20081023, 13:26  #1 
Oct 2008
2^{4} Posts 
Who can give me a proof ?
Let n is an integer which is great then 3,
b=2^(n1)+1 , Is there anybody who can prove that n can't be the factor of b？ I verified that there is no counterexample for n below 300000000. Until now I can only prove:If there is such an n ,then n must be an odd composite number ! Maybe it is a problem which is just like Fermat Last Theorem which is easy to understand but difficult to give a proof .I guess that there may be no such a bright man in the world who can prove it ! I verified it by using Mathematica 6.0. The code is follow: Do[If[Mod[PowerMod[2,n1,n]+1,n]==0,Print[n]],{n,3,3*10^8,2}] 
20090316, 16:02  #2  
Feb 2005
2^{2}·3^{2}·7 Posts 
Quote:
Therefore, we've got that n = m*2^k + 1 is the product of prime factors of the form t*2^(k+1) + 1, implying that n1 is divisible by 2^(k+1). This contradiction proves that the required n does not exist. Last fiddled with by maxal on 20090316 at 16:08 

20190308, 07:45  #3  
Feb 2019
China
59 Posts 
Quote:
2^(2n2)=1(mod n) this told me that ordn(2)=2n2>n 2^eulerphi(n)=1(mod n) this told me that ordn(2)<eulerphi(n)<n, contradict! 

20190308, 08:08  #4  
Feb 2019
China
73_{8} Posts 
Quote:
Code:
Do[If[Mod[PowerMod[3, n  1, n] + 1, n] == 0, Print[{n, FactorInteger[n]}]], {n, 3, 10^6}] 3^(281)+1=272342767321*28 is multiple of 28 i do not know why Last fiddled with by bbb120 on 20190308 at 08:09 

20190309, 16:46  #5  
Feb 2017
Nowhere
3×1,481 Posts 
Quote:
A correct proof has already been posted, which I summarize: If 2^(n1) == 1 (mod n), it follows that n is odd, and valuation(n1,2) < valuation(p1, 2) for every prime factor p of n, where valuation(N,2) is the exact number of times 2 divides N. This is impossible. 

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