Thursday, May 25, 2006
JSH: My only hope
I have mostly given up on trying to solve the factoring problem. I guess my abilities are not limitless.
Without mathematics with that kind of impact to force the issue, my only hope is to push for computer checking of mathematical arguments.
If the computer science of mathematical argument checking can be pushed forward, then when I put forward a proof and it goes through the computer—verified—then I can get my work accepted, as the social forces that so many of you effectively use, are taken out of the picture.
I have concluded then that the best way forward for me is to see what ways can be done to advance computer programming that checks math arguments, with the probable need of de-fanging mathematicians who will fight, fight, fight a system which is long overdue where for a "proof" to be accepted, it has to get through a computer checker, first.
Without mathematics with that kind of impact to force the issue, my only hope is to push for computer checking of mathematical arguments.
If the computer science of mathematical argument checking can be pushed forward, then when I put forward a proof and it goes through the computer—verified—then I can get my work accepted, as the social forces that so many of you effectively use, are taken out of the picture.
I have concluded then that the best way forward for me is to see what ways can be done to advance computer programming that checks math arguments, with the probable need of de-fanging mathematicians who will fight, fight, fight a system which is long overdue where for a "proof" to be accepted, it has to get through a computer checker, first.
# posted by James Harris @ 20:54
Self-serving limitations, math world needs scrutiny
The more I think about the lack of computer checking in the 'pure math' part of the mathematical world, the clearer it is that is must be a deliberate, and important strategy for maintaining the status quo.
Not too long ago a Space Shuttle blew up.
Later another Space Shuttle disintegrated on re-entry.
I am sure plenty of engineers were confident about their work in both cases as human nature is to have an eery confidence on a lot of things—later proven wrong by reality.
Sometimes, no matter how confident you are, or how many people are confident with you, a space shuttle blows up, or a bridge collapses, or people keep dying despite "the cure".
But in our supposedly modern math world, one person presents an argument, supposedly perfect, another looks it over by eye, and says "yea" or "nay" and people go on as if "yea" actually does mean that the argument IS perfect!
And in those areas called pure math, there is no danger of any spacecraft blowing up. No bridges can fall. No patients can look at you with deadening eyes before they die despite your miracle drug.
The one thing that can bring reality testing into those areas of the mathematical field that are impractical—pure as the saying goes—is to have computers check.
But for some reason, computers aren't used to check those arguments.
I suggest to you that the lack of real progress in having computers check is the best evidence that mathematicians know on some level that MUCH of the work being done, would not stand real scrutiny.
Otherwise, why force human beings to continue the tedium of looking over someone else's arguments if a standardized way to do it could be done, so that mathematicians could write arguments proof checker ready? Sure if the checkers says an argument is wrong—humans can check. Or if it's a big deal, and it says something is right, again humans can check.
But why isn't that done already?
I suggest to you that it not being done is a quiet admission by the mathematical community that despite all the talk of mathematical proof, the reality is that human beings screw up, put papers through, or research through that is not correct, but with only other human beings checking, the show can go on.
And they need never worry about a spectacular explosion giving away the errors.
They have, instead, the world's blind trust, and a perfect field, in that you only need to be able to convince that other guy in the field with you.
But why? How could they? Don't mathematicians want to be perfect?
Well, what if more of them are not perfect than are?
The truth can hurt, and computers can push the truth to the front that maybe there aren't that many brilliant mathematical minds on the planet. Maybe most people who get Ph.D's in the math field couldn't hack it if perfection were actually required.
Maybe most of the people in the field could not stay, if their work actually had to be—perfect.
Not too long ago a Space Shuttle blew up.
Later another Space Shuttle disintegrated on re-entry.
I am sure plenty of engineers were confident about their work in both cases as human nature is to have an eery confidence on a lot of things—later proven wrong by reality.
Sometimes, no matter how confident you are, or how many people are confident with you, a space shuttle blows up, or a bridge collapses, or people keep dying despite "the cure".
But in our supposedly modern math world, one person presents an argument, supposedly perfect, another looks it over by eye, and says "yea" or "nay" and people go on as if "yea" actually does mean that the argument IS perfect!
And in those areas called pure math, there is no danger of any spacecraft blowing up. No bridges can fall. No patients can look at you with deadening eyes before they die despite your miracle drug.
The one thing that can bring reality testing into those areas of the mathematical field that are impractical—pure as the saying goes—is to have computers check.
But for some reason, computers aren't used to check those arguments.
I suggest to you that the lack of real progress in having computers check is the best evidence that mathematicians know on some level that MUCH of the work being done, would not stand real scrutiny.
Otherwise, why force human beings to continue the tedium of looking over someone else's arguments if a standardized way to do it could be done, so that mathematicians could write arguments proof checker ready? Sure if the checkers says an argument is wrong—humans can check. Or if it's a big deal, and it says something is right, again humans can check.
But why isn't that done already?
I suggest to you that it not being done is a quiet admission by the mathematical community that despite all the talk of mathematical proof, the reality is that human beings screw up, put papers through, or research through that is not correct, but with only other human beings checking, the show can go on.
And they need never worry about a spectacular explosion giving away the errors.
They have, instead, the world's blind trust, and a perfect field, in that you only need to be able to convince that other guy in the field with you.
But why? How could they? Don't mathematicians want to be perfect?
Well, what if more of them are not perfect than are?
The truth can hurt, and computers can push the truth to the front that maybe there aren't that many brilliant mathematical minds on the planet. Maybe most people who get Ph.D's in the math field couldn't hack it if perfection were actually required.
Maybe most of the people in the field could not stay, if their work actually had to be—perfect.
# posted by James Harris @ 20:49
Sunday, May 21, 2006
JSH: Crowd mentality, consensus, and fraud
If you look at a few things that I've done, it's impossible not to wonder what's going on here that people are getting away with calling me a crackpot, and acting as if I've accomplished nothing.
For instance, I have an open source project. It's a tool for Java programmers that lets them look up class information. I think it's a useful thing for coding.
You can see how it ranks worldwide in its category:
http://www.google.com/Top/Computers/Programming/Languages/Java/Development_Tools/Code_Exploring_and_Managing/
I have a B.Sc. in physics from Vanderbilt University, but you can read people deriding a physics degree in reply to me.
I have an article I wrote for the Wikipedia, giving it the first prime counting function article, now available in the history:
http://en.wikipedia.org/w/index.php?title=Prime_counting_function&oldid=9142249
The prime counting function there while similar in important ways to what was known before me, is also different in crucial ways, and come on, it's one of the smallest bits of mathematics for counting prime numbers ever seen.
Just go look at anything on prime counting and look at what else is out there.
And I got a paper published in a peer reviewed mathematical journal which had been around for over nine years before I came along and some sci.math'ers came along, and showed just how weak the current journal system in mathematics is.
Rationalizations.
Faced with a volume of accomplishments like few others and real mathematical results like nothing ever before seen in the world, people call me names.
And they get away with it.
Why?
Because the current math field has been taken over by people who figured out a long time ago that they can just SAY things, without worrying about actually proving them, as the concept of mathematical proof is an ideal, which people can fail to achieve, yet claim they have done so.
I've been in arguments for years about the distributrive property, where I point out that, you know, well, if you multiply something, it gets multiplied whatever it's value--even if it's a function.
The arguments are absurd in a real way, but they go on because math society is a society of style over substance, where people who cannot achieve real mathematical proof, found out they could claim it, and get rewards.
The cons took over the system.
Some of you need to look up Britney Gallivan. She came up with a nifty mathematical argument that covers paper folding. She got some news articles, mentioned on television, but what has the math community done?
She's been snubbed.
Your world is controlled by people who cannot who are afraid of the people like me who can--or Britney Gallivan.
And they are stupid, or they'd throw a bone here and there, so that I wouldn't be able to step out here and point out the obvious.
So why do people let them lie?
Why did people let Bush lie?
And that's about life and blood.
This is just mathematics.
You people live in your lie as long as you think you can, and when it is revealed and the consequences come down, there will be no mercy because people will read post after post, where people here from the math community, showed their true colors.
Remember with Enron? Remember the taped phone calls?
Well, your posts here are the equivalent.
Your public statements are not only freely available for use by reporters around the world, but they are also admissable in courts of law around the world.
My role here has been, investigator for the prosecution.
My role was to use whatever means were necessary that were ethical and within the bounds of the law to fully reveal the full extent of the corruption within the math field.
The most important thing in any prosecution is an air-tight case.
I needed as much on the record as possible, as the best witnesses are mathematicians themselves.
You are my voice. Your posts here are my work product.
For instance, I have an open source project. It's a tool for Java programmers that lets them look up class information. I think it's a useful thing for coding.
You can see how it ranks worldwide in its category:
http://www.google.com/Top/Computers/Programming/Languages/Java/Development_Tools/Code_Exploring_and_Managing/
I have a B.Sc. in physics from Vanderbilt University, but you can read people deriding a physics degree in reply to me.
I have an article I wrote for the Wikipedia, giving it the first prime counting function article, now available in the history:
http://en.wikipedia.org/w/index.php?title=Prime_counting_function&oldid=9142249
The prime counting function there while similar in important ways to what was known before me, is also different in crucial ways, and come on, it's one of the smallest bits of mathematics for counting prime numbers ever seen.
Just go look at anything on prime counting and look at what else is out there.
And I got a paper published in a peer reviewed mathematical journal which had been around for over nine years before I came along and some sci.math'ers came along, and showed just how weak the current journal system in mathematics is.
Rationalizations.
Faced with a volume of accomplishments like few others and real mathematical results like nothing ever before seen in the world, people call me names.
And they get away with it.
Why?
Because the current math field has been taken over by people who figured out a long time ago that they can just SAY things, without worrying about actually proving them, as the concept of mathematical proof is an ideal, which people can fail to achieve, yet claim they have done so.
I've been in arguments for years about the distributrive property, where I point out that, you know, well, if you multiply something, it gets multiplied whatever it's value--even if it's a function.
The arguments are absurd in a real way, but they go on because math society is a society of style over substance, where people who cannot achieve real mathematical proof, found out they could claim it, and get rewards.
The cons took over the system.
Some of you need to look up Britney Gallivan. She came up with a nifty mathematical argument that covers paper folding. She got some news articles, mentioned on television, but what has the math community done?
She's been snubbed.
Your world is controlled by people who cannot who are afraid of the people like me who can--or Britney Gallivan.
And they are stupid, or they'd throw a bone here and there, so that I wouldn't be able to step out here and point out the obvious.
So why do people let them lie?
Why did people let Bush lie?
And that's about life and blood.
This is just mathematics.
You people live in your lie as long as you think you can, and when it is revealed and the consequences come down, there will be no mercy because people will read post after post, where people here from the math community, showed their true colors.
Remember with Enron? Remember the taped phone calls?
Well, your posts here are the equivalent.
Your public statements are not only freely available for use by reporters around the world, but they are also admissable in courts of law around the world.
My role here has been, investigator for the prosecution.
My role was to use whatever means were necessary that were ethical and within the bounds of the law to fully reveal the full extent of the corruption within the math field.
The most important thing in any prosecution is an air-tight case.
I needed as much on the record as possible, as the best witnesses are mathematicians themselves.
You are my voice. Your posts here are my work product.
# posted by James Harris @ 21:36
SF: Latest still won't work well
So I went ahead and thought more carefully about my latest surrogate factoring equations, and realized why they couldn't work well.
I figured out that to get solutions you needed with
f_1 f_2 = k_2 z^2
and
g_1 g_2 = k_2 z^2 + T
for f_1 + f_2 = g_1 + g_2,
with all the variables rationals, and I chased down that angle and determined that you need to pick k_2 z^2 perfectly, and the odds are slim. So those equations won't work well.
I did notice something weird though, as if you have, with some variable changes for simplicity,
f_1 f_2 = k
and
g_1 g_2 = k + M
with f_1 + f_2 = g_1 + g_2,
it's trival to show that requires that
(2f_1 g_1 - (f_1^2 + k))^2 = (f_1^2 - (k + 2M))^2 - 4M(k + M)
so I thought maybe you could let k be your target, and search for an M that will work to give a rational f_1, but maybe I did some of the algebra wrong as I tried a simple example:
k = 15, M=1
and that gives f_1 = 9, as a possible, which is close, but it'd be nicer if it were f_1 = 3.
In any event, it looks like another failure for the surrogate factoring idea.
The equations could factor, but only if you picked k_2 z^2 precisely and the odds of getting them right are long--unless you already have T factored, so, another failure.
I figured out that to get solutions you needed with
f_1 f_2 = k_2 z^2
and
g_1 g_2 = k_2 z^2 + T
for f_1 + f_2 = g_1 + g_2,
with all the variables rationals, and I chased down that angle and determined that you need to pick k_2 z^2 perfectly, and the odds are slim. So those equations won't work well.
I did notice something weird though, as if you have, with some variable changes for simplicity,
f_1 f_2 = k
and
g_1 g_2 = k + M
with f_1 + f_2 = g_1 + g_2,
it's trival to show that requires that
(2f_1 g_1 - (f_1^2 + k))^2 = (f_1^2 - (k + 2M))^2 - 4M(k + M)
so I thought maybe you could let k be your target, and search for an M that will work to give a rational f_1, but maybe I did some of the algebra wrong as I tried a simple example:
k = 15, M=1
and that gives f_1 = 9, as a possible, which is close, but it'd be nicer if it were f_1 = 3.
In any event, it looks like another failure for the surrogate factoring idea.
The equations could factor, but only if you picked k_2 z^2 precisely and the odds of getting them right are long--unless you already have T factored, so, another failure.
# posted by James Harris @ 01:39
Saturday, May 20, 2006
JSH: Lying about the distributive property
What makes this story even more dramatic though, especially in terms of the fraud involved, is that posters in fighting against my research are willing to lie about the distributive property--and they get away with it.
I can state the argument simply enough.
The distributive property simply enough says that if you multiply a group you multiply the elements within that group:
a*(b + c) = a*b + a*c
and there is no reason not to use functions so you may have
a*(f(x) + b) = a*f(x) + a*b
and here is where it gets really bizarre, as my position that the value of the function has no impact on the distributive property is now key, so I say that if at x=0, f(x)=0, then that is as valid a point as any other AS THE VALUE OF THE FUNCTION DOES NOT MATTER to the operation of the distributive property.
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
where it's still just the distributive property, but with more stuff, and now, like before, I'm going to put in functions, but then I have to be more careful than before:
7*(h(x) + b)*(c + d) = (f(x) + 7*b)*(g(x) + d)
because one of the functions just swallowed the 7 so that it is now invisible.
How can a function do that? Easy. It's a function, so it can be
f(x) = 7*h(x)
so the function swallowing the visibility of the 7 is not a big deal, but maybe though, it didn't so I add the rule that at x=0, f(x)=0 and g(x) = 0, so that I can SEE what is going on at a particular value.
If you had some other convenient value, that would be ok as well.
7*(h(0) + b)*(c + d) = (0 + 7*b)*(0 + d)
and it's clear that the 7 multiplied through like with
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
while the functions just make things a little more complicated to verify, but not impossible.
Now then, by the logical point that the value of the function does not change the distributive property, I know what happened for ANY x, but posters in fighting this argument have proclaimed x=0 to be a "special case", defying the reality of how the distributive property operates.
I've explained, and explained, and explained so that the best conclusion is that posters lie about this argument.
Otherwise they can't understand the basic principle that the value of the function does not change the distributive property, which is a major stretch.
Why is it such a big deal?
Because once the principle is established, I can get some complicated functions that show a problem with the ring of algebraic integers.
Posters in defying what is mathematically correct are just slashing at what they can, and in this case, that means questioning the distributive property, and then claiming they are not doing so, while they claim x=0 is a special case, but if I push them on the point that the value of functions does not affect the distributive property, they
claim they don't disagree!
It's a case of where the lies just keep coming and it shows you how to defy a mathematical proof.
Just claim it's wrong, keep claiming it's wrong, and get enough people to claim it's wrong so that no one believes that it's correct.
And doing that you can block acceptance of mathematical proof.
These people are undermining the discipline of mathematics by showing its true fragility.
It has few defenses against dedicated group lying about mathematical arguments.
I mean, come on! The distributive property! How could people get away with lying about that?
But they have now, for years.
I can state the argument simply enough.
The distributive property simply enough says that if you multiply a group you multiply the elements within that group:
a*(b + c) = a*b + a*c
and there is no reason not to use functions so you may have
a*(f(x) + b) = a*f(x) + a*b
and here is where it gets really bizarre, as my position that the value of the function has no impact on the distributive property is now key, so I say that if at x=0, f(x)=0, then that is as valid a point as any other AS THE VALUE OF THE FUNCTION DOES NOT MATTER to the operation of the distributive property.
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
where it's still just the distributive property, but with more stuff, and now, like before, I'm going to put in functions, but then I have to be more careful than before:
7*(h(x) + b)*(c + d) = (f(x) + 7*b)*(g(x) + d)
because one of the functions just swallowed the 7 so that it is now invisible.
How can a function do that? Easy. It's a function, so it can be
f(x) = 7*h(x)
so the function swallowing the visibility of the 7 is not a big deal, but maybe though, it didn't so I add the rule that at x=0, f(x)=0 and g(x) = 0, so that I can SEE what is going on at a particular value.
If you had some other convenient value, that would be ok as well.
7*(h(0) + b)*(c + d) = (0 + 7*b)*(0 + d)
and it's clear that the 7 multiplied through like with
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
while the functions just make things a little more complicated to verify, but not impossible.
Now then, by the logical point that the value of the function does not change the distributive property, I know what happened for ANY x, but posters in fighting this argument have proclaimed x=0 to be a "special case", defying the reality of how the distributive property operates.
I've explained, and explained, and explained so that the best conclusion is that posters lie about this argument.
Otherwise they can't understand the basic principle that the value of the function does not change the distributive property, which is a major stretch.
Why is it such a big deal?
Because once the principle is established, I can get some complicated functions that show a problem with the ring of algebraic integers.
Posters in defying what is mathematically correct are just slashing at what they can, and in this case, that means questioning the distributive property, and then claiming they are not doing so, while they claim x=0 is a special case, but if I push them on the point that the value of functions does not affect the distributive property, they
claim they don't disagree!
It's a case of where the lies just keep coming and it shows you how to defy a mathematical proof.
Just claim it's wrong, keep claiming it's wrong, and get enough people to claim it's wrong so that no one believes that it's correct.
And doing that you can block acceptance of mathematical proof.
These people are undermining the discipline of mathematics by showing its true fragility.
It has few defenses against dedicated group lying about mathematical arguments.
I mean, come on! The distributive property! How could people get away with lying about that?
But they have now, for years.
# posted by James Harris @ 22:01
Friday, May 19, 2006
JSH: Fraud question is pertinent
Someone asked the question of what fraud is there in the field of mathematics.
I have pointed out before that eerily there are few if any reported cases that you can find.
Some posters tried to toss up my paper as an example, not surprising for this newsgroup.
But the real story is that it is inconceivable that an active and healthy discipline could not have published cases of fraud or attempted fraud.
It's just not possible in the real world.
The best explanation is that fraud in mathematics does not get reported, not that it doesn't happen.
My own research leads me to conclude that errors dominate the modern math field in areas of "pure math" where you have to take people's word for it that mathematical arguments are correct.
It looks like errors came in over a hundred years ago, and have dominated pure math ever since.
One odd clue that mathematicians are at least on some level aware that much of the field is full of errors--the weird inability to get widespread computer checking of mathematical arguments claimed to be proofs.
Computers help medical doctors, expert systems are dominate in all sorts of areas, but in mathematics, for some reason, supposedly, computers are not up to the challenge.
I suggest to you, that computers would show how full of error the field is, so that's why systems will not get developed until the world accepts the obvious.
I suggest to you that mathematicians will always block, one way or another, development of computer checking to protect themselves, until someone steps in and forces it.
I have pointed out before that eerily there are few if any reported cases that you can find.
Some posters tried to toss up my paper as an example, not surprising for this newsgroup.
But the real story is that it is inconceivable that an active and healthy discipline could not have published cases of fraud or attempted fraud.
It's just not possible in the real world.
The best explanation is that fraud in mathematics does not get reported, not that it doesn't happen.
My own research leads me to conclude that errors dominate the modern math field in areas of "pure math" where you have to take people's word for it that mathematical arguments are correct.
It looks like errors came in over a hundred years ago, and have dominated pure math ever since.
One odd clue that mathematicians are at least on some level aware that much of the field is full of errors--the weird inability to get widespread computer checking of mathematical arguments claimed to be proofs.
Computers help medical doctors, expert systems are dominate in all sorts of areas, but in mathematics, for some reason, supposedly, computers are not up to the challenge.
I suggest to you, that computers would show how full of error the field is, so that's why systems will not get developed until the world accepts the obvious.
I suggest to you that mathematicians will always block, one way or another, development of computer checking to protect themselves, until someone steps in and forces it.
# posted by James Harris @ 20:23
SF: Problem is easy implementation, massive impact
The problem I see with surrogate factoring is that it is a powerful idea, that if fully realized is easy to implement, and impossible to control, as I can't see any way that it can be fully developed without it escaping.
I know a lot of you don't take this seriously, but consider that if I am right, we don't get to go back in time, and fix things.
I have procrastinated, threatened, posted incomplete work, work full of error, and hoped, hoped, hoped that some sanity would intervene, but we are at the threshold, and none has.
Here's the scenario I am frightened by.
No matter what I do, like if I succeeded at getting right equations, they worked well, I implemented and solved some RSA numbers, and kept the details in secret, eventually the information would have to come out, one way or another.
Some initial fallout that I project:
The death of eBay, Skype, and Amazon.com among other companies.
A tremendous blow against all the wireless companies.
A drop in stock markets around the world.
Potential for a tremendous drop in the value of the U.S. dollar, followed by spiraling inflation.
That's initial fallout.
At this time, I think that initial falllout is nearly inevitable, but I am kind of sort of still hoping.
What's needed is government intervention earlier than later.
I can't get that without some of you, contacting the right people, sooner than later.
I can't do it. Why would they believe me?
And besides, I DO not have an implementation. I DO not have full proof that this works.
I have been sitting on the basic approach for months, putting out various ideas which clearly have not worked.
But what if this latest does?
I need help now, not later, if it does.
I know a lot of you don't take this seriously, but consider that if I am right, we don't get to go back in time, and fix things.
I have procrastinated, threatened, posted incomplete work, work full of error, and hoped, hoped, hoped that some sanity would intervene, but we are at the threshold, and none has.
Here's the scenario I am frightened by.
No matter what I do, like if I succeeded at getting right equations, they worked well, I implemented and solved some RSA numbers, and kept the details in secret, eventually the information would have to come out, one way or another.
Some initial fallout that I project:
The death of eBay, Skype, and Amazon.com among other companies.
A tremendous blow against all the wireless companies.
A drop in stock markets around the world.
Potential for a tremendous drop in the value of the U.S. dollar, followed by spiraling inflation.
That's initial fallout.
At this time, I think that initial falllout is nearly inevitable, but I am kind of sort of still hoping.
What's needed is government intervention earlier than later.
I can't get that without some of you, contacting the right people, sooner than later.
I can't do it. Why would they believe me?
And besides, I DO not have an implementation. I DO not have full proof that this works.
I have been sitting on the basic approach for months, putting out various ideas which clearly have not worked.
But what if this latest does?
I need help now, not later, if it does.
# posted by James Harris @ 20:10
Thursday, May 18, 2006
SF: The other route, corrected approach
The same reasons for putting this on Usenet are still there, so I changed my mind again, and am now putting up the surrogate factoring equations that follow from a different approach than what I used before:
(2(k_1-2)y - vz)^2 = (9 - 4k_1)z^2 v^2 + 4(k_1 - 2)(k_2 z^2 + T)
where
T = (x+y+vz)(x+2y-vz)
and T is the target composite.
My previous approaches, if you remember, ended up with you picking y, and getting z, but that way doesn't work for some simple reasons, which I call the broken wing problem.
This way doesn't have the same problem.
(2(k_1-2)y - vz)^2 = (9 - 4k_1)z^2 v^2 + 4(k_1 - 2)(k_2 z^2 + T)
where
T = (x+y+vz)(x+2y-vz)
and T is the target composite.
My previous approaches, if you remember, ended up with you picking y, and getting z, but that way doesn't work for some simple reasons, which I call the broken wing problem.
This way doesn't have the same problem.
# posted by James Harris @ 22:17
Sunday, May 14, 2006
JSH: That was useless
Um, so the factoring idea I have is not new, and the solution I gave was just the upper bound for positive x.
I am going to shift strategy a bit as I do not hate the alt.math newsgroup.
I am dropping routinely posting to the alt.math newsgroup.
You people did not mount am email campaign against my paper.
Your posters have not spent years in dedicated efforts to paint me as a crackpot or put up webpages deriding me and my research.
I started posting to alt.math about my settled research like my prime counting function but when nothing happened kept going even when I wandered off into, well, other stuff.
I will continue to post to the sci.math newsgroup, however, as I do hate that newsgroup, and its members have done everything I mentioned above and more.
Until my settled research is accepted I have little to do but piddle around here or there when it comes to math stuff, and make a lot of noise about factoring, as I have already mentioned many times my reticence about putting up a working factoring solution.
I think that would be punishing people who have nothing to do with any of this, who have no clue about these math wars, so I just do not plan on doing it.
But I can keep posting my speculations and dabblings here or there, with it upfront that I have no intentions of ever succeeding at giving a workable solution to the factoring problem.
I am going to shift strategy a bit as I do not hate the alt.math newsgroup.
I am dropping routinely posting to the alt.math newsgroup.
You people did not mount am email campaign against my paper.
Your posters have not spent years in dedicated efforts to paint me as a crackpot or put up webpages deriding me and my research.
I started posting to alt.math about my settled research like my prime counting function but when nothing happened kept going even when I wandered off into, well, other stuff.
I will continue to post to the sci.math newsgroup, however, as I do hate that newsgroup, and its members have done everything I mentioned above and more.
Until my settled research is accepted I have little to do but piddle around here or there when it comes to math stuff, and make a lot of noise about factoring, as I have already mentioned many times my reticence about putting up a working factoring solution.
I think that would be punishing people who have nothing to do with any of this, who have no clue about these math wars, so I just do not plan on doing it.
But I can keep posting my speculations and dabblings here or there, with it upfront that I have no intentions of ever succeeding at giving a workable solution to the factoring problem.
# posted by James Harris @ 12:12
Saturday, May 13, 2006
JSH: Why factoring solution works
I've been going back and forth for days with what I now am certain is a simple solution to the factoring problem, as in you just plug and chug as they say, and out pops the answer a very high percentage of the time, and I can't find a damn thing wrong with the argument.
The math is simple, but the idea that something believed for so long to be very difficult, is not, is very hard.
So I keep going over the idea, over and over again, trying to figure out what must be wrong, because so many people thought the factoring problem was hard.
I can't find anything wrong.
I did something different by looking at solutions to
sqrt(1+8mT)
where T is the target, and m is selected by quadratic residue modulo n, where n is some odd natural coprime to T, of your choosing.
So I have m = kn + c, where c is the residue mod n, and k is a number to be determined to solve the factoring problem.
Then I did something very clever, as I let
k+x = sqrt(1+8T(kn+c))
where x is just some number, and now there is something weird you can do, as though I want c to be a residue, presumably less than n, but greater than 0, that is not mathematically required, so I can have
k'+x' = sqrt(1+8T(k'n + c'))
where k'+x' = k+x, and k'n + c' = kn + c
as I can subtract n from kn, and add it to c, so I have explicitly
(k+1)+(x-1) = sqrt(1+8T((k+1)n + c-n))
and the math is none the wiser, as in it can't just decide that the c I'm using is not something mod n, so this route actually forces the mathematics to show you how to pick only integers, as if you keep incrementing k, you are DECREMENTING x, and, of course, at some point then x=0.
But when you solve for k, you can solve for the x=0 case, getting the answer.
There is a lot of interesting mathematics in there, but the big deal now will not be about the mathematics but about the factoring problem being solved.
The answer is bizarrely simple considering that for over ten years RSA has been in use, when the answer is such a trivial one to factoring numbers:
x = floor((1+8cT + 16n^2 T^2)/8nT)
and
k = 4nT - x + sqrt(1 + 8cT + 16n^2 T^2 - 8nTx)
The one tricky part is that
c = (r-1)*(8*T)^{-1) mod n
so you have to pick n, which has to be odd and coprime to T, for the modular inverse to be guaranteed to exist, and you also have to pick r, a quadratic residue of n.
Who knew that such a seemingly hard problem requiring huge amounts of computer resources and efforts of hundreds of people worldwide could be reduced to something trivial enough to give to a kid to program for homework?
It is amazing how powerful mathematics can be.
Some of you have not a clue what mathematics is, or just how powerful it can be.
I want some of you to consider how weak some people try to make it.
They fight me, but their fight just forced my hand here.
By using social forces they could convince people of things not true, but what happens when the truth is forced?
Things break. And it's the fault not only of the people who lied to you, and convinced you, but your fault for believing, and believing in them, against mathematical proof.
Mathematical proof is not about opinion, politics, or what makes you feel good.
And it doesn't care about your bank account.
The math is simple, but the idea that something believed for so long to be very difficult, is not, is very hard.
So I keep going over the idea, over and over again, trying to figure out what must be wrong, because so many people thought the factoring problem was hard.
I can't find anything wrong.
I did something different by looking at solutions to
sqrt(1+8mT)
where T is the target, and m is selected by quadratic residue modulo n, where n is some odd natural coprime to T, of your choosing.
So I have m = kn + c, where c is the residue mod n, and k is a number to be determined to solve the factoring problem.
Then I did something very clever, as I let
k+x = sqrt(1+8T(kn+c))
where x is just some number, and now there is something weird you can do, as though I want c to be a residue, presumably less than n, but greater than 0, that is not mathematically required, so I can have
k'+x' = sqrt(1+8T(k'n + c'))
where k'+x' = k+x, and k'n + c' = kn + c
as I can subtract n from kn, and add it to c, so I have explicitly
(k+1)+(x-1) = sqrt(1+8T((k+1)n + c-n))
and the math is none the wiser, as in it can't just decide that the c I'm using is not something mod n, so this route actually forces the mathematics to show you how to pick only integers, as if you keep incrementing k, you are DECREMENTING x, and, of course, at some point then x=0.
But when you solve for k, you can solve for the x=0 case, getting the answer.
There is a lot of interesting mathematics in there, but the big deal now will not be about the mathematics but about the factoring problem being solved.
The answer is bizarrely simple considering that for over ten years RSA has been in use, when the answer is such a trivial one to factoring numbers:
x = floor((1+8cT + 16n^2 T^2)/8nT)
and
k = 4nT - x + sqrt(1 + 8cT + 16n^2 T^2 - 8nTx)
The one tricky part is that
c = (r-1)*(8*T)^{-1) mod n
so you have to pick n, which has to be odd and coprime to T, for the modular inverse to be guaranteed to exist, and you also have to pick r, a quadratic residue of n.
Who knew that such a seemingly hard problem requiring huge amounts of computer resources and efforts of hundreds of people worldwide could be reduced to something trivial enough to give to a kid to program for homework?
It is amazing how powerful mathematics can be.
Some of you have not a clue what mathematics is, or just how powerful it can be.
I want some of you to consider how weak some people try to make it.
They fight me, but their fight just forced my hand here.
By using social forces they could convince people of things not true, but what happens when the truth is forced?
Things break. And it's the fault not only of the people who lied to you, and convinced you, but your fault for believing, and believing in them, against mathematical proof.
Mathematical proof is not about opinion, politics, or what makes you feel good.
And it doesn't care about your bank account.
# posted by James Harris @ 23:25
Solution of the factoring problem
Here is a post with the corrected derivation.
I know it will probably seem strange to now live in a world where a problem previously thought extremely difficult has now been shown to be trivially easy, but mathematics is a wonderful discipline because it's not about what you believe, but about what is proven.
It is a world of absolutes, and absolute truth.
It is bizarre beyond belief that one of the supposedly hardest problems in mathematics, is trivially easy--if you know how to solve it--but in this post I will step through a trivial solution to the factoring problem.
Earlier I noted a way to use quadratic residues to find m as a congruence relationship to factor with
sqrt(1 + 8*mT)
where the idea of looking mod some particular n, goes back to Fermat and later Gauss, which shows you how old this way of looking at things is.
Since m = (r-1)*(8*T)^{-1} mod n, where n is an odd natural of your choosing coprime to T--the target to be factored--and r is some quadratic residue of n, where I was just picking r=4.
So you can let
c = (r-1)*(8*T)^{-1) mod n
to have explicitly
m = kn + c
where k is to be determined.
But, it must be true that sqrt(1+8*mT) for a solution is a natural equal to k plus some number, which I will call x, so I have
(k+x)2 = 1 + 8(kn + c)T
and expanding and collecting to the left side I have
k2 + 2kx + x2 - 1 - 8nkT - 8cT = 0
and collecting with respect to the k's I have
k2 + 2(x-4nT)k + x2 - 1 - 8cT = 0
and completing the square gives
k2 + 2(x-4nT)k + (x-4nT)2 + x2 - 1 - 8cT - (x-4nT)2= 0
so
(k + x-4nT)2 + x2 - 1 - 8cT - x2 + 8xnT - 16n2 T2 = 0
and simplifying further and collecting everything outside the square to the right side gives
(k + x-4nT)2 = 1 + 8cT + 16 n2 T2 - 8nTx
but notice that c because it is defined as
c = (r-1)*(8*T)^{-1) mod n
can be an infinite number of possible values based on adding or subtracting multiples of n, but you can change c without changing m, by incrementing k, as
m = kn + c
so if you subtract n from c, you can add 1 to k, so that m does not change.
e.g. m = (k-1)n + c-n
but by not changing m, you don't change k+x, but if you are incrementing k, then you must be decrementing x, and you can do that to the point that x=0.
Logically then the solution to the factoring problem is
x = floor((1+8ct + 16n2 T2)/8nT)
and
k = 4nT - x + sqrt(1 + 8cT + 16n2 T2 - 8nTx)
with N = sqrt(1 + 8(kn + c)T), giving a non-trivial factor of T, from N-1 or N+1, 50% of the time, where remember,
(I am not as certain now as the success rate may be much higher.)
c = (r-1)*(8*T)^{-1) mod n
and r is any quadratic residue modulo n, and n is a natural number of your choosing, so you have a complete solution to the factoring problem.
Amazing but true. It is the answer to the problem stepped out simply enough.
So why must it work?
Well, consider that with some natural number k to have some integer x, it just must be that x can be decremented in the way I described, forcing a 0, so there is this way of figuring out integer solutions.
I found a way where the mathematics looks for integers, as it "knows" that is the answer you want, so it just pops out the answer.
Oddly enough, it also must be true that only one m exists for any given quadratic residue r mod n.
The equations outlined here are a complete solution to the factoring problem and represent my SECOND solution to that problem.
My previous solution is just a lot more complicated using what I call surrogate factoring.
I know it will probably seem strange to now live in a world where a problem previously thought extremely difficult has now been shown to be trivially easy, but mathematics is a wonderful discipline because it's not about what you believe, but about what is proven.
It is a world of absolutes, and absolute truth.
It is bizarre beyond belief that one of the supposedly hardest problems in mathematics, is trivially easy--if you know how to solve it--but in this post I will step through a trivial solution to the factoring problem.
Earlier I noted a way to use quadratic residues to find m as a congruence relationship to factor with
sqrt(1 + 8*mT)
where the idea of looking mod some particular n, goes back to Fermat and later Gauss, which shows you how old this way of looking at things is.
Since m = (r-1)*(8*T)^{-1} mod n, where n is an odd natural of your choosing coprime to T--the target to be factored--and r is some quadratic residue of n, where I was just picking r=4.
So you can let
c = (r-1)*(8*T)^{-1) mod n
to have explicitly
m = kn + c
where k is to be determined.
But, it must be true that sqrt(1+8*mT) for a solution is a natural equal to k plus some number, which I will call x, so I have
(k+x)2 = 1 + 8(kn + c)T
and expanding and collecting to the left side I have
k2 + 2kx + x2 - 1 - 8nkT - 8cT = 0
and collecting with respect to the k's I have
k2 + 2(x-4nT)k + x2 - 1 - 8cT = 0
and completing the square gives
k2 + 2(x-4nT)k + (x-4nT)2 + x2 - 1 - 8cT - (x-4nT)2= 0
so
(k + x-4nT)2 + x2 - 1 - 8cT - x2 + 8xnT - 16n2 T2 = 0
and simplifying further and collecting everything outside the square to the right side gives
(k + x-4nT)2 = 1 + 8cT + 16 n2 T2 - 8nTx
but notice that c because it is defined as
c = (r-1)*(8*T)^{-1) mod n
can be an infinite number of possible values based on adding or subtracting multiples of n, but you can change c without changing m, by incrementing k, as
m = kn + c
so if you subtract n from c, you can add 1 to k, so that m does not change.
e.g. m = (k-1)n + c-n
but by not changing m, you don't change k+x, but if you are incrementing k, then you must be decrementing x, and you can do that to the point that x=0.
Logically then the solution to the factoring problem is
x = floor((1+8ct + 16n2 T2)/8nT)
and
k = 4nT - x + sqrt(1 + 8cT + 16n2 T2 - 8nTx)
with N = sqrt(1 + 8(kn + c)T), giving a non-trivial factor of T, from N-1 or N+1, 50% of the time, where remember,
(I am not as certain now as the success rate may be much higher.)
c = (r-1)*(8*T)^{-1) mod n
and r is any quadratic residue modulo n, and n is a natural number of your choosing, so you have a complete solution to the factoring problem.
Amazing but true. It is the answer to the problem stepped out simply enough.
So why must it work?
Well, consider that with some natural number k to have some integer x, it just must be that x can be decremented in the way I described, forcing a 0, so there is this way of figuring out integer solutions.
I found a way where the mathematics looks for integers, as it "knows" that is the answer you want, so it just pops out the answer.
Oddly enough, it also must be true that only one m exists for any given quadratic residue r mod n.
The equations outlined here are a complete solution to the factoring problem and represent my SECOND solution to that problem.
My previous solution is just a lot more complicated using what I call surrogate factoring.
# posted by James Harris @ 13:39
Solving the factoring problem
It is bizarre beyond belief that one of the supposedly hardest problems in mathematics, is trivially easy--if you know how to solve it--and in this post I will step through a trivial solution to the factoring problem.
Earlier I noted a way to use quadratic residues to find m as a congruence relationship to factor with
sqrt(1 + 8*mT)
where the idea of looking mod some particular n, goes back to Fermat and later Gauss, which shows you how old this way of looking at things is.
Since m = (r-1)*(8*T)^{-1} mod n, where n is an odd natural of your choosing coprime to T--the target to be factored--and r is some quadratic residue of n, where I was just picking r=4.
So you can let
c = (r-1)*(8*T)^{-1) mod n
to have explicitly
m = kn + c
where k is to be determined.
But, it must be true that sqrt(1+8*mT) for a solution is a natural equal to k plus some number, which I will call x, so I have
(k+x)^2 = 1 + 8(kn + c)T
and expanding and collecting to the left side I have
k^2 + 2kx + x^2 - 1 - 8nkT - 8cT = 0
and collecting with respect to the k's I have
k^2 + 2(x-4nT)k + x^2 - 1 - 8cT = 0
and completing the square gives
k^2 + 2(x-4nT)k + (x-4nT)^2 + x^2 - 1 - 8cT - (x-rnT)^2= 0
so
(k + x-4nT)^2 + x^2 - 1 - 8cT - x^2 + 2xnT - n^2 T^2 = 0
and simplifying further and collecting everything outside the square to the right side gives
(k + x-4nT)^2 = 1 + 8cT + n^2 T^2 - 2nTx
but notice that c because it is defined as
c = (r-1)*(8*T)^{-1) mod n
can be an infinite number of possible values based on adding or subtracting multiples of n, but you can change c without changing m, by incrementing k, as
m = kn + c
so if you subtract n from c, you can add 1 to k, so that m does not change.
e.g. m = (k-1)n + c-n
but by not changing m, you don't change k+x, but if you are incrementing k, then you must be decrementing x, and you can do that to the point that x=0.
Logically then the solution to the factoring problem is
x = floor((1+8ct + n^2 T^2)/2nT)
and
k = 4nT - x + sqrt(1 + 8cT + n^2 T^2 - 2nTx)
with N = sqrt(1 + 8(kn + c)T), giving a non-trivial factor of T, from N-1 or N+1, 50% of the time, where remember,
c = (r-1)*(8*T)^{-1) mod n
and r is any quadratic residue modulo n, and n is a natural number of your choosing, so you have a complete solution to the factoring problem.
So why must it work? Well, consider that with some natural number k, to have some integer x, it just must be that x can be decremented in the way I described, forcing a 0, so there is this way of figuring out integer solutions.
Oddly enough, it also must be true that only one m exists for any given quadratic residue r mod n.
The 50% comes in because the mathematics doesn't distinguish between "trivial" or "non-trivial" factorizations.
The equations outlined here are a complete solution to the factoring problem and represent my SECOND solution to that problem.
My previous solution is just a lot more complicated using what I call surrogate factoring.
This second method though, is trivial enough even for humans to understand.
You people have been calling hard, something so easy it almost brings me to tears contemplating it.
I wonder what is wrong with you.
Earlier I noted a way to use quadratic residues to find m as a congruence relationship to factor with
sqrt(1 + 8*mT)
where the idea of looking mod some particular n, goes back to Fermat and later Gauss, which shows you how old this way of looking at things is.
Since m = (r-1)*(8*T)^{-1} mod n, where n is an odd natural of your choosing coprime to T--the target to be factored--and r is some quadratic residue of n, where I was just picking r=4.
So you can let
c = (r-1)*(8*T)^{-1) mod n
to have explicitly
m = kn + c
where k is to be determined.
But, it must be true that sqrt(1+8*mT) for a solution is a natural equal to k plus some number, which I will call x, so I have
(k+x)^2 = 1 + 8(kn + c)T
and expanding and collecting to the left side I have
k^2 + 2kx + x^2 - 1 - 8nkT - 8cT = 0
and collecting with respect to the k's I have
k^2 + 2(x-4nT)k + x^2 - 1 - 8cT = 0
and completing the square gives
k^2 + 2(x-4nT)k + (x-4nT)^2 + x^2 - 1 - 8cT - (x-rnT)^2= 0
so
(k + x-4nT)^2 + x^2 - 1 - 8cT - x^2 + 2xnT - n^2 T^2 = 0
and simplifying further and collecting everything outside the square to the right side gives
(k + x-4nT)^2 = 1 + 8cT + n^2 T^2 - 2nTx
but notice that c because it is defined as
c = (r-1)*(8*T)^{-1) mod n
can be an infinite number of possible values based on adding or subtracting multiples of n, but you can change c without changing m, by incrementing k, as
m = kn + c
so if you subtract n from c, you can add 1 to k, so that m does not change.
e.g. m = (k-1)n + c-n
but by not changing m, you don't change k+x, but if you are incrementing k, then you must be decrementing x, and you can do that to the point that x=0.
Logically then the solution to the factoring problem is
x = floor((1+8ct + n^2 T^2)/2nT)
and
k = 4nT - x + sqrt(1 + 8cT + n^2 T^2 - 2nTx)
with N = sqrt(1 + 8(kn + c)T), giving a non-trivial factor of T, from N-1 or N+1, 50% of the time, where remember,
c = (r-1)*(8*T)^{-1) mod n
and r is any quadratic residue modulo n, and n is a natural number of your choosing, so you have a complete solution to the factoring problem.
So why must it work? Well, consider that with some natural number k, to have some integer x, it just must be that x can be decremented in the way I described, forcing a 0, so there is this way of figuring out integer solutions.
Oddly enough, it also must be true that only one m exists for any given quadratic residue r mod n.
The 50% comes in because the mathematics doesn't distinguish between "trivial" or "non-trivial" factorizations.
The equations outlined here are a complete solution to the factoring problem and represent my SECOND solution to that problem.
My previous solution is just a lot more complicated using what I call surrogate factoring.
This second method though, is trivial enough even for humans to understand.
You people have been calling hard, something so easy it almost brings me to tears contemplating it.
I wonder what is wrong with you.
# posted by James Harris @ 02:38
Wednesday, May 10, 2006
Factoring idea, trying to recover something
I may as well admit that I was really excited about a fairly trivial quadratic residue result, and it just bugs me that I was so certain it was a big deal, when it turned out to be so trivial, so I find myself trying to rescue something from the idea.
Earlier I noted a way to use quadratic residues to find m as a congruence relationship to factor with
sqrt(1 + 8*mT)
where the idea of looking mod some particular n, goes back to Fermat and later Gauss, which shows you how old this way of looking at things is. Sigh. But I find myself working at it anyway, maybe because I'm just looking for something to mull over, no matter how trivial and old it might be.
Since m = (r-1)*(8*T)^{-1} mod n, where n is an odd natural of your choosing coprime to T—the target to be factored—and r is some quadratic residue of n, where I was just picking r=4.
So you can let
c = (r-1)*(8*T)^{-1) mod n
to have explicitly
m = kn + c
where k is to be determined.
But, it must be true that sqrt(1+8*mT) for a solution is a natural equal to k plus some number, which I will call x, so I have
(k+x)^2 = 1 + 8(kn + c)T
and expanding and collecting to the left side I have
k^2 + 2kx + x^2 - 1 - 8nkT - 8cT = 0
and collecting with respect to the k's I have
k^2 + 2(x-4nT)k + x^2 - 1 - 8cT = 0
and completing the square gives
k^2 + 2(x-4nT)k + (x-4nT)^2 + x^2 - 1 - 8cT - (x-rnT)^2= 0
so
(k + x-4nT)^2 + x^2 - 1 - 8cT - x^2 + 2xnT - n^2 T^2 = 0
and simplifying further and collecting everything outside the square to the right side gives
(k + x-4nT)^2 = 1 + 8cT + n^2 T^2 - 2nTx
which is a lot like what I started with, except if x is positive (it could be negative so it's not necessarily a natural) then you're going down mod 2nT, and with n odd and T odd, it looks like x must be divisible by 4, so you're actually searching mod 8nT.
Besides that, I got nothing. I admit, I did it again. Wandered off into some old area and managed to re-discover what was already known.
But it was fun for a while.
Here's something maybe. That c in there is defined as
c = (r-1)*(8*T)^{-1) mod n
so you can make it as negative as you want by subtracting n.
What if you went for a positive minimum? Subtracting such that
1 + 8cT + n^2 T^2 > 0
was the smallest possible? It looks like that would increase k, and shrink x, if x were positive, and it might shrink the search space.
The logic of it is so damn simple it's almost beyond belief, but what I've given in this thread is the simple solution to the factoring problem.
I will make another post explaing it in boring detail.
The gist of it though is that for a solution to m, there must exist a natural number k, but also you can do this weird thing of decrementing x, while incrementing k, without changing m, so that you can find a simple answer.
What's remarkable is that in my post I solved the factoring problem, and did anyone notice?
Earlier I noted a way to use quadratic residues to find m as a congruence relationship to factor with
sqrt(1 + 8*mT)
where the idea of looking mod some particular n, goes back to Fermat and later Gauss, which shows you how old this way of looking at things is. Sigh. But I find myself working at it anyway, maybe because I'm just looking for something to mull over, no matter how trivial and old it might be.
Since m = (r-1)*(8*T)^{-1} mod n, where n is an odd natural of your choosing coprime to T—the target to be factored—and r is some quadratic residue of n, where I was just picking r=4.
So you can let
c = (r-1)*(8*T)^{-1) mod n
to have explicitly
m = kn + c
where k is to be determined.
But, it must be true that sqrt(1+8*mT) for a solution is a natural equal to k plus some number, which I will call x, so I have
(k+x)^2 = 1 + 8(kn + c)T
and expanding and collecting to the left side I have
k^2 + 2kx + x^2 - 1 - 8nkT - 8cT = 0
and collecting with respect to the k's I have
k^2 + 2(x-4nT)k + x^2 - 1 - 8cT = 0
and completing the square gives
k^2 + 2(x-4nT)k + (x-4nT)^2 + x^2 - 1 - 8cT - (x-rnT)^2= 0
so
(k + x-4nT)^2 + x^2 - 1 - 8cT - x^2 + 2xnT - n^2 T^2 = 0
and simplifying further and collecting everything outside the square to the right side gives
(k + x-4nT)^2 = 1 + 8cT + n^2 T^2 - 2nTx
which is a lot like what I started with, except if x is positive (it could be negative so it's not necessarily a natural) then you're going down mod 2nT, and with n odd and T odd, it looks like x must be divisible by 4, so you're actually searching mod 8nT.
Besides that, I got nothing. I admit, I did it again. Wandered off into some old area and managed to re-discover what was already known.
But it was fun for a while.
Here's something maybe. That c in there is defined as
c = (r-1)*(8*T)^{-1) mod n
so you can make it as negative as you want by subtracting n.
What if you went for a positive minimum? Subtracting such that
1 + 8cT + n^2 T^2 > 0
was the smallest possible? It looks like that would increase k, and shrink x, if x were positive, and it might shrink the search space.
The logic of it is so damn simple it's almost beyond belief, but what I've given in this thread is the simple solution to the factoring problem.
I will make another post explaing it in boring detail.
The gist of it though is that for a solution to m, there must exist a natural number k, but also you can do this weird thing of decrementing x, while incrementing k, without changing m, so that you can find a simple answer.
What's remarkable is that in my post I solved the factoring problem, and did anyone notice?
# posted by James Harris @ 20:01
Sunday, May 07, 2006
JSH: Like being constipated
Over the years I've come up with a lot of rationalizations to explain why I spend as much time as I do fiddling with various mathematical ideas, and thinking back now, I think they were all wrong and I just can't help it.
I don't consider myself to be a mathematician. I don't want to be a mathematician.
But I find myself thinking on these mathematical ideas and being compelled to talk about them.
I don't want to carefully study previous mathematical research--unless I just must for some particular idea--get a degree in mathematics, or become a part of math society as in, like, hanging out with mathematicians.
If you hadn't noticed, I don't like mathematicians.
To me one of the more annoying aspects of this entire thing is that I am on USENET where I should be able to talk math. I do at times push for people to critique my ideas, and demand attention for them, but that's part of this entire thing that I don't fully understand.
Regardless of my antagonistic behavior, reality is, you can still ignore me.
I find myself in idle moments wondering about math stuff, trying to figure this or that out, and then talking about it, likesome freaking compulsion.
It's like being constipated, until you get some other idea out, and then I feel a little better.
I HATE some of you because you feel compelled to obsessively track my postings and reply to me, when I don't see a use for you, but hey, maybe you're compelled as well.
Maybe we're all trapped in our obsessions playing out this stupid drama.
My obsession with playing with math ideas is not necessarily an unhealthy one.
It passes the time. Talking about my amateur research is not wrong. And when I think I have major results, my pushing them is not wrong either.
And there have been so many bizarre and wrong happenings from math society that those cannot be denied, like members of this forum managing to break the formal peer review process and that math journal dying.
My finding myself fiddling with math stuff is not a bad thing. My talking about my ideas is not a bad thing. My pushing when I see other people behaving wrongly is not a bad thing.
Creativity is not a bad thing. Brainstorming is not a bad thing.
But some of you are definitely not acting as if I am someone doing what is ok.
You obsess over my postings. You come after me in vicious ways, or go after my work.
There is something wrong with some of you people.
I admit my little puzzling issues, like being compelled to work at math stuff, and not even fully understanding why so that over the years I've come up with all kinds of reasons.
But why are some of you such nasty people who feel compelled to spew vicious insults, day, after day, after day, month, after month, after month, year after year?
What puts that kind of nasty stew into a human heart?
I get upset when people try to block me from my compulsion, or ignore my math results, even when they're wrong, until I find out, then I feel a need to find more, but some of you are just angry and nasty.
What are you in it for?
I don't consider myself to be a mathematician. I don't want to be a mathematician.
But I find myself thinking on these mathematical ideas and being compelled to talk about them.
I don't want to carefully study previous mathematical research--unless I just must for some particular idea--get a degree in mathematics, or become a part of math society as in, like, hanging out with mathematicians.
If you hadn't noticed, I don't like mathematicians.
To me one of the more annoying aspects of this entire thing is that I am on USENET where I should be able to talk math. I do at times push for people to critique my ideas, and demand attention for them, but that's part of this entire thing that I don't fully understand.
Regardless of my antagonistic behavior, reality is, you can still ignore me.
I find myself in idle moments wondering about math stuff, trying to figure this or that out, and then talking about it, likesome freaking compulsion.
It's like being constipated, until you get some other idea out, and then I feel a little better.
I HATE some of you because you feel compelled to obsessively track my postings and reply to me, when I don't see a use for you, but hey, maybe you're compelled as well.
Maybe we're all trapped in our obsessions playing out this stupid drama.
My obsession with playing with math ideas is not necessarily an unhealthy one.
It passes the time. Talking about my amateur research is not wrong. And when I think I have major results, my pushing them is not wrong either.
And there have been so many bizarre and wrong happenings from math society that those cannot be denied, like members of this forum managing to break the formal peer review process and that math journal dying.
My finding myself fiddling with math stuff is not a bad thing. My talking about my ideas is not a bad thing. My pushing when I see other people behaving wrongly is not a bad thing.
Creativity is not a bad thing. Brainstorming is not a bad thing.
But some of you are definitely not acting as if I am someone doing what is ok.
You obsess over my postings. You come after me in vicious ways, or go after my work.
There is something wrong with some of you people.
I admit my little puzzling issues, like being compelled to work at math stuff, and not even fully understanding why so that over the years I've come up with all kinds of reasons.
But why are some of you such nasty people who feel compelled to spew vicious insults, day, after day, after day, month, after month, after month, year after year?
What puts that kind of nasty stew into a human heart?
I get upset when people try to block me from my compulsion, or ignore my math results, even when they're wrong, until I find out, then I feel a need to find more, but some of you are just angry and nasty.
What are you in it for?
# posted by James Harris @ 14:32
JSH: Pure, pure math
I have been working feverishly to find something HUGE that will push things forward with my latest result, but I'm starting to think it's pure, pure math, as in being such a raw result that it's hard to find anything practical with it.
The result, of course, as I like talking about it so much so you probably have seen it if you read my threads, is with natural numbers, n_1, n_2, C, and k, where
C = n_1 + n_2
and k is a difference of factors of 2C, it must be true that
(8C + k^2) mod n_1
is a square modulor n_1, and
(8C + k^2) mod n_2 is a square modulo n_2,
where I have written that differently in the past, as it also means that
(8n_2 + k^2) mod n_1 is a square modulo n_1, and vice versa,
which follows, of course, fomr C = n_1 + n_2.
That is a pure, pure math result relating the factorization of twice a natural to the quadratic residues of ALL natural numbers, except 1, of course, less than that natural number.
In a way it's a bizarre result, as why in the hell is the addition of two numbers related to the factorization of twice the sum in any way at all?
Well, it may seem strange, but that's the mathematics. The mathematical logic of it is perfect but for creatures who think of addition as trivial, it's odd to find out that there is all this mathematical machinery relating all the freaking natural numbers below a given natural to the difference of factors of twice that natural.
Oh, goody, here's another pure, pure result in there, as why twice the natural?
Maybe it has something to do with primeness as if you have a prime, then with just that prime the only possible difference is p-1, looking at absolute value. But with twice it, you have p-2 and 2p-1, so maybe the math needs at least two possibilities for eveything to work right.
I'd like to think this result might have some practicality in determining primeness, but I think I'm reaching again. Freaking pure, pure math results.
I need something a lot less pure, and a lot mroe practical to break the stupid impasse with dumb mathematicians all over the world ignoring my research.
Maybe in a few years this one will get less pure, and more powerful, but until then, I guess the search for a back breaker against the mathematical community, will continue.
I will find it. And I will be madder than ever when I do.
The result, of course, as I like talking about it so much so you probably have seen it if you read my threads, is with natural numbers, n_1, n_2, C, and k, where
C = n_1 + n_2
and k is a difference of factors of 2C, it must be true that
(8C + k^2) mod n_1
is a square modulor n_1, and
(8C + k^2) mod n_2 is a square modulo n_2,
where I have written that differently in the past, as it also means that
(8n_2 + k^2) mod n_1 is a square modulo n_1, and vice versa,
which follows, of course, fomr C = n_1 + n_2.
That is a pure, pure math result relating the factorization of twice a natural to the quadratic residues of ALL natural numbers, except 1, of course, less than that natural number.
In a way it's a bizarre result, as why in the hell is the addition of two numbers related to the factorization of twice the sum in any way at all?
Well, it may seem strange, but that's the mathematics. The mathematical logic of it is perfect but for creatures who think of addition as trivial, it's odd to find out that there is all this mathematical machinery relating all the freaking natural numbers below a given natural to the difference of factors of twice that natural.
Oh, goody, here's another pure, pure result in there, as why twice the natural?
Maybe it has something to do with primeness as if you have a prime, then with just that prime the only possible difference is p-1, looking at absolute value. But with twice it, you have p-2 and 2p-1, so maybe the math needs at least two possibilities for eveything to work right.
I'd like to think this result might have some practicality in determining primeness, but I think I'm reaching again. Freaking pure, pure math results.
I need something a lot less pure, and a lot mroe practical to break the stupid impasse with dumb mathematicians all over the world ignoring my research.
Maybe in a few years this one will get less pure, and more powerful, but until then, I guess the search for a back breaker against the mathematical community, will continue.
I will find it. And I will be madder than ever when I do.
# posted by James Harris @ 11:11
Saturday, May 06, 2006
Another try, another factoring method
Looking over replies to the factoring idea of mine previous to this one, there is still the problem of too much complexity for most.
So it's yet another factoring idea of mine too complicated, but I have another which is simple enough that most of you should be able to understand it. Don't know how well it works though.
I have from my quadratic residue result that
8T + k^2 = r mod n
where T is the target to factor, k is a difference of factors of 2T, n is a natural number such that n<T, and r is a quadratic residue modulo n.
First you find 8T mod n, so let c = 8T mod n, and now you can just find k, such that
c + k^2 is a square,
as then you have a possible difference, as you have another quadratic residue.
Here's a demonstration factoring 65 = 5(13), using n = 9.
(It seems to me that you want an n that is greater than floor(T/8) for obvious reasons, and it has to be less than T.)
8(65) mod 9 = 7, and with c = 7, I have 7+ k^2, and k=3, witll work as that gives
7 + 9 = 16
and 13 - 2(5) = 3, so I found a difference. To check a difference you plug it into
sqrt(k^2 + 8T), and notice that sqrt(9 + 8(65)) = 23.
Then you have 23-3 = 20, giving you 5, with the gcd with 65, and 23+3 = 26, giving you 13.
If your k doesn't work with your c, you can add n to it, and try again.
That is, with
x = sqrt(k^2 + 8T), you have that the gcd of x+k with your target T, will give one factor, and the gcd of x - k with the target will give the other.
Of course you have to factor c to get all the k's, so it's what I call a surrogate factoring method.
With my other idea, I forced k=1, and then you go looking for this m, which may or may not work well, I don't know. Hell, I don't know if this idea works well either, but it may be less frustrating to play with, as there is a lot less tweaking involved.
You see, the problem with a lot of my ideas is that for most people they are too damn complicated to figure out how to get to work. This one should be simple enough though, so you can play with it, and if it does work well, I'll be surprised.
But at least you can get it to work. My ideas are so freaking complicated in their simplicity that the world gets befuddled. Maybe this one will be different.
So it's yet another factoring idea of mine too complicated, but I have another which is simple enough that most of you should be able to understand it. Don't know how well it works though.
I have from my quadratic residue result that
8T + k^2 = r mod n
where T is the target to factor, k is a difference of factors of 2T, n is a natural number such that n<T, and r is a quadratic residue modulo n.
First you find 8T mod n, so let c = 8T mod n, and now you can just find k, such that
c + k^2 is a square,
as then you have a possible difference, as you have another quadratic residue.
Here's a demonstration factoring 65 = 5(13), using n = 9.
(It seems to me that you want an n that is greater than floor(T/8) for obvious reasons, and it has to be less than T.)
8(65) mod 9 = 7, and with c = 7, I have 7+ k^2, and k=3, witll work as that gives
7 + 9 = 16
and 13 - 2(5) = 3, so I found a difference. To check a difference you plug it into
sqrt(k^2 + 8T), and notice that sqrt(9 + 8(65)) = 23.
Then you have 23-3 = 20, giving you 5, with the gcd with 65, and 23+3 = 26, giving you 13.
If your k doesn't work with your c, you can add n to it, and try again.
That is, with
x = sqrt(k^2 + 8T), you have that the gcd of x+k with your target T, will give one factor, and the gcd of x - k with the target will give the other.
Of course you have to factor c to get all the k's, so it's what I call a surrogate factoring method.
With my other idea, I forced k=1, and then you go looking for this m, which may or may not work well, I don't know. Hell, I don't know if this idea works well either, but it may be less frustrating to play with, as there is a lot less tweaking involved.
You see, the problem with a lot of my ideas is that for most people they are too damn complicated to figure out how to get to work. This one should be simple enough though, so you can play with it, and if it does work well, I'll be surprised.
But at least you can get it to work. My ideas are so freaking complicated in their simplicity that the world gets befuddled. Maybe this one will be different.
# posted by James Harris @ 06:05
Friday, May 05, 2006
JSH: Remarkable quadratic residue result
Given a quadratic residue r modulo n, where all are, of course, natural numbers it must be true that 4r is a quadratic residue modulo n.
I have a remarkable little proof.
That is a HUGE result. It is just such a HUGE result.
To check if a number is a quadratic residue then you can just multiply by 4. If it is, then 4 times it will be as well. e.g. 2*4 mod 7 = 1, and 2 is a square modulo 7.
Now notice that if you are not sure of the result, you can just keep multiplying by 4.
For instance, checking 2 modulo 17, I have 2(4) = 8, if you are not sure, multiply by 4 again, and you have 4(8) mod 17 = 15. If you are not sure, 4(15) mod 17 = 9, and as 9 is a square, it is definite that 2 is as well.
What a neat result. If r is a square modulo n, then 4r is a square modulo n.
All quadratic residues are connected then by 4, as they are all products of 4 times some other quadratic residue.
The proof is nifty, and not obvious.
I have a remarkable little proof.
That is a HUGE result. It is just such a HUGE result.
To check if a number is a quadratic residue then you can just multiply by 4. If it is, then 4 times it will be as well. e.g. 2*4 mod 7 = 1, and 2 is a square modulo 7.
Now notice that if you are not sure of the result, you can just keep multiplying by 4.
For instance, checking 2 modulo 17, I have 2(4) = 8, if you are not sure, multiply by 4 again, and you have 4(8) mod 17 = 15. If you are not sure, 4(15) mod 17 = 9, and as 9 is a square, it is definite that 2 is as well.
What a neat result. If r is a square modulo n, then 4r is a square modulo n.
All quadratic residues are connected then by 4, as they are all products of 4 times some other quadratic residue.
The proof is nifty, and not obvious.
# posted by James Harris @ 20:45
Thursday, May 04, 2006
My factoriing idea, theoretical issues
I noted that you can use the result that with naturals n_1, p a prime factor of n_1, n_2,
C = n_1 + n_2, and k a difference of factors of 2*C that
(8*C + k^2) mod n_1 must be a square modulo p
and, of course, that means that it is a square modulo n_1.
So I thought to myself, hey, why not use this result to try and factor?
So I arbitrarily picked k=1, C = mT, so I'd have
sqrt(1 + 8*mT)
where T is the target to be factored, and I solved for m as a congruence relation and found that
m = (r-1)(8*T)^{-1} mod n_1
where is a square modulo n_1, or as I like to say, is a quadratic residue of n_1.
Notice here you can pick n_1 to be ANY natural number you choose. I tossed out floor(T/2) as a possibility, as, well, it was just an idea of the moment.
I also have tended to just pick r=4, because just about any n_1 will have 4 as a quadratic residue, but you can pick something else.
What the mathematics tells you is that ALL solutions for m that will give you that quadratic residue, must fit into the congruence relationship given.
But how likely are you to find a solution?
I don't know. It is possible, amazingly enough, and possibly unfortunately, that as T increases and size, and as you pick an ever large n_1, close to T, that the number of possible m's increases rapidly, as in, faster than linear.
Or, they may drop rapidly. Or, they may stay roughly the same, while the quadratic residues increase linearly. Or the number of m's may increase at roughly the same rate as the number of quadratic residues.
So the theory is not such a big deal as it's all simple.
But the practicality is in question, where maybe, if you pick n_1 correctly, and go to bigger and bigger composites, um, you may find it easier to factor than with small numbers.
So little test examples may give the wrong idea, but I'm still in guess mode.
The problem, with T = p_1 p_2, where the p's are primes, can be considered to be finding the solution set
xp_1 - yp_2 = 1
where m = xy, and it looks to me like it might go as the square, in turns of possible m's so that while you increase T linearly, the number of m's that will work, increase as the square, which if true means we're screwed, and someone could figure out a way to factor with this thing, where it would actually work better with RSA sized numbers than with smaller numbers.
But I'm still just guessing and I'm not seeing posts that address the actual theoretical issues, so I think it's an open question.
Then again, I think I could be reaching, hoping a nice idea that can't be practical can be, and that instead the odds of landing on an m decrease so rapidly that they are vanishingly small with a large T, which I think is the opinion that most would think should apply.
After all, if that weren't true, then I couldn't just be chatting out a solution to the factoring problem on Usenet for DAYS, so it must not be true, right?
C = n_1 + n_2, and k a difference of factors of 2*C that
(8*C + k^2) mod n_1 must be a square modulo p
and, of course, that means that it is a square modulo n_1.
So I thought to myself, hey, why not use this result to try and factor?
So I arbitrarily picked k=1, C = mT, so I'd have
sqrt(1 + 8*mT)
where T is the target to be factored, and I solved for m as a congruence relation and found that
m = (r-1)(8*T)^{-1} mod n_1
where is a square modulo n_1, or as I like to say, is a quadratic residue of n_1.
Notice here you can pick n_1 to be ANY natural number you choose. I tossed out floor(T/2) as a possibility, as, well, it was just an idea of the moment.
I also have tended to just pick r=4, because just about any n_1 will have 4 as a quadratic residue, but you can pick something else.
What the mathematics tells you is that ALL solutions for m that will give you that quadratic residue, must fit into the congruence relationship given.
But how likely are you to find a solution?
I don't know. It is possible, amazingly enough, and possibly unfortunately, that as T increases and size, and as you pick an ever large n_1, close to T, that the number of possible m's increases rapidly, as in, faster than linear.
Or, they may drop rapidly. Or, they may stay roughly the same, while the quadratic residues increase linearly. Or the number of m's may increase at roughly the same rate as the number of quadratic residues.
So the theory is not such a big deal as it's all simple.
But the practicality is in question, where maybe, if you pick n_1 correctly, and go to bigger and bigger composites, um, you may find it easier to factor than with small numbers.
So little test examples may give the wrong idea, but I'm still in guess mode.
The problem, with T = p_1 p_2, where the p's are primes, can be considered to be finding the solution set
xp_1 - yp_2 = 1
where m = xy, and it looks to me like it might go as the square, in turns of possible m's so that while you increase T linearly, the number of m's that will work, increase as the square, which if true means we're screwed, and someone could figure out a way to factor with this thing, where it would actually work better with RSA sized numbers than with smaller numbers.
But I'm still just guessing and I'm not seeing posts that address the actual theoretical issues, so I think it's an open question.
Then again, I think I could be reaching, hoping a nice idea that can't be practical can be, and that instead the odds of landing on an m decrease so rapidly that they are vanishingly small with a large T, which I think is the opinion that most would think should apply.
After all, if that weren't true, then I couldn't just be chatting out a solution to the factoring problem on Usenet for DAYS, so it must not be true, right?
# posted by James Harris @ 20:49
My quadratic residue result may offer the route to a tiny proof of Fermat's Last Theorem.
The result is that given naturals n_1, p a prime factor of n_1, n_2, C = n_1 + n_2, and k a difference of factors of 2*C, it must be true that
(8*n_2 + k^2) is a square modulo p
so with x^n + y^n = z^n, you can let C = z^n, and k is a difference of factors and there is a set of all such possibles.
The thing is, n_1 and n_2 can be ANY naturals that add to give z^n.
It's trivial to show then that
(8*z^n + k^2) would have to be a square modulo p
for all p, primes less than z^n.
A simple check of combinations using f_1 f_2 = z, with n=3, gives an indication why that's impossible, as I think it actually gets blocked by 5, or if not 5, 7, which has only three quadratic residues.
(f_1^3 - f_2^3), (f_1^2 f_2 - f_1 f_2^2), (f_1^3 f_2^3 - 1)
are just some combinations I can think of immediately with n=3, and if FLT were false, for like p=7, it'd have to be true that all of those squared would add to 8*z^2 mod 7 to give a quadratic residue of 7, i.e. give 0, 1, 2 or 4.
This quadratic residue result of mine may be one of the most powerful in number theory, offering routes to proving Goldbach's conjecture, and probably all kinds of Diophantine equations.
A tiny proof of FLT that is just about as trivially easy as you can get. Kind of strange.
The result is that given naturals n_1, p a prime factor of n_1, n_2, C = n_1 + n_2, and k a difference of factors of 2*C, it must be true that
(8*n_2 + k^2) is a square modulo p
so with x^n + y^n = z^n, you can let C = z^n, and k is a difference of factors and there is a set of all such possibles.
The thing is, n_1 and n_2 can be ANY naturals that add to give z^n.
It's trivial to show then that
(8*z^n + k^2) would have to be a square modulo p
for all p, primes less than z^n.
A simple check of combinations using f_1 f_2 = z, with n=3, gives an indication why that's impossible, as I think it actually gets blocked by 5, or if not 5, 7, which has only three quadratic residues.
(f_1^3 - f_2^3), (f_1^2 f_2 - f_1 f_2^2), (f_1^3 f_2^3 - 1)
are just some combinations I can think of immediately with n=3, and if FLT were false, for like p=7, it'd have to be true that all of those squared would add to 8*z^2 mod 7 to give a quadratic residue of 7, i.e. give 0, 1, 2 or 4.
This quadratic residue result of mine may be one of the most powerful in number theory, offering routes to proving Goldbach's conjecture, and probably all kinds of Diophantine equations.
A tiny proof of FLT that is just about as trivially easy as you can get. Kind of strange.
# posted by James Harris @ 09:49
