prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from there to apply the Euler's theorem

well if $2^{6k}4\equiv m \pmod {18}$ then $2^{2^{6k}4}\equiv 2^m \pmod {19}$.
$2^{6k}4 \equiv 0 \pmod 2$ and $\phi(9) = 6$ so $2^6 \equiv 1 \pmod 9$ so $2^{6k}4\equiv 4 \pmod 9$.  So by Chinese remainder theorem there is one solution $\pmod {18}$ to $2^{6k}4\equiv 0 \pmod 2$ and $6^{6k}4\equiv 4 \pmod 9$.
And that unique solution is $4$
And that's that.
$2^{6k}4 \equiv 4 \pmod{18}$.  So there exists an $M$ so that $2^{6k}4=18M + 4$ so $2^{2^{6k}4}=2^{18M + 4} \equiv 2^4 \pmod{19}$.
