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So I was looking at the Fermat Primes. These are primes of the form $2^k+1$ for a natural number $k$, such that I define by $\mathbb{N}:=\big\{1,2,3,\ldots\big\}$ and $0\notin \mathbb{N}$. We denote by $F_1$ the first Fermat Prime; $F_2$ the second Fermat Prime; et cetera up to $F_k$.

Thus far, however, only $5$ Fermat Primes are known. They are the following, respectively: $$3,5,17,257,65537,\ldots$$ I write $(\ldots)$ because I assume that there exist more Fermat Primes, but it is a conjecture that the above Fermat primes are the only ones.

I introduce this to you in the event that you do not know of them, because my question very much relates to these Fermat Primes.

If $2^k+1$ is prime, then must it be true that $k$ is strictly a power of $2$?

I noticed that, $$\begin{align}3&=2^{2^0}+1\\ 5&=2^{2^1}+1 \\ 17&=2^{2^2}+1 \\ 257&=2^{2^3}+1 \\ 65537&=2^{2^4}+1.\end{align}$$

Attempt at Proof:

I proved that $k$ is even, because supposing otherwise implies that $k=2j+1\,\exists j\in\mathbb{N}$ and $$2^k+1=2^{2j+1}+1=2^{2j+1}-(-1)^{2j+1}\stackrel{\centerdot}{:}2-(-1)=3.\tag*{$\bigg(\begin{align}&\text{$a\stackrel{\centerdot}{:}b$ is read as $a$} \\ &\text{is divisible by $b$.}\end{align}\bigg)$}$$ Ergo, $2^k+1$ is not prime if $k$ is odd; $k$ must be even. $\;\bigcirc$

Now if $k$ is even, then $k=2j$ and $$2^k+1=2^{2j}+1=4^j+1.$$ If $j$ is odd, then likewise, as similarly demonstrated before, $$4^j+1=4^j-(-1)^j\stackrel{\centerdot}{:}4-(-1)=5.$$ Ergo, $l\nmid k\,\exists l$ odd; $k$ must be a power of $2$ as it will only then have no odd divisors. $\;\bigcirc$

Is my proof correct? I don't see anything wrong with it, but if it is correct, are there other (better) proofs? Also, is there anything new on Fermat Primes; particularly, are there any other general constraints on $k$? How far has the conjecture been tested for? I checked the first $250$ Fermat Primes and it seems like the conjecture is true.

(If you want, you can go here, type x=2;x=2x-1;c<=250;x and then wait a minute before the first $250$ Fermat Primes appear. I still have to thank one of the users on the MSE who introduced me to the site accessible via the link.)

Thank you in advance.

Mr Pie
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1 Answers1

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A simple proof is based on the factorization of $x^n+1$ when $n$ is odd: $$ x^n+1 = (x+1)(x^{n-1}-x^{n-2}+\cdots+1) $$ Therefore, if $m=nd$ with $n$ odd, then $x^d+1$ divides $x^m+1=(x^d)^n+1$.

In particular, $2^m+1$ is divisible by $2^d+1>1$ and so is not prime.

Thus, if $2^m+1$ is prime, then $m$ has no odd factor and so is a power of $2$.

lhf
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  • $5$ sentences? That is very hard to beat, I presume. $(+1)$. Is my proof correct, though? Can I write it better? – Mr Pie May 24 '18 at 12:18
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    @user477343 Your proof is essentially correct. The idea behind yours and this one are the same. This one is shorter because it jumps quickly to the fact that the exponent can't have an odd factor. You take a while to get there. But good for you for thinking of it. – Ethan Bolker May 24 '18 at 12:22
  • @EthanBolker ok. I will try to form faster proofs next time. I guess it might just take some practise. Thank you for your comment :) – Mr Pie May 24 '18 at 12:27
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    Anyway, congratulations on your answer! $$\color{green}{\text{Accepted.}} \ \ \color{green}{\checkmark}$$ – Mr Pie May 24 '18 at 13:00