<s>[INST] 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?
 [/INST] 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 $7$.
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