Re: JSH: Why factoring solution must work

On Feb 8, 4:06 pm, Tim Smith <reply_in_gr...@xxxxxxxxxxxxxxxx> wrote:
In article
<90173fa6-787a-4973-b630-278667401...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,



 rdec...@xxxxxxxxxxxx wrote:
That's a weird assumption since my original post notes that given

(r(v) + c(v))(r(v) - c(v)) = D(s(v))^2

where all are integer functions of v, that it must be true that if

abs(r(v) + c(v)) < D and abs(r(v) - c(v)) < D

then you MUST non-trivially factor D if D is a composite, which is a
mathematical absolute!

Mathematical absolute.  Total perfection.  Inviolate.  Unchangeable.

Absolute truth.

Wrong, for two reasons. First, finding v such that both

abs(r(v) + c(v)) < D and abs(r(v) - c(v)) < D will not in general

be possible. Second, it doesn't matter, since even if D were small
enough that the two conditions were simultaneously satisfied, it
wouldn't suffice to generate a nontrivial factorization of D.

Wait a second--I don't see how, if those conditions are met, it can fail
to generate a non-trivial factorization.  In general, let N, M, D, C be
positive integers, with

   NM = DC

You are correct, Professor Rick Decker of Hamilton College--since one
poster seemed to wish for me to add more--is wrong.

If (N,D) = (M,D) = 1, then NM = C, and D = 1.  If D > 1, then at least
one of (N,D), (M,D) must be > 1.  If N < D and M < D, then one of (N,D),
(M,D) must be a non-trivial factor of D.

--
--Tim Smith

Yup. I like simple. Quite simply if the minima condition is met and
D is a composite then it absolutely MUST be factored non-trivially.

That brings the argument to the question of coverage: are you covering
every possible solution in rationals to

x^2 - Dy^2 = 1

with these equations?

Otherwise, as I've repeatedly stated, it's a calculus problem.

Which would mean I'd turned the factoring problem into a calculus
problem of finding minima.

A mathematical absolute. There is no rational debate on that point.
And I don't care what college some professor is at, as mathematics
doesn't give a damn either. Your position means nothing. Human stuff
means nothing.

Human needs mean nothing. Mathematics is perfection.

It is mathematically a perfect argument, so now, possibly, posters can
move on to real outstanding issues.


James Harris
.

Relevant Pages

  • Re: JSH: Trivially easy math
    ... > compelled to pick ONE WAY to show the factorization, ... > thinking that functions can force constants, ... > The real story here is not difficulty in understanding the mathematics. ... S= set of all possible polynomials of x with coefficients in S ...
    (sci.math)
  • Re: Galileo (NOT Einstein) is inventor of Second postulate of Relativity
    ... Polynomial factorization can be considered to be one of the more boring areas of modern mathematics because so much of what most people think of in factoring polynomials was figured out centuries ago. ... You, Sharma, are a bottom feeder. ...
    (sci.math)
  • JSH: Playing with infinities
    ... It just so happens that the mathematics works out that the non-trivial ... complete in that it must give a factorization of M. ... That set of rational solutions for y represents the intersection of two ... supposedly mathematical proof is what matters to you. ...
    (sci.math)
  • Re: Playing with infinities
    ... > It just so happens that the mathematics works out that the non-trivial ... > complete in that it must give a factorization of M. ... > mathematical proof, then you are complete frauds, and must be, as ... > supposedly mathematical proof is what matters to you. ...
    (sci.math)
  • Re: JSH: Playing with infinities
    ... > complete in that it must give a factorization of M. ... > mathematical proof, then you are complete frauds, and must be, as ... > supposedly mathematical proof is what matters to you. ... then you'll understand that mathematics is ...
    (sci.math)
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