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anonymous
 3 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
 3 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?

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

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

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

anonymous
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0To me it doesn't make much sense though :/

anonymous
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0One more thing, how do I know that r should be odd?

anonymous
 3 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
 3 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
 3 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it m is not a power of two, it contains at least one factor other than 2 or an odd factor

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we also know that \[x^n+1\] is always factorable for positive odd integers n.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0one 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we conclude that the only possibility that \[\huge2^m+1\] to be prime is that m does not contain an odd factor, or that \[\huge m=2^n \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@frx the proof is by contradiction

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

anonymous
 3 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0one 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
 3 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 :)
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