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Old 2014-04-10, 00:12   #364
science_man_88
 
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Quote:
Originally Posted by chalsall View Post
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.
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Old 2014-04-10, 00:18   #365
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Quote:
Originally Posted by science_man_88 View Post
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.
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Old 2014-04-11, 17:15   #366
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Quote:
Originally Posted by chalsall View Post
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.
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Old 2014-04-11, 17:37   #367
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Quote:
Originally Posted by chalsall View Post
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.
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Old 2014-04-11, 17:58   #368
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Quote:
Originally Posted by R.D. Silverman View Post
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.
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Old 2014-04-11, 18:04   #369
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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.
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Old 2014-04-11, 18:10   #370
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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.
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Old 2014-04-12, 02:15   #371
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That post deserved to be deleted/moved. That was a "content free post" in his opinion only and not many elses.
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Old 2014-04-12, 02:33   #372
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Quote:
Originally Posted by R.D. Silverman View Post
I can be an asshole too.
You don't say ...
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Old 2014-04-15, 01:49   #373
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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"
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Old 2014-04-15, 02:05   #374
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Quote:
Originally Posted by science_man_88 View Post
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
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