Question: If $R$ is a euclidean ring and $\pi\in R$ is irreducible, prove that $\pi\mid\alpha\beta$ implies $\pi\mid\alpha$ or $\pi\mid\beta$.
A solution is to prove all euclidean rings are PIDs, then prove Euclid's Lemma is true in PIDs. However, is there a solution need not to prove $R$ is a PID and use the definition of euclidean ring directly? And the solution may not use $gcd$ because the existence of $gcd$ doesn't confirm without PID's help.
Forgive my poor English.
Hint $ $ If not $\exists$ irred $\,p\!:\ pc = ab,\ p\nmid a,b.\,$ From such counterexamples with min $\,|p|\,$ choose one with min $\,|a|.\,$ Then $\,|a| < |p|,\ $ else $\,a\mapsto \bar a\, =\, a\ {\rm mod}\ p\, =\, a - rp\ne 0\,$ is a counterexample with $\,|\bar a| < |a|.\,$ Now $\,p\nmid b\,\Rightarrow\,a\,$ is a nonunit, so it has an irred factor $\,q\mid a.$ By $\,|p|\,$ minimal and $\,|q|<|p|$ we infer $\,q\mid pc,\ q\nmid p\,\Rightarrow\, q\mid c,\ $ so $\ p(c/q) = (a/q)b\,$ is a counterexample with $\,|a/q| < |a|$ contra hypothesis.