Find $\gcd(3^{20} + 3, 3^{21} +6)$

Question

Find $\gcd(3^{20} + 3, 3^{21} +6)$

I am honestly so confused, I know $3$ divides both terms but am unsure if that's the $\gcd$.

Answer 1

Hint: if $d$ divides $a$ and $b$, it also divides $pa+qb$, for every $p$ and $q$; thus the gcd will divide $$ 3(3^{20}+3)+(-1)(3^{21}+6) $$ Why $p=3$ and $q=-1$? Because doing so will remove the powers of $3$.

Answer 2

Alternatively, without explicitly eliminating the powers of $3$

$\qquad \gcd(\underbrace{ 3^{20}\!+\!3}_{\large a},\, 3(\underbrace{3^{20}\!+\!2)}_{\large a\ -\ 1}) = \gcd(a,3(a\!-\!1)) = \gcd(a,3)=3,\,$ by $\,a,a\!-\!1$ coprime