Re: Proof of fraud, flawed number theory still taught



jstevh@xxxxxxx wrote:
Some years ago I figured out a new technique in algebraic analysis.
The techniques I developed were developed by me to try to prove
Fermat's Last Theorem and I didn't realize that I'd stumbled across
proof of this massive error in the number theory field, though as
people argued and argued with me, I figured that out and came to
understand just how massive it was.

The gist of it is that I've found a flaw that takes out just about the
entire field of modern number theory.

Since I claim to have the correct mathematics it's just common sense
that if I am right then my research results show what the flawed number
theory cannot.

Quite simply, with my mathematical ideas that actually work, I can make
predictions in number theory which are absolutely perfect, where the
flawed number theory is useless or wrong.

I did so years ago.  No one has shown me wrong, but it's easy to do so
if I am, so I thought I'd remind you of how easy it is, in case some of
you are suffering under the delusion that you are actually highly
intelligent and using correct mathematics, when in fact, you are part
of a group that is deliberately refusing to use correct mathematics.

My theories show that given

a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0

with non-zero non-unit integer f and non-zero integer x coprime to f,
it must be true that only two of the roots of the resulting cubic have
f as a factor, while one is coprime to f.

For example, with x=1 and f=7 you get the result

a^3 + 144a^2 - 110593 = 0

and that cubic's roots must follow my theory in that only two have 7 as
a factor while one is coprime to 7, though not in the ring of algebraic
integers.

That theory works without regard to whether or not the roots are
rational or irrational, which is why my ideas are easily testable, as
if I'm wrong, it's just a matter of finding an f and x where you get a
rational root and it doesn't fit with my predictions.

It needs to be a rational solution as it is easy to prove as I have
that it does not matter what is true in the ring of algebraic integers,
so claims that my work is refuted by irrationals that do not have f as
a factor in the ring of algebraic integers are specious.

That is, posters are then relying on the very flaw I've outlined to try
and attack my work, which is the kind of stupid crap that has worked
for years to my amazement.

So RATIONAL solutions are key here.

So you could use some math software, write a script and let your
computer crunch for a while finding f's and x's where you get a cubic
with a rational solution or more than one rational solution and see if
that solution has f as a factor or is coprime to f.

If in a single instance it is not, then I am wrong.

But I know what the math shows so you will not find that single
instance, but you may find people who will reject hundreds or hundreds
of thousands of cases proving I'm right because they themselves are the
problem--they don't care about what's true.

But for some of you, a hundred thousand or more cases showing I'm right
with none showing I'm wrong will mean something, I hope.

Remember, it only takes ONE case to show I'm wrong.

That way to refute my work has been around for years.  Posters shy away
from it.  People criticizing me make damn sure not to talk about it,
and people have lied about it, as it's a test they can't win.

After all, it's mathematics.  What's true is absolutely true.  Since I
am absolutely correct, going to an area where the math behaves
perfectly as I say it does is just a way to lose a social battle, where
people arguing with me, so far, have won, by keeping it social and
hiding from the truth.

I like sci.crypt for this post as you guys are a little more
no-nonsense and of course you do use number theory, a lot of which does
work, but if you have used some of the supposedly advanced number
theory and been baffled by it not working as you'd thought it should,
then now you can find out why--it's wrong.

The test is an easy one, but I'm sure posters will reply to this with
more social crap and distractions as they've done--quite
successfully--for years.

But maybe some of you will get out your math software and run the
scripts and post the truth.  And maybe, it will mean something to
somebody who actually cares about correct mathematics.

Math professors keep teaching the wrong ideas despite my having proven
them wrong years ago, in what I see as an expression of contempt for
their students and their society.  I keep wondering why, but all I can
see is that some people despise the truth--when it hurts their bottom
line.

So for a math professor, what price their student's minds?  Seems to be
a few years salary waiting until I forced the issue, which is rather
cheap to me.

But that's the choice they made.


James Harris


Being that this has nothing to do with cryptology how about taking sci.crypt off the list of recipients?
.




Relevant Pages

  • Re: Proof of fraud, flawed number theory still taught
    ... >> intelligent and using correct mathematics, when in fact, you are part ... >> It needs to be a rational solution as it is easy to prove as I have ... >> so claims that my work is refuted by irrationals that do not have f as ... >> So you could use some math software, write a script and let your ...
    (sci.crypt)
  • Re: All that work, for naught
    ... James Harris wrote: ... > Well, correct mathematics is correct mathematics - be it janitor, ... > The chief editor yanked my paper and emailed me afterwards, ... > That's the math world people. ...
    (sci.crypt)
  • All that work, for naught
    ... It was prime counting that got to me. ... Well, correct mathematics is correct mathematics - be it janitor, ... The chief editor yanked my paper and emailed me afterwards, ... That's the math world people. ...
    (sci.crypt)