AUT_ABooklet_2023

Answer. The only solutions are $(2,2)$, $(3,2)$ and $(5,3)$.

Because of $n!+n>n$, we immediately get $k\ge 2$. We divide both sides of the equation by $n$ and get $$(n-1)!+1=n^{k-1}.$$ Now, we distinguish two cases:

$n$ is not a prime. Since $n$ is clearly not $1$, we can write $n$ as $n=ab$ for integers $a,b$ with $1<a,b<n$ which implies $1<a\le n-1$ and therefore $a\mid (n-1)!$. We conclude that $a>1$ is relatively prime to the left-hand side $(n-1)!+1$, but $a$ divides the right-hand side $n^{k-1}$. This is not possible, so there are no solutions in this case.

$n$ is a prime. We check $n=2,3,5$ and find the solutions $(2,2)$, $(3,2)$ and $(5,3)$. From now on, let $n\ge 7$. We get $$\begin{array}{rl}(n-1)!& =n^{k-1}-1\\ ⟹(n-1)!& =(1+n+n^2+\cdots +n^{k-2})(n-1)\\ ⟹(n-2)!& =1+n+n^2+\cdots +n^{k-2}\end{array}$$ Since $n$ is prime and bigger than $3$, the number $n-1$ is even and not a prime. Furthermore, $n-1$ is not the square of a prime since $4$ is the only even square of a prime and $n-1\ge 6$. Therefore, we get $n-1=ab$ with $1<a,b\le n-1$ and $a\ne b$. We obtain that $(n-2)!$ contains the separate factors $a$ and $b$ and is therefore divisible by $ab=n-1$ which implies $(n-2)!\equiv 0\mathrm{mod}(n-1)$. Furthermore, $n\equiv 1\mathrm{mod}(n-1)$, and therefore $$0\equiv 1+1+1^2+\cdots +1^{k-2}\equiv k-1\mathrm{mod}(n-1).$$ We conclude that $n-1$ divides $k-1$ and we write $k-1=l(n-1)$ for a positive integer $l$. The case $k=1$ and $l=0$ has already been treated. Therefore, we get $k-1\ge n-1$. However, $$(n-1)!=1\cdot 2\cdot 3\cdots (n-1)<\underset{n-1\text{ times}}{\underbrace{(n-1)\cdot (n-1)\cdots (n-1)}}=(n-1{)}^{n-1},$$ and therefore $$n^{k-1}=(n-1)!+1\le (n-1{)}^{n-1}<n^{n-1}\le n^{k-1},$$ giving a contradiction. So there are no further solutions.

(Michael Reitmeir) $\square$