Prove that if both $x$ and $y$ are odd, then $\gcd(x, y)=\gcd\left(\frac{x-y}{2}, y\right)$

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Wasn't able to include this in the title, but it is given that if $x$ is odd and $y$ is even then $\gcd(x, y)= \gcd\left(x, \frac{y}{2}\right)$.

I am struggling to get started with this problem. I would appreciate any help.

What I have so far:

Let $z = \gcd(x, y)$. We know $z\mid x$ and $z\mid y$.

Truly not sure where to go from here.

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I will give you an intuitive explanation and leave the formal proof writing to you.

First of all, note that if $G = \gcd(x, y)$ divides both $x$ and $y$, then $G$ must also divide the difference $x - y$. In other words, $\gcd(x, y) = \gcd(x - y, y)$ is immediate.

Now since both $x$ and $y$ are odd, we have $2\not \mid G$ because neither $x$ nor $y$ are divisible by $2$. However, the difference $x - y$ is even (the difference of two odd numbers is even), so taking out a factor of $2$ from $(x - y)$ is allowed and also won't affect the GCD.