<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] Not with induction:
The residues of each term cycle since $a^{2^{n+1}} = (a^{2^n})^2$
$2^{2^n} \pmod {19}$ with $4, 16, 9, 5, 6, 17, 4, \dots$
$3^{2^n} \pmod {19}$ with $9, 5, 6, 17, 4, 16, 9, \dots$
$5^{2^n} \pmod {19}$ with $6, 17, 4, 16, 9, 5, 6, \dots$
By adding terms pairwise, you can see that the statement is true. This also provides some motivation for induction using $k+6$ rather than $k+1$.
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