Re: JSH: Why factoring solution must work

On Feb 10, 12:02 pm, rdec...@xxxxxxxxxxxx wrote:
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

It was always rationals. That's not a change.

Readers looking at the post that starts this thread will see it state
the field is rationals.

Quote: "In rationals--I'll explain more on that later--given..."

When I'm looking for integers I specify integers.

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.

That is a mathematically false statement.

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.

Nope. That is a mathematically false statement.

I DID do some work with these equation for my D=15 example, which
allows me to dismiss silly stuff immediately.

There are 4 solutions for x and y if you do it right. Since x = r/c,
you have 4 potential solutions if you don't throw away the plus or
minus willy-nilly.

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.

That is a mathematically false statement.

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?

Yet you claim to have solutions for the functions but now don't know
how to solve for a solution to one of them?

Take s(v)/c(v) = 1, or s(v)/c(v) = -1, and solve for v.

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

Oh come on, you didn't even try to do the math honestly.

I challenged you on plus or minus and you just tossed some absolute
value signs around your equations and claimed that fixed things.

It's like you're not even trying.

Weird. Like you are laying down on the tracks in front of the
oncoming train. That is what you are doing, isn't it?

So sad, and so bizarre.


James Harris
.

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