$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ I tried to demonstrate the next equation is divisible by 19:
$$ 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}} $$
When $n$ is $1$:
$$ 2^{2^1} + 3^{2^1} + 5^{2^1} $$
$$ 4 + 9 + 25 = 38 $$
When $n$ is $k$:
$$ 2^{2^k} + 3^{2^k} + 5^{2^k} $$
Finally, when $n$ is $k+1$:
$$ 2^{2^{k+1}} + 3^{2^{k+1}} + 5^{2^{k+1}} $$
I try by expanding, by subtraction, but no solution /:

By Fermat's little theorem, if $n\equiv m\ \ \text{mod}\  (p-1)$, then $a^m \equiv a^n\ \ \text{mod}\ p$.
Now, since any power of two modulo $18$ is one of the six numbers in $S=\{2,4,8,10,14,16\}$, it remains to check that:
$$2^k+3^k+5^k \equiv 0 \ \ \text{mod}\ 19$$
For all $k\in S$.
