20210630, 23:23  #12 
"Curtis"
Feb 2005
Riverside, CA
1658_{16} Posts 

20210630, 23:41  #13  
"Viliam Furík"
Jul 2018
Martin, Slovakia
2^{5}·5^{2} Posts 
Quote:
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 *extremelybigcofactor* = 2^{9999971879}  1. However, it depends on whether I myself have understood it as originally meant. 

20210630, 23:47  #14 
"James Heinrich"
May 2004
exNorthern Ontario
5×7^{2}×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. 
20210630, 23:53  #15 
"Viliam Furík"
Jul 2018
Martin, Slovakia
2^{5}×5^{2} Posts 

20210630, 23:58  #16  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{4}·3·157 Posts 
Quote:
https://www.primepuzzles.net/conjectures/conj_012.htm Last fiddled with by kriesel on 20210701 at 00:24 

20210701, 09:18  #17  
Romulan Interpreter
"name field"
Jun 2011
Thailand
10100000101101_{2} Posts 
Quote:
1. With a big Booo! He didn't say that. And he edited his post after it was replied to, to change the meaning. 

20210701, 09:23  #18 
Romulan Interpreter
"name field"
Jun 2011
Thailand
5×11^{2}×17 Posts 
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.

20210701, 10:19  #19  
"Robert Gerbicz"
Oct 2005
Hungary
1,621 Posts 
Quote:
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^(q1)==1 mod q so with p/(q1)=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^p1. Likely with current search we have 12% of chance to find even a single (p,q) pair. 

20210701, 17:48  #20 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{4}×3×157 Posts 
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=2^{p}1 divisible by q^{2} for prime exponent p) would settle it. Last fiddled with by kriesel on 20210701 at 18:12 
20210701, 19:44  #21  
"James Heinrich"
May 2004
exNorthern Ontario
4165_{10} Posts 
Quote:
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? 

20210701, 20:20  #22  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{4}·3·157 Posts 
Quote:
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 q^{2} 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 20210701 at 20:25 

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