A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Show that if \[2^{m}+1\] is prime then
\[m=2^{n}\] \[n \in \mathbb{N} \]
Guidance: \[m=r*2^{n}\]
So I've started like this: \[2^{m}+1=2^{r*2^{n}} +1 = \left( 2^{2^{n}} \right)^{r}+1\]
I assume that's a somewhat correct beginning but how do I continue?
anonymous
 4 years ago
Show that if \[2^{m}+1\] is prime then \[m=2^{n}\] \[n \in \mathbb{N} \] Guidance: \[m=r*2^{n}\] So I've started like this: \[2^{m}+1=2^{r*2^{n}} +1 = \left( 2^{2^{n}} \right)^{r}+1\] I assume that's a somewhat correct beginning but how do I continue?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would really appreciate some help :) @hba @ganeshie8 @ghazi

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you shown why m has to be a multiple of \(2^n\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That was just the information I got from the start

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup, don't you feel strange that m has to be a multiple of 2 to start with? the question seems to apply for all \(2^m +1\) is prime statements.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well kind of but i guess the idea is to show the fermat primes by construction the fermat number with the mulitiple of r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah. So the question immediately assumes that this is not a proof of fermat primes and instead is asking to proof r=1? lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[r=1\rightarrow 2^{2^{n}}+1  (2^{2^{n}})^{r}+1\] \[r=odd \ge 1 \rightarrow 2^{2^{n}}+1  (2^{2^{n}})^{r}+1\] The second one should be " does not divide" The conlusion is that the trivial dividor is r=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0To me it doesn't make much sense though :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the second one will be out as there is no integer resulting from dividing \(2^{2n} +1\)and \(2^{2nr} +1\) as the number is prime. r=even case is trivial as if r=even, then it's in \(2^n\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh I think I understand, so since the number is prime it can't divide which shows us that the assumption in the beginning statement is correct, that if 2^m+1 is prime then m=2^n, am I right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0One more thing, how do I know that r should be odd?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, since it's \(2^{2nr}\), r cannot be even as that would still mean \(2^{2n}\) so it should be odd

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, I see that you're writing 2^(2n) and it's 2^(2^n), just a writing error or is the explanation you've made based on 2^(2n) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0writing error lol if r is even, then it will either revert to case one or case two. e.g. r=6 will revert to case two and r=8 will revert to case one.

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1it m is not a power of two, it contains at least one factor other than 2 or an odd factor

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1we also know that \[x^n+1\] is always factorable for positive odd integers n.

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1one root of \[\large x^n+1\] for odd n is \[\large x=1\] hence a factor \[\large x+1\] thus \[\large x^n+1=(x+1)(x^{n1}x^{n2}+...x+1)\] or \[\large x^n+1\] is not prime

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1we conclude that the only possibility that \[\huge2^m+1\] to be prime is that m does not contain an odd factor, or that \[\huge m=2^n \]

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1@frx the proof is by contradiction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think i understand it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have the solution now so I need to sit down and learn the procedure.

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1one warning, though. even if m=2^n there is no guarantee that 2^m + 1 is prime. What the problem asserts is that 2^m + 1 is ALWAYS composite if m contains a factor other than 2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That I know, the only primes known are those produced by n=0,1,2,3,4 for 5<=n<=32 the numbers are all composite :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.