TITLE: Prove that if $n$ is a composite and $p \gt \sqrt[3]n$, then $n/p$ is a prime.
QUESTION [5 upvotes]: Also, $p$ is the least prime factor of $n$.  I'm trying to do this by way of contradiction.
Since $n$ is a composite, $n = pq$, for some $q \in \Bbb Z$.  So, we have $p | n$, $q|n$ and $q = \frac np$.
There's a theorem that says $p \le \sqrt n$.  So, I'm trying to make use of that this theorem, but not sure how to proceed.  Is there a way to come to a contradiction that claims $q \lt p$?
EDIT:  I copied the problem wrong, sorry about that.  But problem is to show that $\frac np$ is a prime.

REPLY [2 votes]: In response to Jacob Mayle's request for a explanation of part of vonbrand's solution, we can get to $q^2 < p^2$ by the following (remembering that $n < p^3$):
$$q^2 \le \frac{n}{p} \Rightarrow q^2 p \le n < p^3 \Rightarrow q^2 < p^2$$
(the last implication occuring as we divide through by p).