<s>[INST] 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}$$Using both result: $2^{70} \equiv 10\pmod {13}$I failed again to show that. Any hints or solution will be appreciated.Thanks in advance.
 [/INST] We have
$$2^{6k+1}\equiv 8^{2k}\cdot 2\equiv 2\pmod{9} $$
from which
$$2^{6k+2}\equiv 4\pmod {18} $$
hence by Fermat's little theorem
$$2^{2^{6k+2}}\equiv 2^4\equiv -3\pmod {19} $$
For the second  $2^4\equiv 3\pmod {13} $ and $2^{12}\equiv 1\pmod {13}$ by Fermat little theoren hence 
$$2^{70}+3^{70}\equiv 2^{70}+2^{280}\equiv 2^{10}+2^{4}\equiv 9\cdot 4+3\equiv 0\pmod {13} $$
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