Re: JSH: Why factoring solution must work

On Feb 8, 7:37 pm, rdec...@xxxxxxxxxxxx wrote:
On Feb 8, 8:19 pm, JSH <jst...@xxxxxxxxx> wrote:
<snip>





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.

Tim is correct--my second assertion was flat-out wrong. However, the
first, that it will almost never be the case that you can find a v
for which you'll simultaneously have |r(v)+c(v)| < D and |r(v)-c(v)| <
D
is certainly true.



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.

The point is that for all interesting D, you won't be able to satisfy
your minima condition. Try it and see:

I wish I'd immediately thought of the best counter to that false
claim, which is to note that you can go the other way and solve for r
and c by non-trivially factoring some test composite D, and ALWAYS
find the minima conditions, as you will always find v. And it will be
rational.

So you are trivially wrong. The minima conditions MUST exist, as r
and c exist such that

(r-c)(r+c) = D

for a non-trivial factorization of D if it is a composite, so you can
solve for them, and find v.

So now let's consider your equation claims that follow.

1. Solve for x and y in terms of v and D (Hint: for your factors of
D-1,

   x = -(4(D + 1)v^2 -4(D - 1) + (D - 1)^2) / (D - 1)(4v^2 + 4v - (D -
1))

   y = (8v^2 - 4(D - 1)) / (D - 1)(4v^2 + 4v - (D - 1))

2. Compute r + c and r - c (Hint:

   r + c = -(4(D + 1)v^2 -4(D - 1) + (D - 1)^2) + (D - 1)(4v^2 + 4v -
(D - 1))
         = -2(2v - (D - 1))^2

   r - c = -(4(D + 1)v^2 -4(D - 1) + (D - 1)^2) - (D - 1)(4v^2 + 4v -
(D - 1))
         = -2Dv^2

3. Observe that unless you were very lucky with v, you'll see that r +
c
   will have no factors in common with D and all of D, _unfactored_
will
   be found in the r - c term.

Given the plus or minus in the original equations, your solutions
can't be right as you should have plus or minus in them as well.
Either you did the math wrong or selectively picked plus or minus at
each decision point to get what you wanted.

I think there should be at least 4 potential solutions for x and y
that emerge if you do the algebra right.

Further it is clear now that solving the factoring problem is simpler
than I realized before, as rather than bother finding minima, you can,
if you have solutions with y, where y = s(v)/c(v), simply solve for v
such that s(v) = 1 or -1.

Which guarantees:

(r+c)(r-c) = D

There are 4 potential solutions so I'd guess that for at least one
you'd have r+c or r-c = D or - D, which may be what you deliberately
found to post, but there will also be solutions which factor D.

You posted against a trivially easy solution to the factoring problem
by doing the math wrong.

It is infuriating how people like you quickly close the book in other
people's mind so that they believe something false.

Your duty now is to acknowledge your errors.


James Harris
.

Relevant Pages

  • Re: JSH: Why factoring solution must work
    ... demonstration factorization of 15. ... then you MUST non-trivially factor D if D is a composite, ... Mathematical absolute. ...
    (sci.math)
  • Re: JSH: Why factoring solution must work
    ... wouldn't suffice to generate a nontrivial factorization of D. ... The point is that for all interesting D, you won't be able to satisfy ... Is your reality only a hatred of mathematics? ...
    (sci.math)
  • Re: JSH: Why factoring solution must work
    ... wouldn't suffice to generate a nontrivial factorization of D. ... The point is that for all interesting D, you won't be able to satisfy ... your minima condition. ...
    (sci.math)