Is this a correct demonstration that 13 divides $2^{70} + 3^{70}$? Is it true that: $13|(2^{70} + 3^{70})$ iff 13 mod(n) divides $2^{70} + 3^{70}$ mod(n)?
I assumed that it is the case. Being that correct, then 13 = 1  (mod(2)) and $2^{70} = 0$ (mod(2)); $3^{70} = 3x3x...x3 = 1x1x...x1 $ (mod(2)). Therefore $2^{70} + 3^{70}$ = 0 + 1 = 1 = 13 (mod(2)), and it follows that $13|2^{70} + 3^{70}$.
I missed a lot of classes on modular arithmetic and didn't compensate it at home, so I probably got something wrong. But what I used to conceive this demonstration is that, as far as I know, the set of congruence classes forms a subring of the integers, therefore it satisfies all ring operation properties.

A trickier approach. Since $4$ and $9$ are quadratic residues $\!\!\pmod{13}$, for $p=13$ we have
$$ 4^{\frac{p-1}{2}}\equiv 9^{\frac{p-1}{2}}\equiv 1\pmod{13} $$
hence
$$ 4^{36}\equiv 9^{36}\equiv 1\pmod{13} $$
and
$$ 2^{70}+3^{70} = 4^{35}+9^{35} \equiv 4^{-1}+9^{-1} \equiv 0\pmod{13} $$
since $\frac{1}{4}+\frac{1}{9}=\frac{\color{red}{13}}{36}$. Even simpler: $35$ is odd, hence $13=2^2+3^2$ is a divisor of $2^{2\cdot 35}+3^{2\cdot 35}$.
