(13-1)!+1=13^2 \times 2834329$ Dec 18, 2011 at 9:20
• @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. Dec 18, 2011 at 9:29
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