20080229, 15:41  #45  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
13275_{8} Posts 
Quote:


20080303, 06:25  #46  
5·739 Posts 
Quote:
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 

20080304, 22:39  #47  
Feb 2007
2^{4}×3^{3} Posts 
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:
Last fiddled with by m_f_h on 20080304 at 22:44 

20080305, 07:57  #48  
Jan 2008
2·11 Posts 
... and the demonstration for the particular case j=4 of conjecture 3
Quote:


20080306, 13:25  #49 
Jan 2008
2·11 Posts 
About the possible practical applications
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:
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
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 
20080310, 10:11  #50 
Jan 2008
10110_{2} Posts 
About the possible practical applications(part II)
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^p1 =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 
20080310, 10:29  #51 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6083_{10} Posts 
I'm glad that you decided not to use that ugly TimesRoman font again. It might look nice on paper, but on the screen it is hard to read.

20080311, 07:44  #52  
Jan 2008
26_{8} Posts 
Last Mersenne conjecture solved?
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^{(d1)/(2j)}=1 (mod d)." Best Regards, Olivier Quote:


20080312, 08:22  #53  
Jan 2008
2×11 Posts 
Last Mersenne conjecture solved?
Well, it seems that nobody found an explanation for the equivalence that Pr Sun used in his demonstration, thus the demonstration remain incomplete.
Quote:
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:
In this last conditions x and y are always coprime, j and m are always coprime. Olivier 

20080312, 09:50  #54 
"Nancy"
Aug 2002
Alexandria
4643_{8} Posts 
> 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 
20080312, 10:04  #55 
Jan 2008
10110_{2} Posts 
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.

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