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How can I prove that if $2^x+1$ is prime, then $x$ is a power of $2$? Can someone help me prove this?

hadron
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  • Sorry. Forgot a $+1$ It's edited now –  Sep 06 '16 at 14:45
  • Not equivalent. $2^{2^x}=4^{2^{x-1}}$ –  Sep 06 '16 at 14:46
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    You probably meant to ask about $2^x + 1$ - if that's prime then $x$ is a power of two. – Ethan Bolker Sep 06 '16 at 14:46
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    Not the same. $2^{2x} = 4^x$ and this is very different from $2^{2^x}$. –  Sep 06 '16 at 14:46
  • @Ethan Bolker: Yea you're right... Going to etdit it now –  Sep 06 '16 at 14:47
  • Not true! The first calculation with $x=3$ gives a counterexample. – Piquito Sep 06 '16 at 14:52
  • Why did you post two almost identical questions? – Ross Millikan Sep 06 '16 at 14:52
  • @Piquito Since when has $9$ been a prime? – Jam Sep 06 '16 at 14:53
  • @Ross Milikan; I tried to delete the other one because it was really wrong... I didn't do the job better this time... –  Sep 06 '16 at 14:53
  • I never stop being surprised how triggerhappy people are here to jump in and answer questions that obviously have typos / are super-easily shown to be false. It would be much better if people gave OP some minuttes to clarify and comments are here exactly for this purpose: for asking for clarifications. – Winther Sep 06 '16 at 14:53
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    @Winther The same could be said for people too trigger-happy to edit their questions before posting. – Jam Sep 06 '16 at 14:56
  • @Winther, people often ask questions here that are "super-easily" shown to be false. Super easy for us $\ne$ super easy for them. –  Sep 06 '16 at 15:04

3 Answers3

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Hint. You can probably show that $$ A-B \text{ divides } A^n - B^n . $$

Then try to show that $$ A+B \text{ divides } A^n + B^n $$ when $n$ is odd.

Ethan Bolker
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It is not true. $2^{2^3}+1=2^8+1=257$ is prime. The theorem you are trying to quote is if $2^x+1$ is prime then $x$ is a power of $2$. Here, if $x$ has an odd factor $a$, $2^a+1$ divides $2^x+1$

Ross Millikan
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Write $x = 2^l n$, where $n$ is odd. We have

$$2^x + 1 = (2^{2^l})^n + 1 = (2^{2^l} + 1)(2^{2^l(n-1)} - \cdots + 1).$$

Since $2^x + 1$ is prime, then the above factorization implies $n = 1$.

kobe
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