Application of Fermat's little theorem to check divisibility Using Fermat's little theorem to prove: $(i)19\mid 2^{2^{6k+2}}+3$, where $k=0,1,2.....$$(ii)13\mid 2^{70}+3^{70}$My Approach: I couldn't think of how to go with $(i)$ but i tried $(ii)$ to show $2^{70} \equiv 0\pmod {13}$ and $3^{70} \equiv 0\pmod {13}$.Since,
$$2^{12} \equiv 1 \pmod {13}\Rightarrow (2^{12})^5\equiv 1 \pmod {13}\Rightarrow2^{60} \equiv 1 \pmod {13}$$Again, $$2^4 \equiv 3 \pmod {13}\Rightarrow2^8.2^2 \equiv 10\pmod {13}$$Using both result: $2^{70} \equiv 10\pmod {13}$I failed again to show that. Any hints or solution will be appreciated.Thanks in advance.

In (mod 13): $ 3^{70} = 9^{35} = (-4)^{35} = -(4^{35}) = -(2^{70}) $
So that: $ 2^{70} + 3^{70} = 0 (mod 13) $
The main idea is: $ 3^2 = (-1)2^2 $ (mod 13)
For all odd n: $ 13 $ $|$ $2^{2n} + 3^{2n} $
