Re: JSH: Why factoring solution must work

On Feb 10, 3:45 am, JSH <jst...@xxxxxxxxx> wrote:
On Feb 8, 7:37 pm, rdec...@xxxxxxxxxxxx wrote:

<snip>

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.

I'll grant you this, though you realize that this changes your
search space for v to the rationals. Now it's an even worse
problem than you had when your algorithm (unbeknownst to
you) was trial division over the integers. Now it's
trial division over the rationals. Ugh.

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.

Do the math--changing from j = (x+Dy-1)/D to j = (x+Dy+1)/D
merely changes the signs of x and y and hence has no effect
on the final result, namely that

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

|r - c| = 8D v^2

All that changing the sign in j does is swap the right
hand sides of the two results above.

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

Sure, plus or minus either. No substantial changes.

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.

Heh. How do you propose to do that?

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.

Actually, the math is right. The problem is with your unwillingness
to check it yourself. I completely understand your reluctance is
caused by your belief that you have a groundbreaking factoring
algorithm, but the fact is that while you do indeed have a
factoring algorithm, it is for all practical purposes nothing
but trying numbers until you find a factor of D.


Regards,

Rick
.

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