New Mersenne and Cunningham conjecture
Hi everyone!
If you are interested in Mersenne and Cunningham numbers, I have the pleasure to invited you to read my new web page dedicated to a new theory about the factorization of these numbers: [URL="http://Olivier.Latinne.googlepages.com/home"][COLOR=#810081]http://Olivier.Latinne.googlepages.com/home[/COLOR][/URL] Have a nice reading! This paper contain 3 new conjecture. Try for instance the funny exemple (from table 5 page 6 and notice 1 page 8): Let define M(p) = 2^p1=(2*p*j+1)*(2*p*k+1) for j = 3 , if p = (2*n^2+54*i*(i+1)+13)/3 is prime, with n and i integers, 2*p*j+1 is prime, mod(2*p*j+1,8) = 7, then 2*p*j+1 will always divide M(p) Regards, Olivier Latinne 
[QUOTE=olivier_latinne;124386]Hi everyone!
If you are interested in Mersenne and Cunningham numbers, I have the pleasure to invited you to read my new web page dedicated to a new theory about the factorization of these numbers: [URL="http://Olivier.Latinne.googlepages.com/home"][COLOR=#810081]http://Olivier.Latinne.googlepages.com/home[/COLOR][/URL] Have a nice reading! This paper contain 3 new conjecture. Try for instance the funny exemple (from table 5 page 6 and notice 1 page 8): Let define M(p) = 2^p1=(2*p*j+1)*(2*p*k+1) for j = 3 , if p = (2*n^2+54*i*(i+1)+13)/3 is prime, with n and i integers, 2*p*j+1 is prime, mod(2*p*j+1,8) = 7, then 2*p*j+1 will always divide M(p) Regards, Olivier Latinne[/QUOTE] This "conjecture" is (1) Poorly posed (2) Basically wrong. Allow me to rephrase. By selecting j=3 (in your notation), you are suggesting that if q = 6p+1 is prime, then it will divide 2^p1 if certain other conditions are satisfied. The conditions you state are wrong. The *correct* condition requires that 2^p = 1 mod q, whence 2^(q1)/6 = 1 mod q. This will be true whenever 2 is both a quadratic residue mod q (which from quad. reciprocity requires q = 1 or 1 mod 8) AND 2 is a CUBIC residue mod q. If q = 1 mod 6, then 2 is always a cubic residue (all numbers are cubic residues of primes that are 1 mod 6). If q = 1 mod 6, then 2 will be a cubic residue 1/3 of the time. See [url]http://planetmath.org/encyclopedia/CubicReciprocityLaw.html[/url] for an explanation of cubic reciprocity. 
Mersenne conjecture
Hi!
You should try!! if:[LIST][*]j=3,[*]d=(2*p*j+1) is prime,[*]mod(d,8)=7,[*]p = (2*n^2+54*i*(i+1)+13)/3 is prime, with n and i integers,[/LIST]then 2*p*j+1 will [COLOR=black]always[/COLOR] divide M(p) Please TRY before to says that is wrong. This conjecture is only a very particular case of the conjecture no 2 of my paper. I have done calculations for thousand of hours on a big 56 processors SGI for the 3 conjectures put in my paper. I defy you to find a counterexample. Olivier Latinne [quote=R.D. Silverman;124397]This "conjecture" is (1) Poorly posed (2) Basically wrong. Allow me to rephrase. By selecting j=3 (in your notation), you are suggesting that if q = 6p+1 is prime, then it will divide 2^p1 if certain other conditions are satisfied. The conditions you state are wrong. The *correct* condition requires that 2^p = 1 mod q, whence 2^(q1)/6 = 1 mod q. This will be true whenever 2 is both a quadratic residue mod q (which from quad. reciprocity requires q = 1 or 1 mod 8) AND 2 is a CUBIC residue mod q. If q = 1 mod 6, then 2 is always a cubic residue (all numbers are cubic residues of primes that are 1 mod 6). If q = 1 mod 6, then 2 will be a cubic residue 1/3 of the time. See [URL]http://planetmath.org/encyclopedia/CubicReciprocityLaw.html[/URL] for an explanation of cubic reciprocity.[/quote] 
[QUOTE=olivier_latinne;124400]Hi!
You should try!! if:[LIST][*]j=3,[*]d=(2*p*j+1) is prime,[*]mod(d,8)=7,[*]p = (2*n^2+54*i*(i+1)+13)/3 is prime, with n and i integers,[/LIST]then 2*p*j+1 will [COLOR=black]always[/COLOR] divide M(p) Please TRY before to says that is wrong. This conjecture is only a very particular case of the conjecture no 2 of my paper. I have done calculations for thousand of hours on a big 56 processors SGI for the 3 conjectures put in my paper. I defy you to find a counterexample. Olivier Latinne[/QUOTE] There is nothing to TRY. I already gave a complete explanation for when 6p+1 will divide M_p. Nothing more is needed. (1) Testing such conjectures via computation will not lead to verification. (2) The numbers "54" and "13" in your statements above seem to come from nowhere. Please give the theory behind these numbers. (3) If you want professional review of your work, you must write it using standard mathematical notation. Failure to use standard notation is an an indication that you do not know what it is, or that you are a crank. (4) One can put forth many conjecture about numbers of special form such as M_p that only work because of numerical coincidences among small examples. Trying massive amounts of computation to find counter examples is counter productive because it is often the case that there are only finitely many examples that do work. Please give us some theory to explain why your conjectures should be correct. Please give some theory as to why you believe that more than finitely many examples should exist. Do the theory FIRST, then compute. 
Mersenne conjecture
Please I invite you to read carefully my paper at
[URL="http://Olivier.Latinne.googlepages.com/"][U][COLOR=#0000cc]http://Olivier.Latinne.googlepages.com/[/COLOR][/U][/URL] you will find [U]all[/U] the details and explanations [quote=R.D. Silverman;124407]There is nothing to TRY. I already gave a complete explanation for when 6p+1 will divide M_p. Nothing more is needed. (1) Testing such conjectures via computation will not lead to verification. (2) The numbers "54" and "13" in your statements above seem to come from nowhere. Please give the theory behind these numbers. (3) If you want professional review of your work, you must write it using standard mathematical notation. Failure to use standard notation is an an indication that you do not know what it is, or that you are a crank. (4) One can put forth many conjecture about numbers of special form such as M_p that only work because of numerical coincidences among small examples. Trying massive amounts of computation to find counter examples is counter productive because it is often the case that there are only finitely many examples that do work. Please give us some theory to explain why your conjectures should be correct. Please give some theory as to why you believe that more than finitely many examples should exist. Do the theory FIRST, then compute.[/quote] 
[QUOTE=olivier_latinne;124434]Please I invite you to read carefully my paper at
[URL="http://Olivier.Latinne.googlepages.com/"][U][COLOR=#0000cc]http://Olivier.Latinne.googlepages.com/[/COLOR][/U][/URL] you will find [U]all[/U] the details and explanations[/QUOTE] I looked at it. *Even if* the paper has some worthwhile content, it is so badly presented and the notation so bad that it is painful to read. Hendrik telling you to put it on the web was a polite brushoff. The paper, as written, can not be taken seriously. All it contains is a bunch of numerology. I see no mathematics to suggest why the socalled "conjectures" should be true. There is insufficient justification to even call them conjectures. "open questions" or "wild assed guesses" is closer to the truth. I gave you the mathematics to determine when 6p+1 divides M_p. But of course, you didn't want to hear it. > A Hallmark of a crank is a refusal to learn. 
Mersenne conjecture
The presentation is not very important. It is what there is inside that is important. If this paper would have been publish on a math journal, I would have addapt it to the journal. I recognize that the notation are maybe not of the best but I'm not mathematician of formation (I have a Phd in theoretical physics).
What I have put on this paper is only a fraction of what I actually have discovered. I'm still working for the generalisation to the cae of unspecified [I]p [/I]exponent, Fermat numbers, to the case of composite divisor, square free conjecture, made intensive calculation test about a new detererministic primality test that had only twice the complexity of LL test directly derived from conjecture no 1, ... all of that take me a huge amount of time. If the reader want an explanation I would always be pleased to give it! I'm even ready to give my fortran program to made test. It is not only HW Lenstra that suggest me to put my paper on the Web but also Chris Caldwell (without see the paper). I have only conjecture and no demonstration and any serious math journal would have accept it, whatever the quality of the presentation. [quote=R.D. Silverman;124461]I looked at it. *Even if* the paper has some worthwhile content, it is so badly presented and the notation so bad that it is painful to read. Hendrik telling you to put it on the web was a polite brushoff. The paper, as written, can not be taken seriously. All it contains is a bunch of numerology. I see no mathematics to suggest why the socalled "conjectures" should be true. There is insufficient justification to even call them conjectures. "open questions" or "wild assed guesses" is closer to the truth. I gave you the mathematics to determine when 6p+1 divides M_p. But of course, you didn't want to hear it. > A Hallmark of a crank is a refusal to learn.[/quote] 
Olivier,
what is you background in number theory? Are you familiar with, say, quadratic reciprocity? Alex 
[QUOTE=olivier_latinne;124487]The presentation is not very important. It is what there is inside that is important.
<snip> If the reader want an explanation I would always be pleased to give it! I'm even ready to give my fortran program to made test. It is not only HW Lenstra that suggest me to put my paper on the Web but also Chris Caldwell (without see the paper). I have only conjecture and no demonstration and any serious math journal would have accept it, whatever the quality of the presentation.[/QUOTE] (1) Presentation is *always* important. If you expect someone else to read your scribblings, you must make them readable. Thinking that presentation does not matter is another hallmark of a crank. (2) You are dropping names. This is another hallmark of a crank. (3) You give no supporting explanations for your numerology. This is another hallmark of a crank. You pull magic numbers from somewhere and insert them into your "formulae" without explaining where they come from. (4) Many of the phenomena/relationships under discussion have very sparse density in the integers. Growth rates of such things are often found to be doubly logarithmic. Trying to convince others of the soundness of speculation such as yours MUST depend on more than mere numerical tests. (5) Noone will take you seriously until you present your material in a manner that is cogent AND give some mathematical explanation to suggest why your conjectures should be true. <plonk> 
:crank:

Mersenne conjecture
During the time you spent in not interesting discussion, there is for sure someone that will try to find demonstrations of the conjectures of my work.
Sorry for you, you will miss the train for twice! You will convice me only if you give me a numerical counter exemple. But I see that you are reluctant to made a small program to check it! Or are you worry that I'm most probably on the good way? [quote=R.D. Silverman;124502](1) Presentation is *always* important. If you expect someone else to read your scribblings, you must make them readable. Thinking that presentation does not matter is another hallmark of a crank. (2) You are dropping names. This is another hallmark of a crank. (3) You give no supporting explanations for your numerology. This is another hallmark of a crank. You pull magic numbers from somewhere and insert them into your "formulae" without explaining where they come from. (4) Many of the phenomena/relationships under discussion have very sparse density in the integers. Growth rates of such things are often found to be doubly logarithmic. Trying to convince others of the soundness of speculation such as yours MUST depend on more than mere numerical tests. (5) Noone will take you seriously until you present your material in a manner that is cogent AND give some mathematical explanation to suggest why your conjectures should be true. <plonk>[/quote] 
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