Mathematical Reflections 2011, Issue 5 - Problem S206

Problem:
Find all integers $n \geq 2$ having a prime divisor $p$ such that $n-1$ is divisible by the
exponent of $p$ in $n!$.

Proposed by Tigran Hakobyan.

Solution:
It is not difficult to show that every power of a prime do the trick. Indeed, let $n=p^k$, where $p$ is a prime number and $k$ is a positive integer. Then $$e_p(n)=e_p(p^k)=\sum_{i\geq 1} \left\lfloor \dfrac{p^k}{p^i} \right\rfloor = p^{k-1}+p^{k-2}+\ldots+p+1$$ and $n-1=p^k-1=\displaystyle(p-1)e_p(n)$. We want to show that these are the only integers which satisfy the given conditions. So, suppose that $n$ is not a power of a prime, namely $n=p^k m$, where $\gcd(p,m)=1$, $m > 1$ and $k$ is a positive integer. Suppose that $e_p(n)|(n-1)$.
$$e_p(n) = \sum_{i=1}^{k} \left\lfloor \dfrac{p^k m}{p^i} \right\rfloor + \sum_{i>k} \left\lfloor \dfrac{p^k m}{p^i} \right\rfloor = me_p(p^k) + e_p(m).$$ Since $$mp^{k-1} < me_p(p^k) = m \dfrac{p^k-1}{p-1}$$ and $$0 < e_p(m) < \dfrac{m}{p-1},$$ we obtain $$mp^{k-1} < me_p(p^k)+e_p(m) < \dfrac{mp^k}{p-1},$$ and rewriting $$\dfrac{n}{p} < e_p(n) < \dfrac{n}{p-1}.$$ Since $e_p(n)$ is an integer, we have $\dfrac{n-1}{p} < e_p(n) \leq \dfrac{n-1}{p-1},$ (otherwise $\dfrac{n-1}{p-1} < e_p(n) < \dfrac{n}{p-1}$, a contradiction) which gives $$p-1 \leq \dfrac{n-1}{e_p(n)} < p,$$ i.e. $n-1=(p-1)e_p(n)$. Hence, $$\dfrac{m(p^k-1)+m-1}{p-1}=me_p(p^k)+\dfrac{m-1}{p-1}=e_p(n),$$ so $e_p(m)=\dfrac{m-1}{p-1}$.
But since $\gcd(p,m)=1$, we also have
$$e_p(n) = me_p(p^k) + e_p(m) = me_p(p^k) + e_p(m-1) = \dfrac{n-m}{p-1}+\dfrac{m-2}{p-1} = \dfrac{n-2}{p-1},$$ contradiction.