Prove this number must be a prime number or 1. Show that if the smallest prime factor $p$ of the positive integer
$n$ exceeds $\sqrt[3]{n}$, then $\frac{n}{p}$ must be prime or 1.
I'm stuck trying to prove this. I tried this using contradiction, but I haven't been able to prove it.
My approach:
Let $p$ be the smallest prime factor of the positive integer $n$.
Let $p > \sqrt[3]{n}$
Then we must show that $\frac{n}{p}$ is either a prime number or 1.
We know that $n = pe$ for some $e \in \mathbb{Z}$.
Thus, $\frac{n}{p} = e$
We also know that $p$ > $\sqrt[3]{n}$, so $p^3 > n$, and $p^2 > \frac{n}{p}$
So, assume $\frac{n}{p} = e$ is not prime nor 1, meaning it's a composite number. Let's try to get a contradiction.
If $\frac{n}{p}$ is composite, then $\exists a,b \neq 1: \frac{n}{p} = ab $
Now, here's where I get stuck. What can I do with this info? $\frac{n}{p} = ab $ and $p^2 > \frac{n}{p}$.
 A: You forgot the condition "smallest factor". 
The smallest factor of $n=p^{m} \; {p_1}^{m_1} \; \cdots$ is necessarily $p$.
Consider as you did that 
$$
n = pab\quad \left| \begin{gathered}
  p < a \hfill \\
  p < b \hfill \\
  p < ab < p^{\,2}  \hfill \\ 
\end{gathered}  \right.
$$
then the conditions aside leads to the contradiction $p^2<ab<p^2$.
So you cannot have two terms following $p$, it must be
$$
n = pa\quad \left| {p < a < p^{\,2} } \right.
$$
and $a$ cannot decompose in primes lower than $p$, otherwise
$p$ will not be smallest, and cannot decompose in two primes higher than $p$.
Hence $a$ must be a prime between $p$ and $p^2$.
A: Property :
If the smallest prime factor $p$ of the positive integer $n$ exceeds $\sqrt[3]n$, then $\frac np$ must be prime or $1.$
Proof :
First of all, we consider the prime factors decomposition of $n$ :
$n=p_1\cdot p_2\cdot p_3\cdot\ldots\cdot p_r$
where $\;r\in\mathbb{N}\;,\;$ whereas $\;p_1\leqslant p_2\leqslant p_3\leqslant\ldots\leqslant p_r\;$ are all prime numbers.
So the smallest prime factor of $\;n\;$ is $\;p=p_1\;.$
Now we will prove that $\;r\geqslant3\;$ leads to a contradiction.
If $\;r\geqslant3\;,\;$ we get that
$p>\sqrt[3]n\;\;$ implies $\;\;p_1^3=p^3>n=p_1\cdot p_2\cdot p_3\cdot\ldots\cdot p_r\geqslant p_1^3\;\;$ that is $\;\;p_1^3>p_1^3\;\;$ which is a contradiction.
Hence $\;r=1\;$ or $\;r=2\;.$
If $\;r=1\;,\;$ then $\;\dfrac np=\dfrac{p_1}p=1\;.$
If $\;r=2\;,\;$ then $\;\dfrac np=\dfrac{p_1\cdot p_2}p=p_2\;$ is prime.
Consequently ,
$\dfrac np\;$ must be prime or $1.$
A: Suppose that $\frac{n}{p}$ is different from $1$ and is not prime.
So $\frac{n}{p}=a\cdot b$ with $a,b>1$.
Note that $a,b > \sqrt[3]{n}$, otherwise you have a factor of $n$ smaller or equals to $\sqrt[3]{n}$.
That implies $n=a\cdot b\cdot p > \sqrt[3]{n} \cdot \sqrt[3]{n} \cdot \sqrt[3]{n} > n$, which is an absurd.
A: If $n/p$ is composite, then $n$ is the product of at least three primes, one of which must be less than or equal to $\sqrt[3]n$. So if the smallest prime divisor of $n$ is greater than $\sqrt[3]n$, then $n/p$ cannot be composite, so it's either a prime or $1$.
