Find a GCD of number $A_0,A_1,\cdots,A_{2013}$ if

$A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ where $n=0,1,\cdots,2013$

I have no idea can you help me. Only what I can see that they have the same degree $3n$ and $5=3+2$ but I do not know can that help me somehow.


Note that $A_0 = 35$. This should give you a hint , because $35$ has only four divisors, therefore only one of these can be the GCD. Let us verify via congruences the remainders upon division by $5$ and $7$.

Note that $2^{3n} \equiv 3^n \mod 5$ as $2^3 = 8 \equiv 3 \mod 5$.

Next, for the second term we have : $3^{6n+2} \equiv 9^{3n+1} \equiv (-1)^{3n+1} \equiv (-1)^{n+1}\mod 5$.

Therefore, as a whole we have $A_n \equiv 3^n + (-1)^{n+1} \mod 5$. As can be seen by taking $n = 1$, $A_n \equiv 4 \mod 5$, so is not a multiple of $5$ in general.

Next, we have $2^{3n} = 8^n \equiv 1 \mod 7$. For the second term, $3^{6n+2} \equiv 2^{3n+1} \equiv 2 \times 8^{n} \equiv 2 \mod 7$. For the third term, $5^{6n+2} \equiv 4^{3n+1} \equiv 4 \times 8^{2n} \equiv 4 \mod 7$. Hence,the sum is $1+2+4 = 7 \equiv 0 \mod 7$ for all $n$. Consequently, $A_n$ is a multiple of $7$ for all $n$.

This tells us that the $\gcd$ from $A_0$ to $A_{2013}$ is $7$.


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