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2021-06-30, 23:23   #12
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

165816 Posts

Quote:
 Originally Posted by birtwistlecaleb Good job! You can say that you can multiply 17120076576497274889 enough times to get M9999971879!
No, that's not at all what one can say. 2 is a factor of 50, but I can't multiply 2 enough times to get 50.

2021-06-30, 23:41   #13
Viliam Furik

"Viliam Furík"
Jul 2018
Martin, Slovakia

25·52 Posts

Quote:
 Originally Posted by VBCurtis No, that's not at all what one can say. 2 is a factor of 50, but I can't multiply 2 enough times to get 50.
Sure you can, 25 x 2 = 50.

Even though what he said is not technically correct, depending on how one understands it, it was correct in the way it was meant to be understood, i.e. 17120076576497274889 x *extremely-big-cofactor* = 29999971879 - 1.

However, it depends on whether I myself have understood it as originally meant.

 2021-06-30, 23:47 #14 James Heinrich     "James Heinrich" May 2004 ex-Northern Ontario 5×72×17 Posts You can multiply the found factor by another (possibly composite) factor to get the original number. You cannot multiply the found factor by itself any number of times to get the original number.
2021-06-30, 23:53   #15
Viliam Furik

"Viliam Furík"
Jul 2018
Martin, Slovakia

25×52 Posts

Quote:
 Originally Posted by James Heinrich You can multiply the found factor by another (possibly composite) factor to get the original number. You cannot multiply the found factor by itself any number of times to get the original number.
Exactly.

2021-06-30, 23:58   #16
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

24·3·157 Posts

Quote:
 Originally Posted by birtwistlecaleb Good job! You can say that you can multiply 17120076576497274889 enough times to get M9999971879!
What power of 17120076576497274889 are you claiming M9999971879 is? Can you show that M9999971879 is divisible by 17120076576497274889 more than once? In general, we do not find squares as composite factors. https://www.planetmath.org/tableoffa...ersennenumbers
https://www.primepuzzles.net/conjectures/conj_012.htm

Last fiddled with by kriesel on 2021-07-01 at 00:24

2021-07-01, 09:18   #17
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

101000001011012 Posts

Quote:
 Originally Posted by VBCurtis No, that's not at all what one can say. 2 is a factor of 50, but I can't multiply 2 enough times to get 50.
+1.
Quote:
 Originally Posted by Viliam Furik Sure you can, 25 x 2 = 50.
-1. With a big Booo! He didn't say that. And he edited his post after it was replied to, to change the meaning.

2021-07-01, 09:23   #18
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

5×112×17 Posts

Quote:
 Originally Posted by kriesel In general, we do not find squares as composite factors.
Moreover, this is an open problem ("are all mersenne numbers with prime exponent square free?"), and you can make some money if you prove it one way or another. There are some heuristics and justifications both sides, but generally, an "open problem" means that we don't have any freaking idea if it goes one way or another, as opposite to a conjecture, for which we guess/assume/see the answer being one way or the other but we just don't know how to prove it.

2021-07-01, 10:19   #19
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

1,621 Posts

Quote:
 Originally Posted by LaurV Moreover, this is an open problem ("are all mersenne numbers with prime exponent square free?"), and you can make some money if you prove it one way or another. There are some heuristics and justifications both sides
It is on the edge, but I think that all Mp is squarefree. My heuristic:

For fixed p prime, we need [q=2*k*p+1 prime] q^2 | Mp.
First we have 1/log(q) probability that q is prime.
We know that 2^(q-1)==1 mod q so with p/(q-1)=1/(2*k) probability we have 2^p==1 mod q.
with another 1/q~1/(2*k*p) probability: 2^p==1 mod q^2 is also true.

Sum these for k=1..inf: sum(k=1,inf,1/log(k*p) * 1/(k^2*p))=O(1/(p*log(p)*sum(k=1,inf,1/k^2)))=O(1/(p*log(p)).
[notice that even for k=1 we have that 1/(p*log(p)) probability, so that is the correct magnitude]
that is for a fixed prime, but sum(p,1/(p*log(p))) is convergent, hence there should be finitely many
(p,q) primes for that q^2 | 2^p-1. Likely with current search we have 1-2% of chance to find even a single (p,q) pair.

2021-07-01, 17:48   #20
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

24×3×157 Posts

Quote:
 Originally Posted by LaurV Moreover, this is an open problem.
Yes. I read https://www.primepuzzles.net/conjectures/conj_012.htm before I posted the link in what you responded to. Did you read it?

I wonder how hard it would be for George or James to script database queries for mersenne.org or mersenne.ca for square factor occurrence. (And maybe temporarily plant a fake square entry or two for script test purposes.) One verified counterexample (one Mp=2p-1 divisible by q2 for prime exponent p) would settle it.

Last fiddled with by kriesel on 2021-07-01 at 18:12

2021-07-01, 19:44   #21
James Heinrich

"James Heinrich"
May 2004
ex-Northern Ontario

416510 Posts

Quote:
 Originally Posted by kriesel I wonder how hard it would be for George or James to script database queries for mersenne.org or mersenne.ca for square factor occurrence.
I'm not sure about mersenne.org but mersenne.ca database goes on the assumption that factors are unique.

I suppose I could script something to walk through all the known factors and check if the square of each factor is also a factor. It would probably take a while to run. If that's what you were asking?

2021-07-01, 20:20   #22
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

24·3·157 Posts

Quote:
 Originally Posted by James Heinrich I'm not sure about mersenne.org but mersenne.ca database goes on the assumption that factors are unique. I suppose I could script something to walk through all the known factors and check if the square of each factor is also a factor. It would probably take a while to run. If that's what you were asking?
Thanks for the response.
That's more computer cycles (and James cycles) than I anticipated but would do the job. It would have the advantage of detecting any squares present, even if the factoring run would not have, due to TF limits or whatever. Also if only q is recorded once if q2 is encountered. I think checking factors less than cube root of Mp would be sufficient, but it is probably not worth testing for that.

I had thought that any composite factors would get both / all parts entered. But based on your response I now see the database might contain a list of unique factors per exponent, not recording counts for squares or higher powers, or recording duplicates in any way.

It suggests a possible addition to the factor check code for new result submissions. If the reported factor is found composite, do any of its factors match? If so, that''s notable news.

All in your spare time, if/when so inclined, if it's not too much trouble.

Last fiddled with by kriesel on 2021-07-01 at 20:25

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