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Old 2008-02-29, 15:41   #45
henryzz
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Quote:
Originally Posted by flava View Post
Wrong, mathematics is an exact science (as opposed to physics for instance).
In physics, chemistry etc. we try to come up with a theory that covers most of the observable phenomena... theory which is often prooved wrong when our observation techniques improve. See gravity for instance:
- Newton came up with a theory based on observation. That theory is false.
- Einstein came up with a better theory which stood up quite nicely, but now requires the invention of exotic stuff like "dark matter" and "dark force" to match the observations.
- fill in here the next theory

Mathematics is never based on observation, if it is it's not mathematics. Yes, you can guess something but until a proof is found, it has no value.
i think what is confusing him is that sometimes u can in mathematics make an observation and then prove it once u know what to prove but until it is proved nothing is really helpful in mathematics
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Old 2008-03-03, 06:25   #46
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Quote:
Originally Posted by olivier_latinne View Post
But I'm cautious (not like Dr. Silverman, who jumped strait in the bobby trap), because historically there are plenty of new discoveries that we thought there were only interesting for a theoretical point of view and after a more or less long time we find that there are also very useful for a practical point of view.
I'm new here, and don't expect to be a frequent presence, but I figured I should add a point to this:

I hardly think the idea of this being theoretically vs practically interesting is what Dr Silverman is having trouble with. In math, theoretically interesting is interesting from a practical point of view. Unlike physics where "this is neat" and "this can change the world" present very real differences, "this could change math" and "this is neat" are related statements; the one key difference is "this is neat" means "this could change the way I view math."

The problem is that your math is not very rigorous. In math, it doesn't matter how many observations you make, there's always an infinite more required to verify an assertion.

If you could show that it must follow from our understanding of mathematics (read: if you use rigorous mathematics), then I highly doubt Dr. Silverman will object. As it stands you are throwing around "things"-- what he says (quite correctly) are more accurately "open problems" than conjectures.

A list of numbers means nothing to mathematicians; in fact, I opened your pdf and closed it again within 30 seconds, because the list says nothing.

Also, it would be helpful to see your paper in an accessible format. Perhaps a portable format. As it stands, I can't read half your math, as I do not have MS Office, and Open Office can only go so far in reading Word documents.

Cheers,
Cory
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Old 2008-03-04, 22:39   #47
m_f_h
 
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I think members of this forum should be informed that an explicit proof of the corrected conjecture (in one special case) had been posted some days ago on the NMBRTHRY list.
It indicates well how other cases might be proved as well (along the lines suggested by R.D.S., afaics):

Quote:
Zhi-Wei SUN to NMBRTHRY - Mar 1, 2008

Dear number theorist,

Olivier Latinne conjectured that
d=6p+1 divides 2^p-1 if and only if
p is a prime with 6p+1=7 (mod 8) (i.e., p=1(mod 4))
and d=6p+1 is a prime in the form x^2+27y^2.
(since d=3 (mod 4), x cannot be odd).

Below I prove this conjecture for any prime p.

Proof of the "if" part. Let p=1 (mod 4) be a prime such that d=6p+1 is a
prime in the form x^2+27y^2. By Corollary 9.6.2 of K. Ireland and M.
Rosen's book "A Classical Introduction to Modern Number Theory" [GTM
84, Springer, 1990], 2 is a cubic residue mod d, and hence d divides
2^{(d-1)/3}-1=(2^p-1)(2^p+1). If 2^p=-1 (mod d), then
1^p=(2/d)^p=(-1/d)=-1, a contradiction! So we have 2^p=1 (mod d).

Proof of the "only if" part. Suppose that p is a prime and d=6p+1
divides 2^p-1. If d=6p+1 is composite then it has a prime factor q not
exceeding sqrt(6p+1)<2p+1. But any prime divisor of 2^p-1 is of the form
2pk+1, so d=6p+1 must be a prime.
As d=1 (mod 3) and 2^{(d-1)/3}=2^{2p}=1 (mod d), again by Corollary
9.6.2 of Ireland and Rosen's book, d can be written in the form
x^2+27y^2. Since d=3 (mod 4), x must be even.

Zhi-Wei Sun

Last fiddled with by m_f_h on 2008-03-04 at 22:44
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Old 2008-03-05, 07:57   #48
olivier_latinne
 
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... and the demonstration for the particular case j=4 of conjecture 3

Quote:
Zhi-Wei SUN to NMBRTHRY, me
show details Mar 2 (3 days ago) Reply

Dear number theorists,
Here I give a simple proof of another conjecture of Olivier Latinne.
THEOREM. Let p be a prime, and let d=8p+1. Then d divides 2^p-1
if and only if d is a prime in the form x^2+64y^2 with y odd.
Proof. If d=8p+1 divides 2^p-1 then d cannot be composite since
sqrt(8p+1)<2p+1 and any prime divisor of 2^p-1 is =1 (mod 2p).
Observe that d=9 (mod 16). When d is a prime, by Corollary 7.5.8 in
the book [B.C. Berndt, R.J. Evans and K.S. Williams, Gauss and Jacobi
Sums, Wiley, 1998], d is in the form x^2+64y^2 with x,y odd if and only
if 2 is an octic residue mod d (i.e., 2^p=2^{(d-1)/8}=1 (mod d)). So the
desired result follows.

Zhi-Wei Sun
http://math.nju.edu.cn/~zwsun
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Old 2008-03-06, 13:25   #49
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Hi everyone!

I think my work is not only a theoretical curiosity, but maybe will have some practical interests.
Right now, to find which form has a divisor take more time than simply apply the fast exponentiation algorithm.
I will give an example for which, when we know the form of the divisor, it restricted not one divisor but many.

Conjecture #2 and #3 (they are demonstrated only for j=3 and j=4) of my paper http://Olivier.Latinne.googlepages.com/home are valid for “candidate” divisors prime, but what append if they are not prime?
When a number has the form 6n+1 and is not prime it has not necessary the Euler form x^2+3*y^2.

I have observed that (this is not in my paper) for conjecture #2:
  • When a candidate divisor 2*p*j+1 has the form “forbidden divisor” (table 4 page 5), this number can never divide even if this number is composite
  • When a number 2*p*j+1, is composite, has the form “allowed divisor” (table 4 page 5) and is a divisor of 2^p+-1 then all factors of 2*p+j+1 are resricted:
2*p*j+1=(2*p*j1+1)*(2*p*j2+1)* … * (2*p*jn+1)
and mod(j1,3)=0, mod(j2,3)=0 … mod(jn,3)=0
  • When a candidate divisor 2*p*j+1 is composite and has none of the form “allowed divisor” or “forbidden divisor” (table 4 page 5), there is no (global) restrictions about the ji values.
And it is similar for conjecture #3 but then: mod(j1,4)=0, mod(j2,4)=0 … mod(jn,4)=0

Of course, this example will have probably still not a practical interest, but maybe other people will find a subtle way to find an application.
I think that my work adds huge complexity, and usually when a system is complex, solutions are numerous.

Olivier
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Old 2008-03-10, 10:11   #50
olivier_latinne
 
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My last post is also applicable not only to the divisors but also to Mersenne numbers itself:

Let p be prime and suppose M(p) know to be composite (by LL test for instance)
M(p)= 2^p-1 =1 mod 6,
M(p)=2*p*j+1 => j = (M(p)-1)/(2*p) and in this case mod(j,12)=3,9

So, by applying my remarks of thread #49 and conjecture #2 (table 4 page 5) of http://olivier.latinne.googlepages.com/C.doc, when M(p) has the form 4*n^2 + 3*(3+6*i)^2, if d=2*p*k+1 is a divisor of M(p) then mod(k,3)=0
We can also easily determine that in this case the minimum number factor of M(p) is 3: because mod(M,8)=7 and mod(d,8)=7 =>
mod(M,8)= 7 = 7 X 1 = 7 x (7 x 7)

Olivier
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Old 2008-03-10, 10:29   #51
retina
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I'm glad that you decided not to use that ugly Times-Roman font again. It might look nice on paper, but on the screen it is hard to read.
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Old 2008-03-11, 07:44   #52
olivier_latinne
 
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Hi everyone!

Pr Sun provide a demonstration for conjecture #1 of my paper.
Can anyone explain me in details (it's maybe obvious, but not for me) the following transition of that demonstration:
" ... Observe that x^{2j}=2 (mod d) is solvable if and only if 2^p=2^{(d-1)/(2j)}=1 (mod d)."

Best Regards,

Olivier

Quote:

Dear number theorists,

If (j,m)=1, then 1=aj+bm for some integers a and b. If j is even and
(j,m)=1, then we also have (2j,m)=1. Thus the latest conjecture of
Olivier Latinne can be restated in the following simple form that I will
prove.

THEOREM. Let p be an odd prime and let d=2pj+1 (0<j<2p+2).

(1) When j is odd, d divides 2^p-1 if and only if d is a prime with d=7
(mod 8) and x^j=2 (mod d) is solvable.

(2) When j is even, d divides 2^p-1 if and only if d is a prime with d=1
(mod 8) and x^{2j}=2 (mod d) is solvable.

Proof. Suppose that d|2^p-1. If d is composite, then d has a prime
divisor q not exceeding sqrt(2pj+1)<2p+1 but any prime divisor of 2^p-1
has the form 2pk+1. So d must be a prime. As 2^p=1 (mod d) we have
1=(2/d)^p=(2/d). So d=1 or 7 (mod 8). If j is odd then d=3 (mod 4) and
hence d=7 (mod 8). If j is even, then d=1 (mod 4) and hence d=1 (mod 8).

Now assume that d is a prime with d=(-1)^j (mod 8). Observe that
x^{2j}=2 (mod d) is solvable if and only if 2^p=2^{(d-1)/(2j)}=1 (mod
d). So (2) holds. Note that x^j=2 (mod d) is solvable if and only if
2^{2p}=2^{(d-1)/j}=1 (mod d), i.e., d|(2^p-1)(2^p+1). If j is odd, then
d=7 (mod 8) and (2/d)^p=1 is different from (-1/d)=-1, and thus d does
not divide 2^p+1. So, when j is odd, x^j=2 (mod d) is solvable if and
only if d|2^p-1.

Zhi-Wei Sun

http://math.nju.edu.cn/~zwsun
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Old 2008-03-12, 08:22   #53
olivier_latinne
 
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Well, it seems that nobody found an explanation for the equivalence that Pr Sun used in his demonstration, thus the demonstration remain incomplete.

Quote:
Originally Posted by olivier_latinne View Post
Pr Sun provide a demonstration for conjecture #1 of my paper.
Can anyone explain me in details (it's maybe obvious, but not for me) the following transition of that demonstration:
" ... Observe that x^{2j}=2 (mod d) is solvable if and only if 2^p=2^{(d-1)/(2j)}=1 (mod d)."
Anyway, there is something interesting that Pr Sun has understood:
if you see equations for j odd and j even (see also thread #38)

for j odd: d= (x^j + 2^m * y^j) / k and mod(d,8)=7

for j even: d= (-x^(2*j) + 2^m * y^(2*j) ) / k and mod(d,8)=1 ,

there is four variables x, y, m and k.
Conjecture #1 remains valid if you:
  • fix x, y and you free m, k
  • fix x, m and you free y, k
  • fix y, m and you free x, k
In his incomplete demonstration Pr Sun has fix y=1 and fix m=1.
In this last conditions x and y are always coprime, j and m are always coprime.

Olivier
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Old 2008-03-12, 09:50   #54
akruppa
 
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> Well, it seems that nobody found an explanation for the equivalence that Pr Sun used in his demonstration

Either that, or those who could don't bother posting in this thread any more.

Alex
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Old 2008-03-12, 10:04   #55
olivier_latinne
 
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If you are not interested it is your choice. But it seems that there is a significant fraction of people of the group that are interested, by analysing the number of views for each new post on that subject. Please, have some respect.

Quote:
Originally Posted by akruppa View Post
> Well, it seems that nobody found an explanation for the equivalence that Pr Sun used in his demonstration

Either that, or those who could don't bother posting in this thread any more.

Alex
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