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 2010-01-30, 08:32 #45 blob100   Jan 2010 379 Posts Can anyone tell me how do I connected to that scientist who manipulated?
 2010-01-30, 10:18 #46 xilman Bamboozled!     "πΊππ·π·π­" May 2003 Down not across 100111110100102 Posts Off-topic stuff about standards in the use of English has been moved, as requested. Anyone who wishes to continue can head over to http://www.mersenneforum.org/showthread.php?t=13033 Paul
2010-01-30, 11:30   #47
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

5,879 Posts

Quote:
 Originally Posted by blob100 I would like to know how to write my conjecture here..
Quote:
 Originally Posted by Mini-Geek

 2010-01-30, 18:04 #48 blob100   Jan 2010 379 Posts Yes I know that mini greak told me how to do it, but I didn't understand what he said.
 2010-01-30, 18:26 #49 blob100   Jan 2010 5738 Posts n: natural number. e(m)<: the smallest odd element of m. p: prime element of merssen numbers with an odd exponentiation. e(p-1)<: the smallest odd element that accepts: $2^{e(p-1)}>p$. $2^{e(p-1) The problem for me is to explain what is the p elements. I can say that 7,23,31.. are examples of it. If we put them in p: $2^{3n}\equiv -1\(mod\: 7)$ $2^{11n}\equiv -1\(mod\: 23)$ $2^{5n}\equiv -1\(mod\: 31)$ Other ideas about the elements are: 1)The smallest element of p-1 which accepts $2^{e(p-1)}>p$ is allways odd. 2)If the smallest element of p-1 that accepts $2{e(p-1)}>p$ isn't an odd number, p isn't an element of merssen numbers with an odd exponential. I know, I didn't write everything legal and understandable, but that what I can do here in English. Please write a non petty answer and be nice.. ( I mean, not bad and not neutral). Last fiddled with by blob100 on 2010-01-30 at 18:49
 2010-01-30, 18:58 #50 __HRB__     Dec 2008 Boycotting the Soapbox 2·5·61 Posts Now we got you, you ... you nonkosherophagian gentile! You are attempting to make all israelians, Jews and other geniuses look like total morons, right?
 2010-01-30, 19:00 #51 blob100   Jan 2010 379 Posts what? I just put my conjecture here and want to get a review if it was found before...
 2010-01-30, 20:57 #52 blob100   Jan 2010 5738 Posts Here I published my conjecture and no one helps me :(
2010-01-30, 21:42   #53
R.D. Silverman

Nov 2003

723210 Posts

Quote:
 Originally Posted by blob100 n: natural number. e(m)<: the smallest odd element of m. p: prime element of merssen numbers with an odd exponentiation.
What set is m? Do you mean the set of natural numbers less than m?

Do you mean that p is a prime such that 2^p-1 is also prime i.e. p is
a Mersenne exponent?

Quote:
 e(p-1)<: the smallest odd element that accepts: $2^{e(p-1)}>p$.
What does "<' mean here? What do you mean by "accepts"?
Do you really mean "greater than" in you use of ">"?

Do you mean that e(p-1) is the order of 2 mod p? If you don't know
what the order of an element in a group is, then please look it up.
The order of a mod p is the smallest integer x such that a^x = 1 mod p.

Or do you really mean that e(p-1) is the smallest integer x such that
2^x > p, where ">" really means greater than? This can't be what you
mean. It would make no sense. You only write "> p", and I see
no modular notation here.
Quote:
 $2^{e(p-1) The problem for me is to explain what is the p elements. I can say that 7,23,31.. are examples of it. If we put them in p: $2^{3n}\equiv -1\(mod\: 7)$ $2^{11n}\equiv -1\(mod\: 23)$ $2^{5n}\equiv -1\(mod\: 31)$
The above makes no sense. Take the first expression. 2^(3n)
equals 1 mod 7 for all natural n. It never equals -1 mod 7.
Similarly 2^(11n) = 1 mod 23 for all n and 2^5n = 1 mod 31 for all n.

Quote:
 Other ideas about the elements are: 1)The smallest element of p-1 which accepts $2^{e(p-1)}>p$ is allways odd.
Ah!. Your notation "e(p-1)" seems to be a red herring.
Are you asking whether for all primes p, the smallest integer x such
that 2^x > p always has x odd? The answer to that is trivially no.

Quote:
 2)If the smallest element of p-1 that accepts $2{e(p-1)}>p$ isn't an odd number, p isn't an element of merssen numbers with an odd exponential.
I don't know what you mean by "p isn't an element of merssen numbers with an odd exponential". The exponent for all Mersenne primes except 2^2-1 = 3
is always odd.

notation. I do not know what you mean by "accepts". I think I understand
what you mean by "smallest element of p-1". Do you mean the SET of
integers [1, p-1]? Have you studied sets at all?

Quote:
 I know, I didn't write everything legal and understandable, but that what I can do here in English. Please write a non petty answer and be nice.. ( I mean, not bad and not neutral).
Perhaps you would like to use a different language? My Hebrew is worse
than your English. I have not used it in 40 years. Francais, peut-etre?

2010-01-30, 21:45   #54
R.D. Silverman

Nov 2003

26·113 Posts

Quote:
 Originally Posted by blob100 Here I published my conjecture and no one helps me :(
Your post appeared shortly after 1PM EST. I first read it (and replied)
3 hours later. Learn a little patience.

2010-01-31, 02:05   #55
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

10100000012 Posts

Quote:
 Originally Posted by blob100 n: natural number. e(m)<: the smallest odd element of m. p: prime element of merssen numbers with an odd exponentiation. e(p-1)<: the smallest odd element that accepts: $2^{e(p-1)}>p$. $2^{e(p-1) The problem for me is to explain what is the p elements. I can say that 7,23,31.. are examples of it. If we put them in p: $2^{3n}\equiv -1\(mod\: 7)$ $2^{11n}\equiv -1\(mod\: 23)$ $2^{5n}\equiv -1\(mod\: 31)$ Other ideas about the elements are: 1)The smallest element of p-1 which accepts $2^{e(p-1)}>p$ is allways odd. 2)If the smallest element of p-1 that accepts $2{e(p-1)}>p$ isn't an odd number, p isn't an element of merssen numbers with an odd exponential. I know, I didn't write everything legal and understandable, but that what I can do here in English. Please write a non petty answer and be nice.. ( I mean, not bad and not neutral).
I think when you're saying element, you mean factor.

So, for $p=7,\ e(p-1) = 3$, because $2^3 > 7$?
and for $p = 31,\ e(p-1) = 5$, because $2^5 > 31$?

So is your conjecture that for a prime p, if e is the smallest factor of p-1 such that $2^e > p$ and e is even, then p does not divide $2^q - 1$ for all primes q > 2?

Last fiddled with by Orgasmic Troll on 2010-01-31 at 02:06

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