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: No, your approach is very false. You could take $n=1$ and the congruence would always hold. 
Your statement is equivalent to $2^{70} + 3^{70}$ being congruent to $0$ modulo $13$. 
Thus, what you want to do is compute $2^{70} + 3^{70}$ mod $13$. 
To this end you may wish to recall  $a^{13}$ is $a$ modulo $13$, which is basically Fermat's Little Theorem. 
A: 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 $2^{2\cdot 35}+3^{2\cdot 35}$.
A: ${\rm mod}\ 13\!:\, \left[ 2^{\large 2}\equiv -3^{\large 2}\right]^{\large 35}\!\Rightarrow\, 2^{\large 70}\equiv -3^{\large 70}\,\Rightarrow\, 2^{\large 70}+3^{\large 70}\equiv 0\ $ by the Congruence Power Rule
A: Hint.
we have
$$3^3=27 \equiv 1 (13) \implies$$
$$3^{69}\equiv 1 (13)  \implies$$
$$3^{70} \equiv 3 (13).$$
and
$$2^4 \equiv 3 (13)  \implies$$
$$2^{12} \equiv 1 (13)  \implies$$
$$2^{60} \equiv 1 (13).$$
but
$$2^{10} \equiv 10 (13)$$
then
$$2^{70} \equiv 10 (13)$$
and finally
$$2^{70}+3^{70}\equiv 10+3 (13)$$
or
$$2^{70}+3^{70} \equiv 0 (13)$$
qed.
A: To prove this, compute $2^{70} \mod 13$ and $3^{70} \mod 13$. This can be done using Fermat's Little Theorem. Then add the results. It will be 0 mod 13 since
$$
2^{70} + 3^{70} = 13 \cdot 192550423461109399456637645953021 \, .
$$ 
A: $$2^{70}=2^{12\cdot5+10}=1^5\cdot2^{10}=1024-13\cdot78\equiv 10\pmod{13}\\3^{70}=3^{12\cdot5+10}=1^5\cdot3^{10}=59049-13\cdot4542\equiv 3\pmod{13}$$ It follows $$2^{70}+3^{70}\equiv 0\pmod{13}$$
