Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
frx
Group Title
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?
 one year ago
 one year ago
frx Group Title
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?
 one year ago
 one year ago

This Question is Closed

frx Group TitleBest ResponseYou've already chosen the best response.0
I would really appreciate some help :) @hba @ganeshie8 @ghazi
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
have you shown why m has to be a multiple of \(2^n\)?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
No, not really
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
That was just the information I got from the start
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
yup, 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.
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Well kind of but i guess the idea is to show the fermat primes by construction the fermat number with the mulitiple of r
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
ah. So the question immediately assumes that this is not a proof of fermat primes and instead is asking to proof r=1? lol
 one year ago

frx Group TitleBest 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
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
To me it doesn't make much sense though :/
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
the 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 \(2^n\)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Oh 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?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
One more thing, how do I know that r should be odd?
 one year ago

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

frx Group TitleBest ResponseYou've already chosen the best response.0
Ok, 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) ?
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
writing 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.
 one year ago

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

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

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
one 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
 one year ago

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

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
@frx the proof is by contradiction
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
I think i understand it
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
I have the solution now so I need to sit down and learn the procedure.
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Thank you sirm3d :)
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
one 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.
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
That 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 :)
 one year ago
See more questions >>>
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.