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 /: