If m satisfies Euclid's Lemma does it mean m is prime?

Question

Let $m$ be a natural number larger than 1, and suppose that $m$ satisfies the following property:

For any integers $a$ and $b$, if $m$ divides $ab$, then $m$ divides either $a$ or $b$ (or both).

Show that $m$ must be prime.

I think we should solve this question with a proof by contradiction. So, suppose $m$ is not prime, therefore $\exists$$k$ $\in$ $\mathbb{Z}$ such that $k|m$. We also notice that the property $m$ satisfies is Euclid's Lemma, and from it we can conclude $gcd(a,m) = m$ and $gcd(b,m) = m$ depending on if $m|a$ or $m|b$ or both.

Now from here I don't know how I should proceed further. In fact, I'm not sure if I'm looking in the right direction, so your help would be appreciated.

Answer 1

You are right: suppose $m\in \mathbb{N}$ is not prime and let's fix $(k,l)\in \mathbb{N}^2$ such that $m=kl$ and $ k,l \neq m$ then either $m\mid k$ or $ m\mid l $ which is absurd as both k and l are strictly less than m (and positive)

Answer 2

Hint: $ $ first show: $\,p\,$ is prime $\!\iff\! [\,p \color{#c00}{\bf =} ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\,],\,$ then use $\,p\color{#c00}{\bf =}ab\,\Rightarrow\, p\mid ab$