Re: Factoring, sf, and reasonable requests

jstevh_at_msn.com
Date: 02/19/05 

Date: 19 Feb 2005 04:42:14 -0800

Tim Peters wrote:
> [JSH]
> [...]
> >>> Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
> >>>
> >>> Az= Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
> >>>
> >>> where T = M^2 - j^2
> >>>
> >>> Here you have a two equations defining rationals Ax and Az, where
M
> >>> is the number to be factored and j is an integer you pick to try
and
> >>> factor it.
> >>>
> >>> I noted that by working backwards--using an M for which you have
> >>> the factorization--you can see that for an integer Az that
factors
> >>> M, the rational Ax has a denominator that has the same prime
factors
> >>> as T.
>
> [Rick Decker, tries to make sense of that]
> >> Not in all cases. Let's take an extremely simple example.
> >>
> >> M = 15, j = 2. Thus T = 221 = 13*17. Now pick integers that
factor
> >> M and consider D = sqrt((Az - 2M^2)^2 - 4TM^2)):
> >>
> >> Az = 1. D = sqrt(2701), which is not rational.
> >>
> >> Az = 3. D = sqrt(3^2 * 101), which is not rational.
> >>
> >> Az = 5. D = sqrt(5^2 * (-35)), which isn't even real.
> >>
> >> Az = 15. D = sqrt(15^2 * (-883)), which also isn't real.
> >>
> >> So since none of the possible Az values yield a rational Ax,
> >> your assertion about the denominator of Ax is meaningless.
>
> [JSH, pulls a surprise out of his hat]
> > An integer Az that factors M, is one which has a *single* prime
factor
> > of M.
> >
> > I did not mean an integer Az that is a factor of M.
> >
> > Given an Az with a single prime factor of M, you can get that
factor by
> > using a gcd, and that's what I meant.
> >
> > Ok, so there's no debate about saying what you mean, if it was
> > ambiguous, sorry, but I don't mean that Az is a factor of M.
> >
> > Now go back, use Az's that have a single prime factor of M that do
> > work, and see what happens.
>
> OK, how about Az = 1120 = 2^5*5*7? Then gcd(Az, M) = gcd(1120, 15) =
5. Is
> that of the proper form?
>
> I'm guessing it is. Then (sorry, I'm switching to using "**" for
> exponentiation, so I can paste these equations directly into a Python
shell
> for evaluation):
>
> D = sqrt((Az - 2*M**2)**2 - 4*T*M**2) =
> sqrt((1120 - 2*15**2)**2 - 4*221*15**2) =
> sqrt((1120 - 450)**2 - 198900) =
> sqrt(448900 - 198900) =
> sqrt(250000) =
> 500
>
> That looks rational to me <wink>.
>
> Now I guess that by "denominator" in "Ax has a denominator" you mean
the
> (2j^2 - 2Az) in
>
> Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
>
> Is that right? Or do you mean something else? Assuming that is
right,
>
> 2*j**2 - 2*Az =
> 2*2**2 - 2*1120 =
> 8 - 2240 =
> -2232 =
> - 2**3 * 3**2 * 31
>
> That doesn't have any factors in common with T = 221 = 13*17. But
maybe I
> don't know what "has a denominator that has the same prime factors as
T"
> means.
>

My thesis was that the denominator would have the same prime factors as
T, and it's not an ambiguous statement.

So the denominator didn't. Interesting. That thesis was wrong.

Now then, do you suppose that denominator's factors are *random* or
dependent in some way on mathematical rules?

If they are not random, then do you suppose some way might be found to
figure out how to determine them?

If that way does not itself depend on knowing the prime factors of M,
why would that not be a potential major threat to Internet security?

Now then, for some of you the exchange should be surprising as you're
looking a this as black and white:

a: I have to get everything right or lose

b: If I'm wrong anywhere that proves my ideas are worthless

If you believe that, fine, I don't care, but I thought it worth
pointing out how simplistically some of you may be thinking.

James Harris

Relevant Pages

  • Re: Factoring, sf, and reasonable requests
    ... Tim Peters wrote: ... use Az's that have a single prime factor of M that do ... My thesis was that the denominator would have the same prime factors as ... and it's not an ambiguous statement. ...
    (sci.math)
  • Re: Factoring, sf, and reasonable requests
    ... >> your assertion about the denominator of Ax is meaningless. ... [JSH, pulls a surprise out of his hat] ... > Given an Az with a single prime factor of M, you can get that factor by ... rational the denominator of Ax _never_ has a factor in common with T? ...
    (sci.math)
  • Re: Factoring, sf, and reasonable requests
    ... >> your assertion about the denominator of Ax is meaningless. ... [JSH, pulls a surprise out of his hat] ... > Given an Az with a single prime factor of M, you can get that factor by ... rational the denominator of Ax _never_ has a factor in common with T? ...
    (sci.crypt)
  • Re: (Not quite) Cantors diagonal proof
    ... >> Please read the replies to you made by Tim Peters, Dave Seaman, Daniel ... > Method fails on rational numbers. ... Yes it fails because the rationals are countable, ... Given any number with repeating decimal, we can calculate the integer numerator and denominator. ...
    (sci.logic)
  • Re: (Not quite) Cantors diagonal proof
    ... >> Please read the replies to you made by Tim Peters, Dave Seaman, Daniel ... > Method fails on rational numbers. ... Yes it fails because the rationals are countable, ... Given any number with repeating decimal, we can calculate the integer numerator and denominator. ...
    (sci.math)