Is my proof fully correct? I cannot see if I am missing anything critical so far.
Prove that for all natural numbers $p>2$, $p$ is prime if and only if for all natural numbers $a$ and $b$, if $p|ab$ then $p|a$ or $p|b$.
Solution: Since $p|ab$ there exists an integer $c$ such that $ab = pc.$ Since $p$ is prime, therefore, either $p|a$ or $(p,a)=1.$
If $p|a,$ we are done.
If $(p,a)=1$ then there exist an integer $s$ and $t$ such that $ps + at=1$ Now multiply with $b$ on both sides
$$ psb +(ab)t = b $$
Put $ab=pc$ $\longrightarrow$ $p(sb+tc)=b$ Therefore, $p|b.$
The proposition is a biimplication. You have proved that if $p$ is prime, then whenever $p | ab$ then $p|a$ or $p|b$. You must also prove the converse, namely that if whenever $p | ab$ then $p|a$ or $p|b$, then $p$ is prime.