20100203, 17:13  #166 
Jan 2010
379 Posts 
I want to thank everyone who answered my questions and helped me.
Big thanks Tomer. 
20100203, 17:16  #167 
Nov 2003
2^{2}×5×373 Posts 

20100203, 18:46  #168 
Aug 2006
2·11·271 Posts 

20100204, 11:11  #169 
Jan 2010
17B_{16} Posts 
what is the tiger challenge?

20100204, 15:16  #170  
Aug 2006
2·11·271 Posts 
Quote:


20100204, 21:09  #171 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 

20100204, 22:04  #172 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
If you mean "Year of the Tiger Challenge", it's
[Year of the Tiger] Challenge In the traditional Chinese zodiacal calendar, a Year of the Tiger starts on Feb. 14, 2010. (http://www.springsgreetingcards.com/...asp?pid=250913) Last fiddled with by cheesehead on 20100204 at 22:06 
20100206, 11:47  #173 
Jan 2010
379_{10} Posts 
Here is another conjecture I tought about:
If . Where p is an odd prime number,and A is the set of the odd prime factors of 2^{p}1. . If 2^{p}1 is a prime number, exists. From here, the conjecture "says" that . . . . Which is fermat's little theorem. Last fiddled with by blob100 on 20100206 at 11:56 
20100206, 12:20  #174  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts 
Quote:
This is a result of the known fact that any factor of 2^p1 is of the form 2kp+1. What? If I get what you're saying, that's equivalent to the trivial equation , or "2^{p}1 divides 2^{p}1". And that's certainly not restricted to a prime number, or a number of the form 2^{p}1. Take a look at these two "equivalent" statements and tell me if you see anything off. You've gotta see it. What about where you went wrong? Well? If you can't figure it out, highlight this text to see it: You're saying 2^p2 is equivalent to 2^p1, which is of course not true. The problem is caused by turning 2^p2 into 2(2^p1) instead of 2(2^(p1)1). If you correctly transform that, you get an alternate form of Fermat's little theorem, So you're saying that your conjecture is equivalent to Fermat's little theorem, being a subset of it that adds nothing new? Umm...I think it should be pretty clear that the fact there is already known. Last fiddled with by MiniGeek on 20100206 at 12:21 

20100206, 12:50  #175 
Jan 2010
379 Posts 
[quote=MiniGeek;204717]
This is a result of the known fact that any factor of 2^p1 is of the form 2kp+1. I didn't know about this fact, so I found it by myself (somehow). Thanks for the help. BTW: I didn't see the part about fermat's little theorem was not correct, i meant what you wrote. Last fiddled with by blob100 on 20100206 at 12:52 
20100216, 15:06  #176 
Jan 2010
379 Posts 
I have a new conjecture (it isn't about mersenne numbers),
I conjecture that: between n^2 to n^2+2n there is always a prime number. When n is an odd number. We can show n^2+2n as n(n+2). So, Last fiddled with by blob100 on 20100216 at 15:10 
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