If $2^n+1$ is prime then $n=2^k$ for some positive integer $k$.

Question

This problem came up as an extra problem to do for homework in my discrete mathematics class. I'm really just looking at some sort of a hint on how to proceed because I'm really not sure what to do. I figure I probably want to assume that n is not equal to $2^k$ but I don't know how to express this in a way that I can use it in the proof.

score 1 · Answer 1 · answered Sep 17 '19 at 16:53

1

Hint: Suppose $n$ has an odd factor $m$. If $m$ is odd, then $$ a^m + 1 = (a+1)(a^{m-1}-a^{m-2}+...+a^2-a+1) $$ Use this formula with $a=2$.

answered Sep 17 '19 at 16:53

Prism

11,782

How is this a "hint"? Isn't this a full solution, with a few minor details left to the reader? – Arthur Sep 17 '19 at 16:54
@Arthur: Well, you are probably right. It is more than a hint. – Prism Sep 17 '19 at 16:58
@Arthur but less than a full solution as one has to sue $a=2^{n/m}$ for the general case ... – Hagen von Eitzen Sep 17 '19 at 17:20

score 0 · Answer 2 · answered Sep 17 '19 at 16:53

0

If $n$ is odd the $a^n + b^n$ is divisible by $a+b$. Hence if $n$ is odd, $2^n + 1$ is divisible by $3$. Can you complete the proof that $n = 2^k$ from here?

answered Sep 17 '19 at 16:53

Nilotpal Sinha

24,086

Yeah I think I can, thank you this is definitely very helpful. – Bryce Smith Sep 17 '19 at 23:39

If $2^n+1$ is prime then $n=2^k$ for some positive integer $k$.

2 Answers2