Prove that if $n$ is a composite and $p \gt \sqrt[3]n$, then $n/p$ is a prime. 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.
 A: If I understand correctly, you want to prove that if $p$ is the smallest prime factor of $n$, and $p > \sqrt[3]{n}$, then $\frac{n}{p}$ is prime.
The proof is by contradiction. Suppose $\frac{n}{p}$ is composite. Then it has a least prime divisor $q$, and $q \le \sqrt{\frac{n}{p}}$, i.e., $q^2 \le \frac{n}{p}$. But $n < p^3$, so $q^2 < p^2$ and $q < p$, contradicting $p$ is $n$'s smallest prime divisor.
A: This is false. For instance, consider $n = 101 \times 103$.
Clearly, $101$ and $103$ are primes and both are greater than $\sqrt[3]{n}$. But $\dfrac{n}{101}$ and $\dfrac{n}{103}$ are both primes.
A: 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).
A: For the new problem, if $n$ is composite, $p$ is the least prime factor of $n$ and $p \gt \sqrt[3]n$, then either $n=p^2$, in which case $\frac np=p$, prime or $n=pq$ with $p \lt q \lt p^2$.  $q$ must be prime because otherwise one factor would be less than $p$ and $\frac np=q$
A: Let $n$ be 323 = 17$\times$19.  Then $p=19$ is greater than $n^{1/3} \approx 6.86$, but $n/p = 17$, which is not composite.  
Seems that the statement is wrong?!
A: While the copied version is false as other answers have demonstrated, if we require $p>\sqrt[2]n$ then it is true.
Suppose that $\frac np$ was prime, by our assumption $p^2>n$ and therefore $p^2>p\cdot\frac np$, and therefore $p>\frac np$. But we assumed that $p$ was the least prime divisor of $n$, which is a contradiction.
