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-   -   New Mersenne and Cunningham conjecture (https://www.mersenneforum.org/showthread.php?t=9945)

 olivier_latinne 2008-01-31 09:52

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:

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^p-1=(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

 R.D. Silverman 2008-01-31 14:53

[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:

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^p-1=(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^p-1 if certain other
conditions are satisfied.

The conditions you state are wrong. The *correct* condition
requires that 2^p = 1 mod q, whence 2^(q-1)/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.

 olivier_latinne 2008-01-31 15:20

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 counter-example.

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^p-1 if certain other
conditions are satisfied.

The conditions you state are wrong. The *correct* condition
requires that 2^p = 1 mod q, whence 2^(q-1)/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]

 R.D. Silverman 2008-01-31 17:28

[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 counter-example.

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.

 olivier_latinne 2008-01-31 21:43

Mersenne conjecture

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]

 R.D. Silverman 2008-01-31 23:47

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
brush-off. 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 so-called "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.

 olivier_latinne 2008-02-01 06:45

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!
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
brush-off. 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 so-called "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]

 akruppa 2008-02-01 09:49

Olivier,

what is you background in number theory? Are you familiar with, say, quadratic reciprocity?

Alex

 R.D. Silverman 2008-02-01 12:51

[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!

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
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>

 Mini-Geek 2008-02-01 13:55

:crank:

 olivier_latinne 2008-02-01 23:33

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
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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