Show that 13 divides $2^{70}+3^{70}$ 
Show that $13$ divides $2^{70} + 3^{70}$.

My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?
Thanks!

So by FlT, you know that $2^{12}\equiv 1\pmod{13}$ and $3^{12}\equiv 1\pmod{13}$. So 
$$
2^{70}+3^{70}\equiv (2^{12})^5\cdot 2^{10}+(3^{12})^5\cdot 3^{10}\equiv 2^{10}+3^{10}\pmod{13}.
$$
However, again by FlT, 
$$
2^{12}\equiv 2^{10}\cdot 2^{2}\equiv 1\pmod{13}\implies 2^{10}\equiv 2^{-2}\pmod{13}.
$$
That is to say, $2^{10}$ is congruent to the multiplicative inverse of $4$ modulo $13$. You can do the same for $3^{10}$, except you will need to find the multiplicative inverse to $9$ modulo $13$. You'll see that the sum of these inverses is $0$ modulo $13$, to get your answer.
