Can I directly apply the theorem 1=(a/d,b/d) for this? Let a and b be two positive even integers. Prove that (a, b) = 2(a/2, b/2)

Question

Let d=(a,b)
a and b are even integers
$→$ $2|a$ $and$ $2|b$
$→$ $2|ax + by$ ($x,y$ are some integers)
$→$ $2|d$
Then we can express,
$a =2p,$ $b=2q$, $d= 2k$ ($k,p,q$ are some integers)
then $d=(a,b)=(2p,2q)$
We know that $(a/d,b/d)=1$
Then,
$1 = (2p/2k,2q/2k)$
$1 = (p/k,q/k)$
$k = (p,q)$
$d/2 = (a/2,b/2)$ (since $k=d/2,p=a/2,q=b/2$)
$d = 2(a/2,b/2)$
My problem is, Can I directly apply the theorem $(a/d,b/d)=1$ to prove this?
Please, Can anyone offer some assistance, please?

Answer 1

Yes, you can directly apply the theorem, but you don't need to. See this.

Here's a proof for the theorem:

Let $d=(a,b)$, then there exists $x,y\in\mathbb Z$ with $a=dx$ and $b=dy$.

We want to show that $(x,y)=1$. Suppose that $(x,y)=d'>1$. Then there exist integers $m,n$ with $x=d'm$ and $y=d'n$, but then we have $a=dx=dd'm$ and $b=dy=dd'n$, so $(a,b)>d$, which is a contradiction since $d=(a,b)$. Thus we have $(x,y)=1$.

In other words, since the greatest common divisor, $d$, is the largest positive integer that divides $a$ and $b$, if $(\frac{a}{d},\frac{b}{d})>1,$ then $d$ will not be the greatest common divisor of $a$ and $b$ because $\frac{a}{d}$ and $\frac{b}{d}$ have a common divisor bigger than $1$.