If $\operatorname{gcd}(a,b)=1$ then $\operatorname{gcd}(3a+b,a+3b)=$?

Question

If $\operatorname{gcd}(a,b)=1$ then $\operatorname{gcd}(3a+b,a+3b)=$?

I try to solve it.

We are two cases :

1. If $a$ be even and $b$ is odd number. Then $\operatorname{gcd}(3a+b,a+3b)=1$

2. $a$ and $b$ be odd numbers.then $\operatorname{gcd}(3a+b,a+3b)$ divided 8...

I guess $\operatorname{gcd}(3a+b,a+3b)=2$.

Answer 1

If $d$ divides $3a+b, a+3b$

d must divide

• $3(3a+b)-(a+3b)=8a$

• $3(a+3b)-(3a+b)=8b$

$\implies d$ must divide $(8a,8b)=8(a,b)=8$

So, $g=(3a+b,a+3b)\in\{1,2,4,8\}$

$g=1$ if $3a+b$ is odd $\iff a+b$ is odd i.e., $a,b$ have opposite parities

Again $g=8$ if $3a+b\equiv0\pmod8\iff b\equiv-3a\pmod8\iff a\equiv-3b\pmod8$

What about $g=2,4?$