If $m$ is a divisor of $k$ and $n$ is a divisor of $k$ and $m$ and $n$ prime to each other then $mn$ is divisor of $k$.

Question

If $m$ is a divisor of $k$ and $n$ is a divisor of $k$ and $m$ and $n$ prime to each other then $mn$ is divisor of $k$.

I tried but somehow I didnt manipulate.

As $m$ is divisor of $k$ then $k=mp$ for some $p$ and similarly $k=nq$ for some $q$ and as $gcd(m, n)=1$ there exist integer say $u$ and $v$ such that $mu+nv=1$

After that how I proceed?, Please help.

Answer 1

Write $k=mq$. Since $\gcd (m,n)=1$,by Gauß's lemma $n\mid q$. So $q=rn\implies k=rmn$. Thus $mn\mid k$.

Note: Thanks are due to @Bernard for enlightening me about Gauß's lemma...

Answer 2

Think of it in terms of the fundamental theorem of arithmetic. Let $m=\prod p_i$. Since $m\mid k, k=\prod p_ir$. Similarly, $n=\prod q_j$. But since $m$ and $n$ are relatively prime, $p_i\neq q_j$ for all $i,j$. So to say that $n\mid k$ is to say that $n\mid r$.

So, $r=\prod q_js$ and $k=\prod p_i \prod q_js$. This reduces to $k=mns$ which is plainly divisible by $mn$.

Answer 3

If $m\mid k \Rightarrow k=mt\ \wedge \ n\mid k\Rightarrow k=nl$. Using the Bezout's Identity $(m,n)=1\Rightarrow mx_0+ny_0=1,$ with $x_0,y_0\in \mathbb{Z}$.

$mx_0+ny_0=1 \Leftrightarrow mx_0k+ny_0k=k$

$\Leftrightarrow mx_0(nl)+ny_0(mt)=k$

$\Leftrightarrow mn(x_0l)+mn(y_0t)=k$

Therefore $mn\mid k$