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Find all natural numbers $n$ and $m$ for which $$n^m=(n-1)!+1.$$

By the Wilson theorem, $n$ is prime. Obviously, $m < n$.

For example, some solutions $(n, m)$ are $(2, 1)$, $(3, 1)$ and $(5, 2)$.

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    $\begingroup$ If $n > 4$, then $(n-1)|(n-2)!=\frac{n^m-1}{n-1}=1+n+\ldots+n^{m-1}$. $\endgroup$
    – Aphelli
    Feb 20, 2019 at 17:25
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    $\begingroup$ $1^1\ne 0!+1$? So $(1,1)$ is not a solution. $\endgroup$
    – Peter Foreman
    Feb 20, 2019 at 17:26
  • $\begingroup$ @PeterForeman, I fixed, thank you. $\endgroup$
    – Anastasia K.
    Feb 20, 2019 at 17:32

1 Answer 1

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You have found all the solutions.

Suppose there is a solution with $n>5.$ As you have noted, by Wilson's theorem, $n$ is prime. We have $$n^m-1=(n-1)!,$$ and dividing both sides by $n-1,$ $$(n-2)!=1+n+\cdots+n^{m-1}\tag{1}$$ Since $n>5$ is prime, $n-1$ is composite and $(n-1)|(n-2)!$ (Prove this.)

The preceding is Mindlack's hint, but I gather you haven't seen where to go from there.

Reducing both sides of $(1)$ modulo $n-1$ gives $0\equiv m\pmod{n-1}.$ Since $n-1$ divides $m$, we have $m\ge n-1.$

Now argue that $n^{n-1}>1+(n-1)!$

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  • $\begingroup$ Not getting how $1+n+\cdots +n^{m-1}\equiv m\pmod{n-1}$! Probably missing something trivial.. $\endgroup$
    – tarit goswami
    Mar 29, 2019 at 16:40
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    $\begingroup$ @tarit There are $m$ terms, each of which is congruent to $1$ mod $n-1$ $\endgroup$
    – saulspatz
    Mar 29, 2019 at 16:43
  • $\begingroup$ Oops. Nice solution, can you share how you think what is necessary while solving a particular problem?,e.g; this one $\endgroup$
    – tarit goswami
    Mar 29, 2019 at 16:52
  • $\begingroup$ @taritgoswami I wish I could. Actually, this problem appears in Tom Apostol's undergraduate Analytic Number Theory book. Many years ago, somebody showed me how to do it after I failed to solve it by myself. I was sure I'd found all the solutions, and that the exponent would be too big for there to be others. I think the key technique is to realize that you have to start by assuming $n$ is somewhat large, since we know there are some solutions. Examine closely how the argument uses $n>5.$ $\endgroup$
    – saulspatz
    Mar 29, 2019 at 17:01
  • $\begingroup$ Thanks for sharing :) $\endgroup$
    – tarit goswami
    Mar 29, 2019 at 17:41

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