Prove this number must be a prime number or 1.

Question

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 $p^2 > \frac{n}{p}$.

Answer 1

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 $p^2$.

Answer 2

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 $1.$

Answer 3

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.

Answer 4

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 $1$.