Proof that every positive integer has at most one prime factor greater than its square root?

Question

I read the statement in the title somewhere but I can't find any proof.

For a positive integer $n$, why can't there be 4 numbers $a, b, c, d$ ($b$ and $d$ are prime) for which

$a < \sqrt{n} < b$, where $a \cdot b = n$

and

$c < \sqrt{n} < d$, where $c \cdot d = n$

Answer 1

If $p$ and $q$ are primes that are $>\sqrt{n}$, and $n$ is divisible by $p$ and $q$, then $n$ is divisible by $pq$. But $pq>n$, which is a contradiction.

Answer 2

Assume by contradiction that there exists a number $n\in\mathbb{N}$ with two prime factors $p,q>\sqrt{n}$:

• $p$ and $q$ are prime factors of ${n}\implies\color\red{{p}\cdot{q}\leq{n}}$
• $p,q>\sqrt{n}\implies{p}\cdot{q}>\sqrt{n}\cdot\sqrt{n}\implies\color\red{{p}\cdot{q}>{n}}$