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I understand that this is part of Euclid's Lemma but how do I prove that a must be prime?

Jia Yuan Ng
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    I presume "prime" in this context means that $a$ cannot be written as a product $a=bc$ with $b,c>1$ (which is what is called "irreducible" in algebra). If that's the case, just note that you can take $m=b, n=c$ because $a\mid a=bc$ but $a\not\mid b$ and $a\not\mid c$. –  Oct 27 '20 at 09:24
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    By showing that if it is not a prime that you can construct a counter example. That is, you can find $m,n$ such that $a|mn$ but $a$ does not divide either $m$ or $n$. The prime factorization theorem is your friend. – user2661923 Oct 27 '20 at 09:24

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Suppose, $a=u\cdot v$ with integers $u,v>1$. Then, $a\mid uv$ , but $a\nmid u$ and $a\nmid v$

Peter
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