Prove that if $\gcd (m,n)=1$ and $m\mid x$ and $n\mid x$, then $mn\mid x$. I've come across the statement that if $\gcd (m,n)=1$ and $m\mid x$ and $n\mid x$, then $mn\mid x$. (This is needed for a proof of the correctness of RSA that I have been given.)
I can't see how to prove that is the case. Can anyone either show me how, or give me a clue?
(NB: gcd = greatest common divisor = highest common factor = hcf)
 A: $x=k\cdot m$ and $n$ divides $k\cdot m$.
From Euclid's lemma $n\mid k$ so $k=c\cdot n$
Replacing we have $x=c\cdot n \cdot m$
A: From the definition of L.C.M., $m|x, n|x$ implies L.C.M.of ${m,n}$ divides $x$. If $(m,n)=1$ then their L.C.M is $mn$.
A: Solve $mu+nv=1$ then multiply by $x$.
So $mxu+nxv=x$. 
But $n\mid x$ means $mn\mid mx\mid mxu$ and $m\mid x$ means $mn\mid xn\mid xnv$. So $mn\mid (mxu+nxv)=x$.
A: Hint using ring theory:
$m\mathbb{Z}$$\cap$  $n\mathbb{Z}$ $=$ $lcm(m,n)\mathbb{Z}$ where $m\mathbb{Z}$ denotes ideal of $\mathbb{Z}$   generated by  $m$.
A: From Bezout's identity, $\gcd(m,n) = 1$ implies that there exist integer $a$ and $b$ where $am + bn = 1$.  $m|x$ means there exists integer c such that $cm = x$, and likewise $dn = x$.
$$\begin{align}
&& am + bn = 1\\
\Rightarrow && amx + bnx = x\\
\iff && am(dn) + bn(cm) = x\\
\iff && (ad + bc)mn = x\\
\Rightarrow &&mn | x
\end{align}$$
A: Since $\Bbb Z$ is a Euclidean domain, least common multiples (common multiples that divide all other common multiples) always exist there. Also $\gcd(m,n)=1$ implies that no proper divisor $d$ of $mn$ is common multiple of $m,n$ (if it were, $\frac{mn}q$ would be common divisor of $m,n$). Now $m\mid x$ and $n\mid x$ (i.e., $x$ is a common multiple of $m,n$) together imply that $x$ is a multiple of (the least common multiple) $mn$.
