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

  • $\begingroup$ Can you show $\gcd(2k_1,2k_2-1)|\gcd(k_1,2k_2-1)$ and $gcd(k_1,2k_2-1)|\gcd(2k_1,2k_2-1)$ $\endgroup$ Nov 8 '20 at 21:21

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 $k_1$

  • 1
    $\begingroup$ You mean a common divisor or a prime one. $\endgroup$ Nov 8 '20 at 21:23
  • $\begingroup$ but why is p a prime $\endgroup$
    – hello_word
    Nov 8 '20 at 21:29
  • $\begingroup$ If $p$ is not prime, the same argument works. $\endgroup$ Nov 8 '20 at 21:30
  • $\begingroup$ Im sorry but I still dont get it. Could you explain it with more details plz? $\endgroup$
    – hello_word
    Nov 8 '20 at 21:32
  • 1
    $\begingroup$ if the pair $(2k_1,2k_2-1)$ and $(k_1,2k_2-1)$ have the same divisors, they have the same $gcd$. $\endgroup$ Nov 8 '20 at 21:39

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