proof writing - Proving that if $a,b$ are even, then $gcd(a,b) = 2 gcd(a/2, b/2)$

Prove that if $a, b$ are both even then $\gcd(a,b) = 2\cdot\gcd(a/2,b/2)$.

Little confused here. I have tried the following but it's basically just repeating the proof unfortunately:

• $a = 2 \cdot a_1$ and $b = 2 \cdot b_1$ (Take factor of 2 out)


• $d = \gcd(a,b)$



• $d = 2\gcd(a_1, b_1)$


• $a/2 = a_1$ and $b/2 = b_1$

I've even confused myself! Could someone help!

Answer

Following the first bullet, we let

• $a = 2a_1,\;b = 2b_1$, since we are given that $a, b$ are even.

Then why don't write $\gcd(a, b)$ in terms of the above equalities:
$$\begin{align} \gcd(a, b) & = \gcd(2a_1, 2b_1) \tag{1}\\ \\ & = 2\gcd(a_1, a_2) \tag{2} \end{align}$$

Now, by our bullet, solving for $a_1, a_2$, we get $a_1 = \dfrac{a}{2},\;\;b_1 = \dfrac b2$ (which is your last bullet!)

Now, the only step left is to just substitute these values of $a_1, b_1$ into $(2)$, and you're done!