<s>[INST] $19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ 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 /:
 [/INST] Hint:
$$2^{n+6}+3^{n+6}+5^{n+6}=2^6\cdot2^n+3^6\cdot 3^n+5^6\cdot5^n\equiv7\cdot2^n+7\cdot3^n+7\cdot5^n\ \ (\text{mod }19)$$
What can $2^k\text{ mod }6$ be?
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