I'm supposed to use Bezout's identity to prove this, but I don't really know how. Using purely what is given if $p$ is a prime then it's divisible by only 1 and itself so that means if $p\mid ab$ then $ab=px$, which means that $p=a$ or $p=b$. If $p=b$ then $(p,a)=1$ then $px+ay=1$. Then I get stuck.
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Your second statement is false. $4 \cdot 3 = 12 = 2x$ but $2 \neq 3$ and $2 \neq 4$. – Derek Luna Oct 05 '19 at 04:47
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$p\mid ab\implies ab=pk$ for some integer $k$. Now if $p\mid a$ then you are through. If $p\not\mid a$ then we have by Bezout's identity $$px+ay=1$$ Now mulplying $b$ to both sides of the above equation we get $$pbx+aby=b \implies pbx+pky=b\implies p(bx+ky)=b$$. As $b,x,k,y$ are all integers, $p\mid b$.
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@Dylan Y, yes. Thanks for pointing out. Also $p\not\mid a\implies (p,a)=1$ as p is a prime. Therefore we can use Bezout's identity. – Oct 05 '19 at 04:53