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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

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  • $\begingroup$ what is relation between n and k? $\endgroup$ – sirous Jan 15 '18 at 14:54
  • $\begingroup$ no relation Thanks $\endgroup$ – communnites Jan 15 '18 at 15:39
  • $\begingroup$ @sirous it took me a few readings to figure it out. $n$ is just a variable to define what the term "good postive integers numbers" means. So the OP is asking to prove there is a prime $p$ so that $3^k*p\not \mid 2^{3^k*p} + 1$ but $3^k*p|2^{2^{3^k*p}+1} + 1$. $\endgroup$ – fleablood Jan 17 '18 at 6:43
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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.

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  • $\begingroup$ Hello,the $2^{19}+1$ how to get it? $\endgroup$ – communnites Jan 16 '18 at 4:03
  • $\begingroup$ Tracing backwards helps..... $\endgroup$ – subham saha Jan 17 '18 at 5:59

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