How can I solve this Diophantine equation: $$(n-1)!+1=n^m$$ with $n,m$ positive integers?

From Wilson's theorem we can note that $n$ is a prime number. I proved to rewriting the equation as:$$(n-2)!=n^{m-1}+n^{m-2}+....+1$$ but in vain. I proved to solve also through the theorem LTE bur I analysed only some cases without obtain a general solution.


1 Answer 1


If $1\le n\le 5$, then $(n,m)=(2,1),(3,1),(5,2)$. Let $n\ge 6$.


$$\iff (n-2)!=\frac{n^m-1}{n-1}=$$


$$\iff (n-2)!-m=$$


$2$, $\frac{n-1}{2}$ are different and $\le n-2$, so

$2\cdot \frac{n-1}{2}=n-1\mid (n-2)!$.

For all $a,b\in\mathbb R$, $k\ge 2$, $k\in\mathbb Z$, $$a^k-b^k=(a-b)(a^{k-1}b^0+a^{k-2}b^1+\cdots+a^0b^{k-1})$$

If $k$ is odd, $c=-b$, then


Let $a=n$, $b=1$. Then $n-1\mid n^k-1$ for all $k\ge 0$, $k\in\mathbb Z$ because $n-1\mid n^0-1=1-1=0$ and $n-1\mid n^1-1=n-1$.

$n-1\mid m$, so $m\ge n-1$ (because $m>0$) and $$ n^m = (n-1)! + 1 = 1 \cdot 2 \cdots (n-1) + 1 < $$

$$<\underbrace{(n-1)(n-1)\cdots (n-1)}_{\text{n-1 times}}=$$

$$ = (n-1)^{n-1} < n^{n-1}, $$

contradiction. Answer: $(n,m)=(2,1),(3,1),(5,2)$.


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