Proof about property of the gcd if $m|a$ and $m|b$ then $(a/m,b/m)=(a,b)/m$
proof
show $(a/m,b/m)\leq (a,b)/m$ and $(a/m,b/m) \geq (a,b)/m$
Let $(a,b)=d$, so by bezout's identity there exists intergers x,y such that ax+by=d
$$ax+by=d$$
$$ax/m+by/m=d/m$$
This implies that $gcd(a/m,b/m)|d/m$, so $gcd(a/m,b/m)\leq d/m$
I just could show that direction. Need help on the other one.
 A: Hint $\ $  You can unify both directions and simplify and generalize the proof as follows
$$ d\mid (a,b)/m \!\iff\!  dm\mid (a,b) \!\iff\! dm\mid a,b \!\iff\! d\mid a/m,b/m \!\iff\! d\mid(a/m,b/m)$$
So $\ (a,b)/m\, =\, (a/m,b/m)\ $ since they have same divisors $\,d,\,$ so they divide each other.
Remark $\ $ See here for a  few more proofs of this gcd Distributive Law.
A: First, We need to prove that $\gcd(ka,kb)=k\gcd(a,b)$. Set $d=\gcd(a,b)$. since $d\mid a$ and $d\mid b$, then $kd\mid ka$ and $kd\mid kb$, so $kd\mid \gcd(ka,kb)$.
conversely, $\gcd(ka,kb)\mid ka$ and $\gcd(ka,kb)\mid kb$, so $\gcd(ka,kb)\mid kax+kby$ for every $x,y\in Z$. Let $x_0,y_0\in Z$ such that $d=ax_0+by_0$, so in particular, $\gcd(ka,kb)\mid kax_0+kby_0$, that is 
$\gcd(ka,kb)\mid kd$. 
Now, your claim follows almost immediately, since
$$
m\gcd(a/m,b/m)=\gcd(a,b)
$$
A: The other direction is almost the same:
By Bézout identity, there are $x, y$ such that $(a/m)x+(b/m)y=\gcd(a/m,b/m).$ Then multiply the equation by $m$ and obtain $$ax+by=m\cdot\gcd(a/m,b/m).$$ Hence $\gcd(a,b)\mid m\cdot\gcd(a/m,b/m).$
Combined with what you proved, this proves the statement.  
Hope this helps.
