if $n\nmid 2^n+1, n|2^{2^n+1}+1$ show that the $3^k\cdot p$ is good postive integers numbers

Question

Let say $n$ is good postive integers numbers, if such $$n\nmid 2^n+1, ~~~~~~~n|2^{2^n+1}+1$$

show that there exsit prime number $p>3$ such for any postive integers $k\ge 2$ the $3^k\cdot p$ is good postive integers numbers

I think maybe use Femat Fermat's little theorem solve it,But I can't

Answer 1

Take any prime factor $p \neq 3$ of $2^{19} +1$, this p works.

Because, $o_{p} (2) = 38$, easy to see, so, $ p \nmid 2^{{3^k}.p} + 1$, for any positive integer $k$, also, by LTE, $3^k \mid 2^{2^{3^k.p}+1}+1$ as $3^{k-1}\mid 2^{3^k.p}+1$ for any positive integer $k$.

Also, as $ p \mid 2^{19} +1$, we have $ p \mid 2^{3^9 +1}+1$, thus, $ p \mid 2^{3^{9.p} +1}+1$ and thus, done.