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 2013-09-27, 09:40 #1 jnml   Feb 2012 Prague, Czech Republ 22·32·5 Posts The beer conjecture For every Mersenne prime Mp there exists at least one Mersenne prime Mq such that Mq-1 is divisible by each prime up to and including p [but not by the next prime]. With/without bracketed text == strong/weak version.
 2013-09-27, 11:57 #2 LaurV Romulan Interpreter     Jun 2011 Thailand 24·13·47 Posts This is trivial, because for every odd number n, there exists a p such as 2^p-1 is divisible by that n (and it is easy to find!). All multiples of p have the same property. Just take n to be x# (the primorial of your mersenne prime). ex: 2^x-1 is divisible by: 3 for x=2, 4, 6, 8, 10, etc 5 for x=4, 8, 12... 7 for x=3, 6, 9... 9 for x=6, 12, 18... 11 for x=10, 20, 30... 13 for x=12, 24, 36... ... 23 for x=11, 22, 33... ... 75 for x=20, 40, 60... .... 256641 for x=3300, 6600, etc and not for other smaller x'es edit: 100280245065 (product of the first 10 odd primes) for x=27720 Last fiddled with by LaurV on 2013-09-27 at 12:16
2013-09-27, 12:01   #3
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

22·11·97 Posts

Quote:
 Originally Posted by jnml For every Mersenne prime Mp there exists at least one Mersenne prime Mq such that Mq-1 is divisible by each prime up to and including p [but not by the next prime]. With/without bracketed text == strong/weak version.
(Mq)-1 is a power of 2. It can't have any prime factors besides 2. M(q-1) is one less than a power of 2, it can't have 2 as a prime factor. So, basically, no.

2013-09-27, 12:07   #4
jnml

Feb 2012
Prague, Czech Republ

B416 Posts

Quote:
 Originally Posted by Mini-Geek (Mq)-1 is a power of 2. It can't have any prime factors besides 2. M(q-1) is one less than a power of 2, it can't have 2 as a prime factor. So, basically, no.
I'm not sure if I understood the above correctly, but AFAICS, Mq-1 is a power of 2 iff q is 2.

2013-09-27, 12:13   #5
jnml

Feb 2012
Prague, Czech Republ

22·32·5 Posts

Quote:
 Originally Posted by LaurV This is trivial, because for every odd number n, there exists a p such as 2^p-1 is divisible by that n (and it is easy to find!). All multiples of p have the same property. Just take n to be x# (the primorial of your mersenne prime). ex: 2^x-1 is divisible by: 3 for x=2, 4, 6, 8, 10, etc 5 for x=4, 8, 12... 7 for x=3, 6, 9... 9 for x=6, 12, 18... 11 for x=10, 20, 30... ... 75 for x=20,40,60... .... 256641 for x=3300, 6600, etc and not for other smaller x'es edit: 100280245065 (product of the first 10 odd primes) for x=27720
Mq-1 is (Mq)-1 is (2^q-1)-1 is 2^q-2. I don't know how numbers of such form relate to your analysis of numbers of the form 2^x-1.

 2013-09-27, 12:43 #6 LaurV Romulan Interpreter     Jun 2011 Thailand 24×13×47 Posts What Mp has to do here? Your gibberish can be formulated as "for any p, there is a Mq such as Mq-1 is divisible by all primes up to p". That is what I shown you, and not only for primes but for ANY ODD number. Mq-1 is 2 multiplied with an odd number. This is still very trivial. Last fiddled with by LaurV on 2013-09-27 at 12:44
2013-09-27, 13:22   #7
jnml

Feb 2012
Prague, Czech Republ

22·32·5 Posts

Quote:
 Originally Posted by LaurV What Mp has to do here? Your gibberish can be formulated as "for any p, there is a Mq such as Mq-1 is divisible by all primes up to p". That is what I shown you, and not only for primes but for ANY ODD number. Mq-1 is 2 multiplied with an odd number. This is still very trivial.
I'm sorry you find something gibberish in a simple sentence of mine. However, I have no idea where the gibberish is to be found. Can you please elaborate a bit? BTW, are you sure you're considering the word 'prime' in the OP? Occurs three times, actually. IOW, both Mp and Mq are prime numbers and that's, being said explicitly, the only domain of the conjecture.

Anyway, wrt "What Mp has to do here?":

First of all, the strong version hints that there may be no solution (no such q) for eg. p == 11 because M11 in not prime. (And indeed no known solution exists ATM for p == 11 in the strong version).

Secondly, but perhaps more importantly: For p in {2, 3, 5}, smallest q's are {2, 3, 5}. For p > 5, smallest q > p (eg. p = 7 -> q -> 13, etc). IOW, the conjecture should be equivalent to "There are infinitely many Mersenne primes." because for every Mersenne prime above M5 the conjecture "generates"/claim existence of at least one another, but larger Mersenne prime than is the "generating" one. Apply recursively

And to be clear, by Mx-1 I mean (Mx)-1, but up to now I thought there's no ambiguity in that. My teachers and WolframAlpha seem to agree on this [0].

[0]: http://www.wolframalpha.com/input/?i...+%282%5Ex%29-1

PS: Preemptive disclaimer: Conjecture implies unproven, right?

Last fiddled with by jnml on 2013-09-27 at 13:34 Reason: s/twice/three times/

2013-09-27, 16:35   #8
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts

Quote:
 Originally Posted by LaurV This is trivial, because for every odd number n, there exists a p such as 2^p-1 is divisible by that n (and it is easy to find!). All multiples of p have the same property. Just take n to be x# (the primorial of your mersenne prime). ex: 2^x-1 is divisible by: 3 for x=2, 4, 6, 8, 10, etc 5 for x=4, 8, 12... 7 for x=3, 6, 9... 9 for x=6, 12, 18... 11 for x=10, 20, 30... 13 for x=12, 24, 36... ... 23 for x=11, 22, 33... ... 75 for x=20, 40, 60... .... 256641 for x=3300, 6600, etc and not for other smaller x'es edit: 100280245065 (product of the first 10 odd primes) for x=27720

I think what jnml is saying is assume the Mersenne Prime exponents are an infinite set S:

ie. S= {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951,...}

for p and q both in S the following relation holds by conjecture:

q#|(2^q-2) for at least one s,q pair. so p in your example would have to be a Mersenne prime exponent to fit jnml's conjecture.

Last fiddled with by science_man_88 on 2013-09-27 at 16:36

2013-09-27, 21:52   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by science_man_88 q#|(2^q-2) for at least one s,q pair. so p in your example would have to be a Mersenne prime exponent to fit jnml's conjecture.
sorry the first q should be s.

2013-09-28, 10:04   #10
Brian-E

"Brian"
Jul 2007
The Netherlands

327010 Posts

Quote:
 Originally Posted by jnml For every Mersenne prime Mp there exists at least one Mersenne prime Mq such that Mq-1 is divisible by each prime up to and including p [but not by the next prime]. With/without bracketed text == strong/weak version.
I found the original statement of your conjecture perfectly clear, and it's surprising that it had to be clarified by science_man_88.

However I'm intrigued to know why you write this conjecture. There must be all sorts of such strong statements about mersenne prime exponents whose truth or falsehood is out of reach of being established. Do you have reason to believe this particular statement?

Quote:
 Secondly, but perhaps more importantly: For p in {2, 3, 5}, smallest q's are {2, 3, 5}. For p > 5, smallest q > p (eg. p = 7 -> q -> 13, etc). IOW, the conjecture should be equivalent to "There are infinitely many Mersenne primes." because for every Mersenne prime above M5 the conjecture "generates"/claim existence of at least one another, but larger Mersenne prime than is the "generating" one. Apply recursively
This is perhaps your intended justification, but I see no reason why this should work for all p.

Last fiddled with by Brian-E on 2013-09-28 at 10:14

2013-09-28, 12:08   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Brian-E I found the original statement of your conjecture perfectly clear, and it's surprising that it had to be clarified by science_man_88. However I'm intrigued to know why you write this conjecture. There must be all sorts of such strong statements about mersenne prime exponents whose truth or falsehood is out of reach of being established. Do you have reason to believe this particular statement? This is perhaps your intended justification, but I see no reason why this should work for all p.
if I did get the explanation correct I believe I found a counterexample.

Code:
? for(w=1,10,d=prod(x=1,w,prime(x));forprime(y=2,1200*w^2,if((2^y-2)%d==0,print(w","y);break())))
1,2
2,3
3,5
4,13
5,61
6,61
7,241
8,1801
9,19801
10,55441
this is not in the accepted list. edit: nevermind removing the break finds 2281 for 7

Last fiddled with by science_man_88 on 2013-09-28 at 12:14

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