Number Theory Proof Involving Primes

Question

Let $S$ be the set of positive integers of the form $6k+1$ for some integer $k$.

Find an irreducible $p \in S$ such that $p|ab$ for some $a,b \in S$ but $p \nmid a$ and $p \nmid b$.

I know that if p is prime and $p|ab$, the $p|a$ or $p|b$.

Answer 1

Hint: Since $p$ being a "regular" prime means that $p|ab \implies p|a $ or $p|b$, your irreducible must not be a "regular" prime. Can you find a number of the form $6k+1$ that has no factors that are $1\bmod 6$, but isn't prime?

Answer 2

Doesn't Euclid's Lemma state that if $p$ is a prime and $p\mid ab$ then $p\mid a$ or $p\mid b$ or both. That would mean the answers doesn't exists right? Maybe I'm missing something here...

Answer 3

Hint: If $a$ and $b$ are of the form $6k-1$, then $ab \in S$. Pick $a$ and $b$ regular primes. Then $ab$ will be an irreducible element of $S$.