20201009, 07:02  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
1010011010_{2} Posts 
Why does a prime p divide a Fermat Number?
Hi all,
There is an interesting article in Mathematics Magazine (Vol 92, No 4, October 2020). The title of the article is "Why does a prime p divide a Fermat Number?" They reference fermatsearch.org. They state that a Fermat prime divisor divides one and only one Fermat number. Regards, Matt 
20201009, 08:27  #2 
Romulan Interpreter
Jun 2011
Thailand
9275_{10} Posts 
Because he can, and he wants to do so.
The "unicity" of factors is known, and trivial to prove. Last fiddled with by LaurV on 20201009 at 08:28 
20201009, 08:52  #3 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}×19^{2} Posts 

20201009, 16:11  #4 
Romulan Interpreter
Jun 2011
Thailand
5^{2}×7×53 Posts 
Yep, that's what wikipedia used to say for a while

20201009, 20:12  #5 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}·19^{2} Posts 
Really not read, in general I know the shortest proofs. Another way:
If 2^(2^m)==1 mod d then for n>m taking this to the 2^(nm)th power: 2^(2^n)==1 mod d from this F(n)==2 mod d and the rest is the same. This could be even shorter proof when you write down, but needs more knowledge. 
20201009, 22:01  #6 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3·5·311 Posts 

20201013, 15:31  #7 
Romulan Interpreter
Jun 2011
Thailand
10010000111011_{2} Posts 
That's not a "show off" for the fact I know the math. I may know it, I may not know it. I was barking more in that direction that people come and ask questions whose answers (and more details about subject) could be found by a simple search.
Last fiddled with by LaurV on 20201013 at 15:31 
20201013, 22:50  #8 
"Matthew Anderson"
Dec 2010
Oregon, USA
2×3^{2}×37 Posts 
We know that the Fermat numbers are defined as
F(n) = 2^(2^n) + 1. Also, F(0) to F(4) are prime numbers. F(5) to F(32) are composite numbers. F(33) to F(35) are of unknown character. F(36) is composite. This data is available at  http://www.prothsearch.com/fermat.html I spent several years using trial division programs trying to find the smallest prime factor of F(34). So far we know that the smallest prime divisor of F(34) is greater than 7*10^17. Maybe someone else wants to work on this. The program (mmff) can be downloaded from fermatsearch.org Regards, Matt 
20201014, 01:05  #9  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×3×5×311 Posts 
Quote:
The stats show that you mostly reserved ranges for N=36. This means that you can just as well find a factor of F(33) or F(32), if you keep at it. You are not guaranteed to find the factor of F(34)! 

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