<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] $n|a + b$ does not mean that $n|a$ and $n|b$.  It means that if $a \equiv k \mod n$ then $b \equiv -k \mod n$.
So if $2^{70}\equiv 10 \mod 13$ (which it is) then $13|2^{70} + 3^{70}$ if $3^{70} \equiv -10 \mod 13$.
Does $3^{70}\equiv -10 \mod 13$?
Well If $0 < a < 13$ then $a^{12} \equiv 1 \mod 13$ and $a^{60} = (a^{12})^5\equiv 1 \mod 13$ so $a^{70} \equiv a^{10}\mod 13$.
And $2^{10} \equiv 10 \mod 13$.
And $3^3 \equiv 27 \equiv 1 \mod 13$ so $3^9 \equiv 1^3 \mod 13$ and $3^{10} \equiv 3 \equiv -10 \mod 13$.
So $2^{70} + 3^{70}\equiv 2^{10} + 3^{10} \equiv 10 + (-10)\equiv 0 \mod 13$.
i) is a lot harder but the idea is that as $2^{18} \equiv 1 \mod 19$ then if $2^{6k + 2}\equiv m \mod 18$ then $2^{2^{6k+2}}\equiv 2^m \mod 19$.
$2^{6k+2} = 64^k*4 = (7*9 + 1)^k*4$.  Note $(7*9 + 1)^k$ will be $1$ more than a multiple of $9$.  So $(7*9 + 1)^k*4$ will be $4$ more than a multiple of $4*9 = 2*18$.  So $2^{6k+2} \equiv 4 \mod 18$.
And $2^{2^{6k + 2}}\equiv 2^4 \equiv 16 \mod 19$.
So $2^{2^{6k+2}}+3 \equiv 16 + 3 \equiv 0 \mod 19$. 
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