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there exists a prime in the interval...

i'm coming up with a delightfully convoluted way of solving a problem, and now need to prove the following:

show that for n>0, there exists a prime in the set {2^n + 1, 2^n+2, ... , 2^(n+1) }

i have not a clue how to do this, though it feels it should be true...

Reply 1

SimonM

This is weaker than Bertrand's postulate (there is a prime n<p<2n for all n), but I think the most 'natural' way to prove it would be using the same proof. (If you know it, that is)

What's the original question?

Reply 2

Nytram12

surely just an example would do?

eg when n=2 2^n + 1 = 5

which is prime

hmm maybe not... doesn't prove it, just shows it is true sometimes

Reply 3

Chewwy

Nytram12

surely just an example would do?

eg when n=2 2^n + 1 = 5

which is prime

yeah, but i'm struggling to find an example for the n=353421523 case

Reply 4

Nytram12

Chewwy

yeah, but i'm struggling to find an example for the n=353421523 case

well thats just easy. can't see why ur stuck there

Reply 5

DFranklin

SimonM

This is weaker than Betrand's postulate (there is a prime n<p<2n for all n), but I think the most 'natural' way to prove it would be using the same proof. (If you know it, that is)I agree, but it's Bertrand (as I'm sure you know).

I'm not sure if it's enough weaker than Bertrand's postulate that there will be an easier proof - my gut feeling is no, but the proof of Bertrand seems a bit tough to set as a problem question.

Reply 6

SimonM

DFranklin

I'm not sure if it's enough weaker than Bertrand's postulate that there will be an easier proof - my gut feeling is no, but the proof of Bertrand seems a bit tough to set as a problem question.

Agreed, hence why I asked for the original question and suggested the 'standard' proof. (Thanks for the typo catch)

Reply 7

Chewwy

ok, this is the original question. prepare to be underwhelmed:

use the fundamental theorem of arithmetic to show

x \leq \big ( 1 + \frac{logx}{log2}\big ) ^{\pi (x)}

Reply 8

DFranklin

I haven't gone through the details, but I'm pretty sure you can get this out by noting that every integer N <= x (as well as some bigger than x), can be written in the form