# Finding a greatest common divisor

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 $$7$$.