20131120, 12:16  #12 
Feb 2010
Sweden
173 Posts 
10613583595427 is also a factor of 2^1968721+1

20131120, 13:58  #13 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×7×349 Posts 
2^{2k+1}=2 (mod 3), so 2^{2k+1}+1 will always be divisible by 3. Same as 2^{2k}1, the last one is (2^{k}1)(2^{k}+1), i.e. a product of two consecutive odd numbers, so one of them is divisible by 3.
So what? 
20131120, 22:55  #14 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,277 Posts 
I have discovered a fantastic new way to factorise numbers using absolutely no compute power whatsoever. Just post is here and give it some name like "Unwilling Number" and within a few hours it will magically be factorised for you. I hope no one has patented this new method yet.

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Finding multiples of a real number that are close to a whole number  mickfrancis  Math  16  20170301 07:17 
Estimating the number of primes in a partiallyfactored number  CRGreathouse  Probability & Probabilistic Number Theory  15  20140813 18:46 
Number 59649589127497217 is a factor of Fermat number F7  literka  Miscellaneous Math  73  20131117 10:33 
Number of distinct prime factors of a Double Mersenne number  aketilander  Operazione Doppi Mersennes  1  20121109 21:16 
Fermat number F6=18446744073709551617 is a composite number. Proof.  literka  Factoring  5  20120130 12:28 