$\gcd(2k_1,2k_2−1)=\gcd(k_1,2k_2−1)$

Question

Prove $\gcd(2k_1,2k_2−1)=\gcd(k_1,2k_2−1)$

Both $k_1$ and $k_2$ are positive integers.

I tried to start from the fact that: $\gcd(2,2k-1) = 1$ because they are relatively prime. But I am not sure how to expand my idea

Answer 1

If $p$ is a divisor of $2k_1$ and $2k_2-1$, $p$ is odd since $2k_2-1$ is odd, this implies $p$ divides $k_1$