# Let $p>1$. Assume: if $p \mid ab$, then $p \mid a$ or $p \mid b.$ Prove that $p$ must be a prime number. [closed]

Let $$p>1$$ belong to the natural numbers. Assume: if $$a$$ and $$b$$ are natural with $$p \mid ab$$, then $$p \mid a$$ or $$p \mid b.$$ Prove that $$p$$ must be a prime number.

I know how to prove Euclid's lemma assuming $$p$$ is prime, but I do not know how to prove the title statement without that initial assumption.

• What are you trying to prove (Euclid's lemma or the title statement)? And what is your definition of prime? Oct 8, 2021 at 0:38
• You may enjoy this math.stackexchange.com/questions/1162373/…
– Stan
Oct 8, 2021 at 0:45
• Im trying to prove the title statement. Oct 8, 2021 at 0:56
• Consider $4|4\cdot4$. Oct 8, 2021 at 10:29

Suppose $$p=xy$$ for $$x,y>1$$. Then $$p|xy$$ but $$p\not\mid x$$ and $$p\not\mid y$$.
• Which part? It's clear why we can write $p=xy$ and why $p|xy$, right? The other observation is that $1<x<p$, so $x$ can't be a multiple of $p$. Oct 8, 2021 at 1:01