Show that: and
Let
Case 1: d|a or d|b
If d|a then d|b (Why?)
Forcing (Why?)
Case 2: d neither divides a nor b
Observe that
By case hypothesis, d|b-a which means d|2b. And similarly d|2a.
Now only the following conditions can be true(since (a,b) = 1)
Thus we have showed that d can only be 1 or 2.
Now to show that there are numbers that achieve both gcd:
Choose
and
Choose