if m, n are coprime which divide k then mn|k

Question

Suppose $m,n$ are coprime numbers such that $m|k$ and $n|k$ for some integer $k$. then prove that $mn|k$ as well.

I am trying to work directly with the definition, so:
$k = ma$,
$k = nb$,
$k^2 = mnab$

How do I take it from here? I have tried to prove that $k|ab$ with no success.

Answer 1

$m|k$ so $k = ma$ for some integer $a$. Now, $n|k \iff n|ma$ and since $(n:m) = 1$, $n|a$ so $a = nj$. Finally,

$$ k = ma = mnj$$

which implies $mn|k$.

Edit: I'm adding the proof of why if $ n|ma$ and $n$ and $m$ are coprime, $n|a$ to make this a bit more self-contained.

If $(n:m) = 1$, there exist $c, \ d \in \mathbb{Z}$ such that

$$ cn + dm = 1 $$

and therefore

$$ cna + dma = a$$

Since $n|cna$ and by hypothesis $n|ma$, we get $n|dma$ so $n|(cna + dma)$ which is exactly the same as $n|a$.

Answer 2

Outline for your proof.

Note that $\gcd(m,n)=1$ is most important.

This means that $m,n$ have no common prime factors between themselves.

Suppose we do not know if $m,n$ are primes or not.If they are primes then its not a big deal.If they are not primes then they can be uniquely factorised by Fundamental Theorem of Arithmetic.Also,consequently,all the prime factors of $m,n$ are contained in $k$.Hence $k$ is divisible by both $m,n$ simultaneously.

Answer 3

You have $k = ma$ and $n|k$, this implies $n|ma$. But $n,m$ are coprimes so $n|a$, then $a=nr$ which implies $k=mnr$. Then you conclude that $mn|k$.