elementary number theory - If $gcd(m,n)=1$ then $gcd(2m-n-1,m-1)=2$?

How to prove following statement:

If $\gcd(m,n)=1$, then $\gcd(2m-n-1,m-1)=2$, where $m,n$ are odd numbers and $m>n$.

Since $m=2k_1+1$ and $n=2k_2+1$ we may write:

$2m-n-1=2(2k_1+1)-2k_2-1-1=4k_1-2k_2=2(2k_1-k_2)$

$m-1=2k_1+1-1=2k_1$

So we may conclude that common divisor is $2$ but how to prove that it is greatest common divisor of those two numbers ?

Answer

If $m=9$ and $n=5$ then $\gcd(9,5) =1$ but $\gcd(2\times 9-5-1,9-1) = \gcd(12,8) =4$ so it is not true.

From your previous work, if $k_1$ and $k_2$ have a common factor but $2k_1+1$ and $2k_2+1$ do not then you will find a counterexample.