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Homework exercise: prove that there exists such a natural number sequence $a_1,a_2,\ldots$, such that $GCD(a_n,2^{a_n}+1)\rightarrow\infty$ when $n\rightarrow\infty$. First of all I tried to just find any such sequence that works for at least the first few terms and quickly came up with $a_n=3^n$. This seems to work up until $n=19$, the limit of WolframAlpha. The $GCD(3^n,2^{3^n}+1)$ seems to be $3^n$. I am not sure why this is the case, but since it seems to hold for the first 19 terms, I am conjecturing that it holds for all $n$. Proving (or disproving) this claim didn't prove to be easy. I tried applying Euler's theorem: $$\phi(3^n)=3^n-3^{n-1}=2\cdot3^{n-1}$$ $$2^{3^n}+1\equiv2^{3^{n-1}+2\cdot3^{n-1}}+1\equiv2^{3^{n-1}}+1 \ (\textrm{mod}\ 3^n)$$ but it got me nowhere. This is the point where I ran out of ideas. I didn't manage to find another sequence which seems to answer the original question. Am I on the right path? Should the proof even be constructive? Please help. Thank you.

DDomjosa
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    Hint: $2$ is a primitive root $\pmod {3^n}$ for all $n$, see this – lulu Sep 18 '22 at 20:53
  • @lulu I can't really see how that is useful... – DDomjosa Sep 18 '22 at 21:02
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    Well...you are trying to prove that $2^{3^n}\equiv -1 \pmod {3^n}$ or something close to that and what I claim proves that the order of $2\pmod {3^n}$ is $2\times 3^{n-1}$. Pretty close. – lulu Sep 18 '22 at 21:16
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    Hint $#2$ : what is $2^{3^{n-1}} \pmod {3^n}$? – lulu Sep 18 '22 at 21:17
  • @lulu OOOOOH, gotcha. Whatever $2^{3^{n-1}}$ is, it squared would have to be congruent to 1, yet it itself cannot equal 1, since 2 is a primitive root. So, -1. Thank you!! – DDomjosa Sep 19 '22 at 03:40
  • A little more work is needed...since $3^n$ is not prime (for $n>1$) you need to argue that $\pm 1 $ are the only square roots of $1\pmod {3^n}$. After all, $\pmod {15}$ there are four square roots of $1$. But just thinking about $x^2-1=(x+1)(x-1)$ modulo odd prime powers should be enough. – lulu Sep 19 '22 at 11:05

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