In what cases is $\gcd(a,2b) = 2\gcd(a,b)$?

Question

I know that $\gcd(a,2b) = 2\gcd(a,b)$ is false, counter examples abound.

I am hoping for some pointers to help get a related statement that is provable, that is, conditions on $a$ and $b$ such that :

$$ \gcd(a,2b) = 2\gcd(a,b) $$

For instance, I know that if the $\gcd(a,2b) = 1$, then this is always false.

Or actually, that the $\gcd(a,2b)$ must be even, and in fact $a$ must be even.

Answer 1

We have

$$\gcd(a,2b) = 2\gcd(a,b) \quad\Leftrightarrow\quad \nu_2(a)>\nu_2(b)$$

because

$$\gcd(a,b) = 2^{\min(\nu_2(a),\nu_2(b))} \gcd(a',b')$$

where $a'=a/2^{\nu_2(a)}$ and $b'=b/2^{\nu_2(b)}$.

Answer 2

Thank you all so much!

So if I understand this correctly, $gcd(a,2b) = 2gcd(a,b)$ as long as $a$ has more factors of $2$ than $b$.

It seems so simple now!