Prime number lemma proof needed

Question

I would like to prove that the maximum divisor of a number $n$ (excluding $n$ itself), is always divisible by the minimum prime factor of $n$ if and only if $n$ itself is divisible by the square of it's minimum prime factor.

Answer 1

Let $k$ be the maximum divisor of $n$. Then $n=kr$, where $r$ is the minimum prime factor. Now $r\mid k\iff k=sr$ for some $s \iff n=sr^2\iff r^2\mid n$

Answer 2

$n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.