Greatest common divisor relatively prime integers proof Let $gcd(c,m)=g.$
Show that if $kc+lm=g$, then $gcd(k,l)=1$
I can see that its true with this example 
Let $c=5, m=15$. We have $gcd(5,15)=5$, then let $k=-2$ and $l=1$ $gcd(-2,1)=1$ but I'm not sure how to generalize it.

We have
$$
\left\{ \begin{gathered}
  \gcd (c,m) = g \hfill \\
  kc + lm = g \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \left\{ \begin{gathered}
  \gcd (c',m') = 1 \hfill \\
  kc' + lm' = 1 \hfill \\ 
\end{gathered}  \right.
$$
Suppose $ \gcd (c,m) \ne 1$ , then you would get the contradiction
$$
\begin{gathered}
  \gcd \left( {k,l} \right) = q \ne 1\quad  \Rightarrow \quad k'c' + l'm' = \frac{1}
{q}\quad  \Rightarrow \quad  \hfill \\
   \Rightarrow \quad \text{at}\,\text{least}\,\text{one}\,\text{of}\,k',l',c',m'\text{not}\,\text{integer} \hfill \\ 
\end{gathered} 
$$
