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to prove that a is a prime number, I need to show that a only has 2 divisors: 1 and itself. My approach to this question is to use the definition of divisibility and the hint in the question that since $a = k(mn)$, a will either be $a = i(m)$ or $a = j(n)$ (where i, j, k are constants) to show that either m and n either has to be a or 1. I'm not sure if the approach I'm taking is a correct/logical one.

Jenson Mutton
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You're on the right track, thinking about the divisors of $a$. Let me provide a proof by contradiction. Let $a\in\mathbb{Z}_{\geq 2}$ have the property given in the question, and assume $a$ is not prime. Well, then $a$ is composite, and so there exist integers $b,c$ greater than or equal to $2$ with $a=bc$. But then clearly, $a|a$ and so $a|bc$, but also clearly, $a\not|b$ and $a\not|c$. So something is wrong; $a$ can't be composite and it must be prime.

tromben
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  • Does just simply giving a counter-example like $4|6*6$ but $4\not|6$ be incorrect? – Leon Nov 02 '20 at 23:41
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    @Leon It would be incorrect, yes. All your counter-example shows is that there is some composite number ($4$ in this case) that doesn't have the property. But it says nothing about any other composite numbers, and we need to prove that there are no composite numbers which have the desired property, i.e. the only numbers with the property we want are prime. :) – tromben Nov 02 '20 at 23:44