Originally Posted by

**ThePerfectHacker** Note we can write,

gcd[a+3b,a+3b-(2b-2a)]=d

Then,

d|(a+3b) and d|(a+3b-(2b-2a)|

By rules of divisibility,

d|(2b-2a) thus, d|2(b-a)

If d is odd then,

d|(b-a)

But then, d|(a+3b+b-a) thus, d|4a thus, d|a

But then from d|(b-a) we have d|b thus, d=1

Thus, the only odd value that can be is d=1

Say d is even and larger then 2.

Then d|2(b-a) yields d'|(b-a) that is d'=d/2

Using the same argument we see that d'|b and d'|a thus, d'=1

Thus, the only possible values are d=1,2