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!
 A: 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!
A: With the notation $\rm\ (x,y) := gcd(x,y),\ $ we have $\rm\:2(m,n)\mid 2m,2n\:\Rightarrow\:\color{#c00}{2(m,n)\mid (2m,2n)},\:$
$\rm\ (2m,2n)\mid 2m,2n\:\Rightarrow\: (2m,2n)/2\mid m,n\:\Rightarrow\:(2n,2m)/2\mid (m,n)\:\Rightarrow\:\color{#c00}{(2n,2m)\mid 2\,(m,n)}.$
Remark $\ $ Note that $\rm\:(2m,2n)/2\in\Bbb Z\:$ since $\rm\: 2\mid 2n,2m\:\Rightarrow\:2\mid (2n,2m)$
The proof exploits to the hilt the fundamental universal property of the gcd i.e.
$$\rm a\mid b,c\iff a\mid(b,c)$$
Note that the same proof works if we replace $2$ by any integer $\ne 0,\,$ yielding the
GCD Distributive Law $\rm\ \ \ (ac,bc)\, =\, (a,b)c$
See here for a few proofs (by linearity (Bezout); universality; prime factorization), e.g. the following more efficient version of the original proof above.
Theorem $\rm\ \ (a,b)\,c\, =\, (ac,bc)$
Proof $\rm\quad\: d\mid(a,b)\iff d\ |\ a,b \iff dc\ |\ ac,bc \iff dc\ |\ (ac,bc) \iff d\ |\ (ac,bc)/c$
