Prove: Let ∈ ℤ and , be distinct primes. Then p|x and q|x if and only if pq|x

Question

Given this theorem: For every prime and all integers a, b, if p|ab, then p|a or p|b Prove: Let ∈ ℤ and , be distinct primes. Then p|x and q|x if and only if pq|x

I have already proven if p|x and q|x then pq|x.

Now I need to prove it the other way around, if pq|x then p|x and q|x

I have this: We have pq|x which means x = pqn, where n is some integer

Where do I go from here?

Answer 1

Here is a short proof which "goes in both directions", but which doesn't directly use our OP Atlecx's "theorem", mentioned the beginning of the question:

It is relatively easy to see that

$pq \mid x \Longrightarrow p \mid x, q \mid x, \tag 1$

since we have

$p \mid pq \mid x, \; q \mid pq \mid x; \tag 2$

going the other way, we note that since $p$ and $q$ are distinct primes, they are relatively prime:

$\gcd(p, q) = 1; \tag 3$

thus via Bezout's identity we affirm

$\exists a, b \in \Bbb Z, \; ap + bq = 1; \tag 4$

we multiply by $x$ and obtain

$apx + bqx = x; \tag 5$

now $q \mid x$ implies

$pq \mid apx, \tag 6$

and likewise, from $p \mid x$,

$pq \mid bqx; \tag 7$

we conclude that

$pq \mid (apx + bqx) = x. \tag 8$

$OE\Delta$.

Answer 2

You've proven the hard part. The other part is trivial.

If $pq|x$ then there is an integer $n$ so that $pqn = x$. So $p(nq)=x$ and $q(np)=x$. And $nq$ and $np$ are both integers, $p|x$ and $q|x$.