For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{16} \cdot 2a + 10 \ \land d \ \vert \ 3^{17} \cdot a + 66$
$\Rightarrow d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 3 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 2$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 30 \ \land \ d \ \vert \ 3^{16} \cdot 6a + 132$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 132 - (3^{16} \cdot 6a + 30) \ = \ 102 \ = 
 \ 2 \cdot 3 \cdot 17$
Also, $d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 33 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 5$
$\Rightarrow d \ \vert \ 3^{17} \cdot 17a$ (almost with the same method as before)
So I get $d \ \vert \ 102 \ \land d \ \vert \ 3^{17} \cdot 17a$
After this, I can't see how to continue.
 A: From $d|102$ you have $d\in\{1,2,3,17,34,51,102\}$ and because $d|3^{16}2a+10$ you have $d\notin \{3,51,102\}$. Clearly $2|d$ iff $2|a$. And with $17$ you have a lot of different subcases to deal with. If $17|a$ then $d\ne 17$...
A: Hint $\ \ \ (d,102) = (d,\,2\cdot 3\cdot 17) = (d,2)\,(d,3)\,(d,17).\, $ Computing these gcds by Euclid & Fermat
$\qquad\quad\ \begin{align} (d,2) &= (\color{#c00}{3^{16}(2a)\! +\! 10}, \ \color{#0a0}{3^{17} a\! +\! 66},\,2) = (\color{#c00}0,\color{#0a0}a,2) = (a,2)\\[.2em]
(d,3) &= (\color{#c00}1,\ \color{#0a0}0,\ 3) = 1,\ \text{   and, finally, using $\,\color{#c00}{3^{16}}\equiv 1\!\!\!\!\pmod{\!17}\ $ we have}\\[.2em]
(d,17) &= (\color{#c00}{2a\!+\!10},\,\color{#0a0}{3a\!-\!2},17) = (2a\!+\!10,a\!+\!5,17) = (0,a\!+\!5,17) 
\end{align}$
Hence $\, (d,102) = (a,2)\,(a\!+\!5,17)$

Remark $\, $ The elimination that you employed to deduce that $\,d\mid 102\,$ is a special case of the determinant criterion presented here. Namely, applying that with $\,b=3^{16}\,$ yields
$\ (ab,1)\mapsto (2ab\!+\!10,3ab\!+\!66)\,$ has $\,\det = 2(66)\!-\!10(3) = 102\ $ so $\ d\mid 102(ab,1) = 102$ 
A: The prime factorization of $102$ is $2 \times 3 \times 17$.  When do $2$, $3$ and $17$ divide your numbers?
A: $$E=\gcd(3^{16}2a+10,3^{17}a+66)=\gcd(b,c)=\gcd(b,c,3c-2b)=\gcd(b,c,102)$$
$$E=\gcd(b \mod 102,b\mod 102)=\gcd(36 a+10,3a+66)=\gcd(d,e)$$
$$E=\gcd(d-12e \mod 102,e)=\gcd(34,e)=\gcd(34,3(a+22))=\gcd(34,a+22)$$
because $34 \mod 3 \neq 0$
so $E=\gcd(a,2)\times \gcd(a+5,17)$
