Re: JSH: Understanding the situation

On Mar 29, 4:59 pm, pumpledumplek...@xxxxxxxxx wrote:
On Mar 29, 1:49 pm, Tim Smith <reply_in_gr...@xxxxxxxxxxxxxxxx> wrote:



In article
<a4e91f95-3e94-485e-b2db-aeb16ddc2...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

 JSH <jst...@xxxxxxxxx> wrote:
That is, given x^2 - Dy^2 = 1, it is EASIER to find solutions by
instead using:

j^2 - Dk^2 = -1, j^2 - Dk^2 = -2, or j^2 - Dk^2 = 2

which are required to exist for D a prime, where only one will be
valid when D is an odd prime, but one must.

They are easier as the first j that solves is roughly the square root
of the first x that solves.

Interestingly, for all prime D < 10000, if you take the convergents of
the continued fraction for sqrt(D) to get your candidate x and y, and
compute x^2-Dy^2, you hit -1, 2, or -2 before you hit 1.

I haven't looked to see if there is some theoretical reason for that.  
Is there?

...

I can see from posts I'm still getting that some of you not only have
no clue about the reality of the situation, you don't realize what
you're revealing to readers who do.

The reality of the situation is that you are already running from your
result.  I see at least two requests for you to clarify by showing how
it works for a particular D, and you running away from those.

...

I am the world's latest major mathematical discoverer.

And you have the number one result for a Google search on

   flawed definition of mathematical proof

--
--Tim Smith

Hi Tim,

The reason that the -1, 2 or -2 show up before the 1 is that,
if we define

p_k^2- D q_k^2 = w_k

for p_k/q_k the kth convergent in the (simple) continued
fraction expansion to sqrt(D), then the w_k is periodic
(and the "end" of the period is when it hits 1). For all
but the smallest D, solutions to x^2 - D y^2 in {-1, 2, -2}
have (x,y)=(p_k,q_k) for some k.

What our little 21st century Newton has rediscovered is,
unsurprisingly, a special cases of an old result (which
goes back, at least, to Nagell in the 1940s) that, for
given square-free D > 1, precisely 2 equations of the shape

a x^2 - b y^2 = 2^d, d in { 0, 1}, ab = D

are soluble in integers....

de P

Regardless of the earlier result though is the direct connection to
Pell's Equation.

Now it is possible to solve for Pell's Equation more simply using the
alternates.

Oh yeah, do you know how to determine WHICH of the alternates will
work for a given prime D?

I do. The way is given in my general solution to binary quadratic
Diophantine equations.

What is happening here is that THIS result can't be talked down
because the literature doesn't reflect it, but some of you are still
trying which is important for me to shred funding for mathematicians
worldwide.

I repeat that so you understand what you are doing.

The game is over. The Math Wars are over.

All you can do now is help me put more of your fellow mathematicians
out of funding.

Thank you for your efforts in that regard.


James Harris
.

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