If positive integer $n$ is not power two and $n^2+n+1 \mid 4^n+2^n+1$, then what sequence of $n$?

Source problem from Prove that there are infinitely many integers $n>0$ such that $n^2+n+1$ divides $4^n+2^n+1$.

We have two first term of sequence: 215,3692374808.

Let $m=n^2+n+1$. For both first terms $m$ is prime.

Question 1: can be $m$ is not prime for next terms?

Suppose 1: $m$ is prime $\overset{FLT}{\implies} 2^{n(n+1)}\equiv 1 \pmod{m} \implies 2^{3n}\equiv 1 \pmod{m}$.

Question 2: how prove if $m$ is prime then $n\equiv2\pmod{3}$?

Note: from post Modular arithmetic with Legendre symbol implies $n\not\equiv1\pmod3$.

Suppose 2: $n\equiv2\pmod{3} \implies 2^{\frac{n(n+1)}{3}}\equiv3^{\frac{n(n+1)}{3}}\equiv n^x\pmod{m}$, where some integer $x\ge0$.

Question 3: what other conditions for $n$ we can seen for speedup find next terms?

  • 1
    $\begingroup$ Please define all data here, without reference to the "other post", which is itself hard to compile. So we have "two term"(s) of which sequence? Please start the question in the text not (only) in the title. So what is $n$, that then determines some $m$. "Next terms" are the terms of which sequence? Abbreviations like FLT over arrows make the question even less readable. What do we suppose? That $m$ (still undefined) is prime? That first implication arrow holds? That both implications hold? Moreover, here we have a lot of questions, but no word on motivation and own tries, not on this site. $\endgroup$
    – dan_fulea
    Jun 20, 2019 at 18:25
  • $\begingroup$ The latter is a form of the former evaluated at $2^n$ . $\endgroup$
    – user645636
    Jun 21, 2019 at 1:13


You must log in to answer this question.