1. ## Prime numbers

Let $\displaystyle n \ge 2$ be a number that is not a prime.
Show that there is a divisor $\displaystyle p$ of $\displaystyle n$, with $\displaystyle p \ge 2$, so that $\displaystyle n \ge p^2$

On my paper, I've written down n=rs where n and s are any natural numbers >1. Not sure if that's needed, though.

2. Originally Posted by Joel24
Let $\displaystyle n \ge 2$ be a number that is not a prime.
Show that there is a divisor $\displaystyle p$ of $\displaystyle n$, with $\displaystyle p \ge 2$, so that $\displaystyle n \ge p^2$

On my paper, I've written down n=rs where n and s are any natural numbers >1. Not sure if that's needed, though.
If n>2 is not prime, then it has a factor p>=2, so there also exists a q>=2 such that pq=n. Now if both n<p^2 and n<q^2, then pq<n a contradiction.

Hence every composite number >2 has a factor whose square is less than or equal to the number.

RonL

3. Ah, a proof (or sorts) by contradiction. I would've never have thought of that.

Excellent. Thank you so much!