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I am in a basic proofs class, and am having trouble with the following question:

Let $a \in \mathbb{Z}$ and let $p$ and $q$ be distinct primes. Show that if $p|a$ and $q|a$, then $pq|a$.

Since we're just starting, we're really only allowed to use the properties of primes and divisibility. However, after trying to prove it directly, through the contrapositive, and through contradiction, I was unable to get to any conclusion; noticeably, I'm not sure where to use the fact that p and q are primes in the question. On a related note: when trying to prove the contrapositive, I have: $$ a \ne pq*k$$ (for any $k \in \mathbb{Z}$). Since $a \ne p(qk)$, why does this not imply that $p$ does not divide $a$ (which is what we want to complete the contrapositive)?

Any and all help is appreciated. Thank you kindly!

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    Hint: Use the fact that there is unique factorization into prime elements in $\mathbb{Z}$. – 2017-01-31
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    @mathma I'd say that OP is not allowed to use that. – 2017-01-31
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    Are you allowed to use the Bezout's identity? Or maybe the four numbers theorem? – 2017-01-31
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    You need to tell us what methods you have available after "just starting". E.g. do you know Euclid's Lemma, or the Bezout gcd identity, or the Prime Divisor Property $p\mid ab\,\Rightarrow p\mid a$ or $p\mid b,\,$ or gcd laws, etc? – 2017-01-31
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    @Gizmo. You need something more than the basic divisibility rules. The reason is that there are rings (like $\Bbb Z[\sqrt{-5}]$) where this is not true. – 2017-01-31

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How about the following. $p|a,$ so $a=kp$ for some $k\in \mathbb{N}$. Now, $q|a$, so $q|kp$. But $p$ is prime, so $q$ can not divide $p$ because $p\ne q$. Then it must be the case that $q|k$. Then $k=jq$ for some $j\in \mathbb{N}$. Combining these facts, $a=jpq$, and we see that $pq|a$.

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    You are using the Prime Divisor Property, and the OP has not yet confirmed knowledge of such - see the comments. – 2017-01-31
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    You may be right. I'll leave it here in case it helps. – 2017-01-31
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    No, it is impossible to prove using just divisibility properties because it fails in more general domains. – 2017-01-31
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    But we're only considering $\mathbb{Z}$, no? – 2017-01-31
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    Any proof using only general divisibility properties (i.e. those true in any domain) would imply that it was true in any domain. But it's not. The proof requires using special properties of $\,\Bbb Z,\,$ e.g. that is has (Euclidean) Division with remainder. – 2017-01-31
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    @BillDubuque Wouldn't be this correct also in general since he is using prime elements, not irreducibles? – 2017-01-31
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    @Bill Dubuque: Yet there is a proof of *Euclid's lemma* using only the definition of divisibility and well-order on $\mathbf N$. – 2017-01-31
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    @Mathma But that amounts to assuming that $p,q$ satisfy the Prime Divisor Property. – 2017-01-31
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    @Bernard Of course, and you can find many variations of that in some of my posts. But those proofs do in fact (implicitly) use the (Euclidean) Division with Remainder algorithm, though the use is often hidden in some induction (or descent). Those proofs essentially are assembly-language versions of higher-level proofs, where higher-level concepts (like denominator ideals), have been completely eliminated. We can *mechanically* generate those proofs from the higher-level versions by "compiling" them into (arithmetical) assembly language. – 2017-01-31
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    But the fundamental fact is well-order. Euclidean division is only a direct consequence of this property. – 2017-01-31
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    @Bernad Those approaches all essentially exploit that $\,\Bbb Z\,$ is a Euclidean domain. Any attempt to denigrate this fact is pedagogically misleading. If you want to understand the essence, you need to ask what relationships hold between these divisibility properties in *weaker* rings, e.g. in domains. You can't compare them in a domain like $\,\Bbb Z\,$ where they are all equivalent. – 2017-01-31
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    @Bernard You can find comparisons of the strength of many closely related divisibility properties in [this answer.](http://math.stackexchange.com/a/164589/242) – 2017-01-31
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    Extremely sorry for the delay in response;this is most likely fine (I will have to double check) but I will "accept" it. Thank you very much! – 2017-01-31