Let $n$ be a natural number greater or equal than $2$ and $p(n)$ be the smallest factor of $n$ that is not $1$. Then $n$ is a prime number if and only if $p(n) > \sqrt{n}$.
$\Rightarrow$: If $n$ is a prime numbers then its only factors are $1$ and $n$ and since $p(n)\neq1$, $p(n)=n>\sqrt{n}$ for every $n\geq2$.
I know that the reverse direction has been asked similarly on here as in "Show that a number can have at most one prime factor bigger than its square root" but I'm looking for a proof that could fit my question where I need to show that this number is a prime number. It seems to involve some sort of proof by contradiction? Thanks!
