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Old 2010-01-30, 08:32   #45
blob100
 
Jan 2010

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Can anyone tell me how do I connected to that scientist who manipulated?
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Old 2010-01-30, 10:18   #46
xilman
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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
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Old 2010-01-30, 11:30   #47
retina
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Quote:
Originally Posted by blob100 View Post
I would like to know how to write my conjecture here..
Mini-Geek already told you.
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Old 2010-01-30, 18:04   #48
blob100
 
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Yes I know that mini greak told me how to do it, but I didn't understand what he said.
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Old 2010-01-30, 18:26   #49
blob100
 
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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)<n}\equiv -1\(mod\: p)

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
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Old 2010-01-30, 18:58   #50
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Now we got you, you ... you nonkosherophagian gentile! You are attempting to make all israelians, Jews and other geniuses look like total morons, right?
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Old 2010-01-30, 19:00   #51
blob100
 
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what?
I just put my conjecture here and want to get a review if it was found before...
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Old 2010-01-30, 20:57   #52
blob100
 
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Here I published my conjecture and no one helps me :(
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Old 2010-01-30, 21:42   #53
R.D. Silverman
 
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Quote:
Originally Posted by blob100 View Post
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)<n}\equiv -1\(mod\: p)

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.

Please, please please, read abook on number theory. Follow and use its
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?
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Old 2010-01-30, 21:45   #54
R.D. Silverman
 
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Quote:
Originally Posted by blob100 View Post
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.
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Old 2010-01-31, 02:05   #55
Orgasmic Troll
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Jul 2003

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Quote:
Originally Posted by blob100 View Post
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)<n}\equiv -1\(mod\: p)

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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