Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$ Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$
Is there some intuitive method or formula for finding GCD of $n$ integers?
 A: $n=0$ gives $1+9+25 = 35$, so it's either $1, 5, 7$ or $35$. 
$n=1$ gives $8+ 3^8 + 5^8 = 397194 = 2 \cdot 3 \cdot 7^3 \cdot 193$ so it could be $1$ or $7$.
Let's see what happens mod  $7$ (this means we look at all numbers ignoring powers of $7$, so basically any number becomes $0,1,2,3,4,5$ or $6$ according to the rest of its division by $7$)
$2^{3n} = 8^n = 1^n = 1$ 
$3^{6n+2} = (9)(3^{6n}) = 2(2^{3n}) = 2(1)^n = 2$ 
$5^{6n+2} = (25)(5^{6n}) = 4(4^{3n}) = 4(64^n) = 4(1)^n = 4$ 
$A_n = 1 + 2 +  4 = 7 = 0$ 
So any of them is divisible by $7$, so it's $7$.
A: Fisrtly, $A_0 = 35$, so $d:=\gcd(A_0\ldots A_{2015})\mid 35$, that means that $d \in \{1,5,7,35\}$.
Now let's try to factorize the $\gcd$. Starting with $5$, notice that $2^{4n} \equiv 3^{4n} \equiv 1\pmod{5}$ for every $n\geqslant 0$ (Fermat's little theorem). So:
\begin{align}A_n = 2^{3n}+3^{6n+2}+5^{6n+2} \equiv 2^{3n}+ 3^{2n+2} \pmod{5}\end{align}
By the previous observation, one can see that $A_{1} \equiv 3+1 \equiv 4\pmod{5}$, so $d\in\{1,7\}$. Again, having Fermat's little thm in mind:
\begin{align}A_n = 2^{3n}+3^{6n+2}+5^{6n+2} &\equiv 8^n+ 9\cdot 3^{6n} + 25\cdot 5^{6n} \pmod{7} \\ &\equiv 1^n + 2\cdot 1^n + 4\cdot 1^{n} \pmod{7} \\ &\equiv 7 \pmod{7} \\ &\equiv 0\pmod{7}\end{align}
So $7\mid A_n,~\forall n\geqslant 0$, thus $d$ must be $7$.
