20080317, 02:59  #1 
"Jason Goatcher"
Mar 2005
3·7·167 Posts 
Are Legendre symbols proven to be defective?
It's not going to surprise me if this gets put in Miscellaneous Math, or hurt my feelings, but I'm going to try to phrase things in order to avoid getting this thread switched over. I know I'm way out of my league with these questions, but a friend of mine, who I very much respect, brought these to my attention. He swore me to secrecy, for reasons I'm not prepared to disclose, so I'm going to be asking questions that may not seem to be connected.
(1) The Legendre symbol method of sieving,(sorry if this is nonsensical, I'm hoping I can make myself understood), is it 100% proven, or is it simply BELIEVED to be 100% effective? (2) The idea that a Fermat prime cannot have odd prime factors in the exponent, is it proven, or just BELIEVED to be effective? If I give a number, like 2^68544+1, can you give me an actual factor with that information, or is it mathematics that makes you believe it has a factor? Are there ANY gaps in the reasoning? (3) How many of the top5000 primes have been tested via integer based math? After that twin prime fiasco with Intel, wouldn't it be better to test numbers with an integer based test after the normal test is run? As I said, there's a lot of stuff that I'm suppressing on request. My friend has told me(approximate quote),"A lot of the stuff that people accept as fact is not true, soandso(note: I'm not telling who or what was mentioned, but it isn't George :) ) intentionally stacked the deck so that they would find the first 10million digit prime. Because of this, a lot of primes on the top5000 list are actually composite. You(jasong) have the necessary skills to figure it out with a good amount of work." From the way he said it(as I said, it's an approximate quote), it sounds like people have discovered what he discovered in the past(further than you would believe) and it is being suppressed. Lastly, is there an integer based prime testing program that would work on an x86 LInux box? If I could get my hands on one and use it, I could establish 100% in my own mind that I wasn't simply being taken for a ride. I don't think my friend is a liar, but there are only two possibilities left, delusional and genius. I badly want to believe it's the latter, but I don't have the education to have a worthwhile opinion. Right now, it's totally up in the air. 
20080317, 03:02  #2 
Dec 2007
2×17 Posts 
Aren't conspiracy theories so much fun?

20080317, 03:22  #4  
"Jason Goatcher"
Mar 2005
3×7×167 Posts 
Quote:
As an example, what form of government is in effect in the United States? Most people would say "democracy" without even thinking about it, but the word democracy isn't in any state Constitution, nor is it in any of the major documents of the US government. Democracy is just a feelgood term that the politicians employ to control our emotions. If you look it up on the web, there's a good chance you'll receive erroneous information, even from a socalled expert. We are a constitutional republic. If the mathematics is being suppressed, there's a good chance that Wikipedia is part of that. It would simply be a matter of a government employee working from home, so that there's no .gov extension on the name. That being said, my IQ is extremely high, I took my last IQ test while I was badly delusional, and still managed to stun my advisor. If I wasn't a paranoid schizophrenic, I'd probably be another superconceited R.D. Silverman type. It's only recently, with the help of a medication called Invega, that I've gotten to a place where I can try to get back to a normal life. My point is that if there's any hankypanky going on anywhere, and from a valuebased standpoint I'll tackle any form of lying out there, I'm a good person to find it. Okay, go ahead and place this in Miscellaneous Math, or probably more appropriately, Soap Box, :) since I've probably lost all credibility in this thread. :) Edit: On second thought, can we leave it here, and split off any Conspiracy theory stuff? Last fiddled with by jasong on 20080317 at 03:32 

20080317, 04:48  #5 
"Jason Goatcher"
Mar 2005
3·7·167 Posts 
If I wanted to take all the numbers that NewPGen threw out after sieving to a billion, from the 66 Riesel ks that are left, and test them independently for the factor given, would anyone be willing to help me do that?
I say Riesel ks, but I'm willing to consider other possibilities. :) Edit: try sieving, with newpgen, with k*2^n+1, for k=1 to 5, and n=3355583. And then, if possible, tell me the factor for 2*2^3355583+1, with newpgen if you can manage it, but some other way if newpgen won't tell you and you know another way. Last fiddled with by jasong on 20080317 at 04:56 
20080317, 09:50  #6 
"Brian"
Jul 2007
The Netherlands
2·3·5·109 Posts 
A couple of comments:
(1) I cannot answer jasong's questions because I am not a specialist. However I am very interested in reading answers from others here who are specialists in the field. Considering that he has asked very specific questions  certainly his questions 2 and 3 are specific, and probably 1 is as well but I don't know what the "Legendre symbol method" means  I think he deserves specific answers instead of being referred to Google. One advantage of wellposed questions being answered here is that we generally know the person who is answering, any inaccuracies that are made can be corrected by others, and it might generate further interesting discussion. Lots of us would like to read that. (2) Jasong, I cannot myself believe in any conspiracy theory because the mathematics involved is public knowledge and testable independently by specialists all over the world using their own independently written software. Coverups can only occur if the information is secret or computationally inaccessible to all but the insiders. In the case of the top 5000 primes, the huge effort needed was in discovering them in the first place, not in verifying their prime status. 
20080317, 11:24  #7 
May 2005
111100_{2} Posts 
Jasong,
Write n = e*o, where e is even and o is odd. Then 1 + 2^n = 1 + 2^(e*o) = 1 + (2^e)^o = 1 + ((2^e+1)  1)^o = 1 + (1)^o = 0 mod (2^e+1). Thus 2^e+1 divides 1 + 2^n. Regards, Joseph 
20080317, 13:26  #8  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Your questions consist largely of "word salad". (1) What does it mean for a theorem to be "100% effective"? I know how mathematicians use the word "effective", but that is clearly not what you mean here. Analytic number theorists use the word effective to describe a theorem where the implied constants in asymptotic (and other) estimates are explicitly known or can be computed. What do YOU mean????? (2) What do you mean by the "Legendre symbol method of sieving"? Please specify the exact sieve algorithm and its context. AFAIK, there is no such method of sieving. One might use Legendre symbols in a sieve precomputation, but I doubt if they are used during the actual sieve; their computation would be too slow. (3) For your second question, Fermat numbers are of the form 2^2^n + 1. It follows immediately from their definition that the exponent is even. What does "belief" have to do with it? And the word "effective" is again word salad. And the rest of the question is gibberish. (4) As for how many of the top 5000 primes have been tested by purely integer methods, I doubt anyone knows. What difference does it make? The prior Intel FP bug is now a red herring. The problem has been corrected. People have thoroughly checked the current IA32 chips for their FP correctness. (5) Who is this "friend"? He is clearly clueless. (6) I suggest that you consult a psychiatrist for your unreasoned paranoia. 

20080317, 15:21  #9  
"William"
May 2003
New Haven
943_{16} Posts 
Quote:
Hence the following are all algebraic factors of 2^{68544}+1: 2^{64}+1 2^{64*3}+1 2^{64*7}+1 2^{64*9}+1 2^{64*17}+1 2^{64*3*7}+1 2^{64*3*17}+1 2^{64*7*17}+1 2^{64*9*7}+1 2^{64*9*17}+1 2^{64*3*7*17}+1 

20080317, 15:54  #10  
Nov 2003
16444_{8} Posts 
Quote:
x^ab + 1 is divisible by x^a+1 when b is odd. I would have expected even Jason to know this. 

20080317, 16:19  #11  
"Jason Goatcher"
Mar 2005
DB3_{16} Posts 
[QUOTE=R.D. Silverman;128985]I would answer your questions if I could figure out what you were asking.
Your questions consist largely of "word salad". Quote:
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