MM127

[Stan]

I believe my proof that MM₁₂₇ is valid but part of my argument is debateable.
Attached is a copy of my reasoning.
I would appreciate any input as to any possible modifications.
Many thanks for your support.
Stan.

[axn]

I am not sure I understand your theory. Are you saying that:

M(p) with p==3 (mod 4) is prime if the smallest prime of the form (2kp+1) == 7 (mod 8) does not divide M(p)?

If so, there are plenty of counterexamples.

If not, you need to clarify your theory.

[science_man_88]

Originally Posted by Stan

I would appreciate any input as to any possible modifications.

the number a=19396094914493492417412352623610788052879 you give for MM127 doesn't work:

(09:14) gp > Mod(2,19396094914493492417412352623610788052879)^(1<<127-1)
%6 = Mod(12878846122774700542942977398640678236038, 19396094914493492417412352623610788052879)

edit: I also misread sorry.

edit2:

your condition that 2*p+1 is always composite in this case fails though:

(09:24) gp > w=parselect(v->v%8==7 && isprime(2*v+1),primes(100))
%7 = [23, 191, 239, 359, 431]

edit3: confused again but:

(09:24) gp > w=parselect(v->v%4==3 && isprime(2*v+1),primes(100))
%8 = [3, 11, 23, 83, 131, 179, 191, 239, 251, 359, 419, 431, 443, 491]

leads to even more exceptions.

edit4: okay I keep confusing myself for some reason but if then which destroys the contradiction your claim is fielded on.

[Stan]

Originally Posted by axn

I am not sure I understand your theory. Are you saying that:

M(p) with p==3 (mod 4) is prime if the smallest prime of the form (2kp+1) == 7 (mod 8) does not divide M(p)?

If so, there are plenty of counterexamples.

If not, you need to clarify your theory.

Please supply a counterexample with k = 4n + 1 and I will have to think again.

[science_man_88]

Originally Posted by Stan

Please supply a counterexample with k = 4n + 1 and I will have to think again.

I know I'm failing to comprehend something but most of these things are possible even when p=1 mod 4, 2^p-1 with p>2 will always be 7 mod 8, therefore the factor setup other than form it takes on is the same I guess I'm still missing what leads to the contradiction part of it for some reason. if we could figure out for when p= 1 mod 4 and then both types of p but with 2*p+1 prime though we know in one case it divides the mersenne number and the other case of those last two it divides the generalized fermat number above a mersenne I think

[science_man_88]

Originally Posted by Stan

Please supply a counterexample with k = 4n + 1 and I will have to think again.

you forgot to state n>0 like in your conditions otherwise 2p+1 being prime is a counter, just thought I'd say that much. edit: why is my brain not working.

[ATH]

It seems you think all factors are of the form 8up+2p+1. Normally we write factors on the form: 2kp+1 (proof), so in your case k=4u+1, but k can take other values besides those.

Many of the factors for this exponent does not have k=4u+1:
http://www.mersenne.org/report_expon...448359&exp_hi=

You also seem to say that there cannot be 2 factors of the form 8up+2p+1 for the same mersenne number Mp, but 23790831079 and 10605046177199 from the exponent above both fit that form.

[science_man_88]

Originally Posted by ATH

It seems you think all factors are of the form 8up+2p+1. Normally we write factors on the form: 2kp+1 (proof), so in your case k=4u+1, but k can take other values besides those.

Many of the factors for this exponent does not have k=4u+1:
http://www.mersenne.org/report_expon...448359&exp_hi=

they are looking for the one factor that's 7 mod 8 because Mp mod 8 is 7 there has to be one for it to be composite they then know the form of a 7 mod 8 factor is 2*k*p+1 with k=4*n+1 ( aka 8np+2p+1) assuming the factor is prime they seem to show it divides Mp and to show it's the only one they say lets say it's the first that works like that lets make another one and then show that if the two of that form both exist that they both can't work because a contradiction says basically that 2*k for the higher k says it must divide 2*k for the lower k and there's no way a higher number divides into a lower one. though I don't quite know how they get the last part. edit: so negating the fact that they say first it's basically like saying if you find one you won't find any more because conditions stop them from being possible.

[ATH]

Originally Posted by science_man_88

they are looking for the one factor that's 7 mod 8 because Mp mod 8 is 7 there has to be one for it to be composite they then know the form of a 7 mod 8 factor is 2*k*p+1 with k=4*n+1 ( aka 8np+2p+1) ......

But k is not always 4n+1. For example the factor 45315868721 of the exponent 566448359 I linked:
45315868721 = 2*40*566448359 + 1, so k=40 which is not 4n+1, and several of the other factors of that exponent does not fit k=4n+1 either.

In fact the only k's that never work for any prime exponent p are of the form k=4n+2, because those leads to factors 2kp+1 which is not +/1 (mod 8).

[science_man_88]

Originally Posted by ATH

But k is not always 4n+1. For example the factor 45315868721 of the exponent 566448359 I linked:
45315868721 = 2*40*566448359 + 1, so k=40 which is not 4n+1, and several of the other factors of that exponent does not fit k=4n+1 either.

In fact the only k's that never work for any prime exponent p are of the form k=4n+2, because those leads to factors 2kp+1 which is not +/1 (mod 8).

that factor is 1 mod 8 according to my PARI,we are looking at 2*k*(4n+3)+1 = 8*y+7-> 8*k*n+6*k+1 = 8*y+7 -> 6*k+1=8*x+7 to give a 7 mod 8 factor we can prove that they follow a pattern of possible solution with k= 4n+1 in this case.

[Batalov]

Originally Posted by science_man_88

that factor is 1 mod 8 according to my PARI,we are looking at 2*k*(4n+3)+1 = 8*y+7-> 8*k*n+6*k+1 = 8*y+7 -> 6*k+1=8*x+7 to give a 7 mod 8 factor we can prove that they follow a pattern of possible solution with k= 4n+1 in this case.