mersenneforum.org theory on Mersenne primes ?
 Register FAQ Search Today's Posts Mark Forums Read

2014-04-10, 00:12   #364
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by chalsall In all honesty, I can't tell if you are profoundly stupid, or profoundly brilliant. I suspect the former.
I did just accidentally change from q to p for variables so I'll agree.

2014-04-10, 00:18   #365
chalsall
If I May

"Chris Halsall"
Sep 2002

245E16 Posts

Quote:
 Originally Posted by science_man_88 I did just accidentally change from q to p for variables so I'll agree.
Self-effacing humor goes a long way. Learn how to use it.

2014-04-11, 17:15   #366
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts

Quote:
 Originally Posted by chalsall Self-effacing humor goes a long way. Learn how to use it.
what I really mean't to say was something along the lines of $4^{q-1}= ({2^2})^{q-1} = 2^{2q-2} = 2^{2^{p+1}-4}= 2^{(2(2kp+1))-2} = 2^{4kp} = \text 1 mod q$ it's personally the last one and the third to last one I like the most right now but I can't figure more about them out yet.

2014-04-11, 17:37   #367
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by chalsall In all honesty, I can't tell if you are profoundly stupid, or profoundly brilliant. I suspect the former.
Typical moderator hypocricy. I label a content free post in the math sub-forum as such, and my post gets deleted.

You openly insult someone as being stupid. I call your hypocricy the open
abuse of power by an asshole.

2014-04-11, 17:58   #368
chalsall
If I May

"Chris Halsall"
Sep 2002

931010 Posts

Quote:
 Originally Posted by R.D. Silverman Typical moderator hypocricy. I label a content free post in the math sub-forum as such, and my post gets deleted. You openly insult someone as being stupid. I call your hypocricy the open abuse of power by an asshole.
I didn't delete your post, sir. But you are free to call me anything you'd like -- you may very well be correct.

2014-04-11, 18:04   #369
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by chalsall I didn't delete your post, sir. But you are free to call me anything you'd like -- you may very well be correct.
I can be an asshole too. But I don't delete posts or practice censorship.
Censorship is contemptible.

2014-04-11, 18:10   #370
Xyzzy

"Mike"
Aug 2002

11110000010112 Posts

Quote:
 Typical moderator hypocricy. I label a content free post in the math sub-forum as such, and my post gets deleted. You openly insult someone as being stupid. I call your hypocricy the open abuse of power by an asshole.

 2014-04-12, 02:15 #371 kracker ἀβουλία     "Mr. Meeseeks" Jan 2012 California, USA 5×433 Posts That post deserved to be deleted/moved. That was a "content free post" in his opinion only and not many elses.
2014-04-12, 02:33   #372
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

203008 Posts

Quote:
 Originally Posted by R.D. Silverman I can be an asshole too.
You don't say ...

 2014-04-15, 01:49 #373 science_man_88     "Forget I exist" Jul 2009 Dumbassville 838410 Posts better ready to express my idea ? Fermat's little theorem states that $a^{p-1}\eq 1\text{ mod p}$ my idea was to use the fact that 4,10, and 14, are all used in different parts of the lucas lehmer test variations and to use them as a we get $14^{p-1}\eq 1\text{ mod p} == (4+10)^{p-1} == \sum_{k=0}^{p-1} {{p-1} \choose{k}} 4^k * 10^{p-({k+1})}$ which can be reduced to a sum of powers of four by adding something with log(10) into each exponent allowing use to get a sum of powers of 4 that must be 1 mod p for p=2^q-1 to have a chance of being prime. edit: $4^{p-1}$, and $10^{p-1}$ should both be 1 mod p so they could be taken off the ends and I meant $log_4(10)$ Last fiddled with by science_man_88 on 2014-04-15 at 02:03 Reason: adding words "variations", "and"
2014-04-15, 02:05   #374
tapion64

Apr 2014

5×17 Posts

Quote:
 Originally Posted by science_man_88 Fermat's little theorem states that $a^{p-1}\eq 1\text{ mod p}$ my idea was to use the fact that 4,10, and 14, are all used in different parts of the lucas lehmer test variations and to use them as a we get $14^{p-1}\eq 1\text{ mod p} == (4+10)^{p-1} == \sum_{k=0}^{p-1} {{p-1} \choose{k}} 4^k * 10^{p-({k+1})}$ which can be reduced to a sum of powers of four by adding something with log(10) into each exponent allowing use to get a sum of powers of 4 that must be 1 mod p for p=2^q-1 to have a chance of being prime. edit: $4^{p-1}$, and $10^{p-1}$ should both be 1 mod p so they could be taken off the ends and I meant $log_4(10)$
How are you going to add log(10) into the exponent? K runs across integers and is defined anyways, it's not a variable. But for the heck of it, what if you made it (2+8) and did binomial again under the sum, then you have powers of 2 :p

EDIT: Also, its' probably not a good idea to rely on Fermat's little theorem, since we don't have converse true, and as we saw in the other thread, Lehmer's theorem to close the gap there really only works for certain forms of numbers.

Last fiddled with by tapion64 on 2014-04-15 at 02:08

 Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Miscellaneous Math 3 2017-08-10 13:47 emily Math 34 2017-07-16 18:44 Nick Math 4 2017-04-01 16:26 Nick Number Theory Discussion Group 0 2016-12-03 11:42 optim PrimeNet 13 2004-07-09 13:51

All times are UTC. The time now is 15:39.

Tue Sep 22 15:39:54 UTC 2020 up 12 days, 12:50, 1 user, load averages: 1.76, 1.70, 1.66