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!

Okay, I'm on a little different wavelength so I'll turn my comment into an answer.. if $n$ is odd the polynomial $x + y$ divides $x^n + y^n$. So letting $x = 2^2, y = 3^2,$ and $n = 35$ you get that $13 = 2^2 + 3^2$ divides $2^{70} + 3^{70}$.

Compute $2^{70}$ and $3^{70}$ modulo $13$ separately (e.g., using Fermat's Little Theorem). If $2^{70}\equiv a\pmod{13}$ and $3^{70}\equiv b\pmod{13}$, then what is $2^{70}+3^{70}$ congruent to modulo 13?

$2^{12} \equiv 1 \pmod{13}$ and $3^{12} \equiv 1 \pmod{13}$ by Fermat's Little Theorem.

Hence, $2^{72} \equiv 1 \pmod{13}$ and $3^{72} \equiv 1 \pmod{13}$

$2^{72} \equiv 1 \pmod{13} \Rightarrow 2^{72} \equiv 40 \pmod{13} \Rightarrow 2^{70} \equiv 10 \pmod{13}$

$3^{72} \equiv 1 \pmod{13} \Rightarrow 3^{72} \equiv 27 \pmod{13} \Rightarrow 3^{70} \equiv 3 \pmod{13}$

Hence, you get the result.

HINT $\rm\: $ Little Fermat shows $\rm\ mod\ 13:\ \ 2^{70} + 3^{70}\ \equiv\ 2^{-2} + 3^{-2}\ \equiv\ 1/4+1/9\ \equiv 13/36\ \equiv\ 0$

NOTE $\ $ For a slight generalization see here.

The "quick" way to do this is as follows:

$2^{6}=64=-1$(mod 13), $3^{3}=1$ (mod 13)

Hence you have $2^{70}=(2^{6})^{11}*2^{4}=(-1)*(3)=-3$ (mod 13)

And $3^{70}=(3^{3})^{13}*3=3$ (mod 13)

Therefore the result is $-3+3=0$ (mod 13).

Fermat's little theorem implies $a^{p-1}\cong 1$ (mod p), but this may be more slick, I am not sure. Mind that Fermat's little theorem is not optimal in many cases.

The "ultra quick" way may be $2^{70}+3^{70}=3^{70}*(1+(2/3)^{70})$. Now notice $2/3=2*9=5$. And $5^2=-1$, hence $(1+(5^{2})^{25})=1+-1=0$. Therefore $2^{70}+3^{70}=0$.