If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$.

My attempt at answering the question:

\begin{align*} x &\equiv 0 \pmod{a}\\\ &\Longrightarrow x\text{ is divisible by $a$}\\\ &\Longrightarrow x = ma\text{ for some integer $m$}\\\ \ \\\ x &\equiv 0 \pmod{b}\\\ &\Longrightarrow x\text{ is divisible by $b$}\\\ &\Longrightarrow x = mb\text{ for some integer $m$}\\\ \ \\\ x^2 &= (ma)(mb)\\\ x^2 &= (m^2)(ab)\\\ x &= \sqrt{m^2ab}\\\ x &= m\sqrt{a}\sqrt{b} \end{align*} Let $m$ be $k\sqrt{a}\sqrt{b}$. Then \begin{align*} x &= kab\\\ &\Longrightarrow x \equiv 0 \pmod{ab} \end{align*}

Is this correct, if not can someone point me in the right direction?