Re: JSH: One of the great relations?

JSH wrote :

On Sep 27, 4:43 pm, "Achava Nakhash, the Loving
Snake"
<ach...@xxxxxxxxxxx> wrote:
On Sep 27, 10:02 am, JSH <jst...@xxxxxxxxx> wrote:

I found it quite profound that now I know that in
the ring of
integers, given

x^2 + Dy^2 = F

I also know that

z^2 + D(x+y)^2 = F*(D+1)

This is interesting (to me at least).  It relates
the solutions of one
Diophantine equation to the solutions of another
Diophantine
equation.  I would find it easier to understand
what you are doing if
you explained what z was instead of leaving it to
the reader to figure
out.  As for its significance, see below.

which is the start of what I call a Diophantine
chain, where if the
original equation has an infinite number of
solutions for x and y,
then you have an infinite chain, but otherwise
you have a finite
chain.

Have you proved the above claim?

Conjectured.

One poster has already replied derisively giving
z explicitly in terms
of x, y and D which kind of puzzled me as it just
looks like such a
beautiful result to me.

Like with D=-2, you have:

x^2 - 2y^2 = 1, followed by z^2 - 2(x+y)^2 = -1

and the next in the series is

w^2 - 2(x+y+z)^2 = 1

so you just get this flipping back and forth, and
with one solution at
the start you can get the solutions that follow,
so with x=3, and y=2,
you have next that

z^2 - 2(5)^2 = -1, so z^2 = 49, so z=7, and then
you have

w^2 - 2(3+2+7)^2 = 1, so w^2 - 2(144) = 1, so w^2
= 289 and w=17.

And you can do that forever.

So in this particular case, you can generate
solutions to a related
Pell equation and maybe more solutions to the
original Pell equation.
You still have to show that you get different
solutions, although this
is probably not difficult.  If you played around
with the solutions of
x^2 - Dy^2 = 1 a great deal (here D is a positive
square-free
integer), you would find that the x's are the
values of a particular 2-
term linear recursion and the y's are values of
exactly the same
linear recursion.  This would give you another way
of generating new
solutions from old.  At some point the mathematical
world realized
that representing a solution (x, y) as (x +
y*sqrt(D)) gave you the
really fascinating discovery that the solutions in
this form were
closed under multiplication, and that you could
generate all solutions
as powers of the solution with the smallest value
that is still larger
than 1.  At some point along the way, it was
demonstrated that every D
(remember positive and square-free) had solutions.
 The mathematical
world was then poised to make siginificant
generalizations.  By the
way, the solutions to x^2 - Dy^2 = F, represented
in the same way as
above, can be given more solutions by multiplying
them by the
solutions to the equation with F = 1.  I am leaving
out the role
played by letting F = 4, -1, and -4 for some values
of D, but the
ideas are pretty much the same.



I play with equations now like

7^2 - 5(3)^2 = 4

so I know that

z^2 - 5(10)^2 = 4*(-4)

and can get z then is 22, and notice that 4
divides off so I have

11^2 - 5(5)^2 = -4, so

w^2 - 5(16)^2 = 16, so w=36, and you can divide
off 16, and get

9^2 - 5(4)^2 = 1.

So is it one of the great relations?

Sorry.  Not even close.

Really? What if I told you that it's fairly easy to
give a general
method now for SOLVING

x^2 - Dy^2 = F

using it?

One that just requires the Chinese Remainder Theorem?

x^2 + Dy^2 = F

requires that

z^2 + D(x+y)^2 = F*(D+1).

Or am I just in love with one more of my
discoveries.  More of my
math?

Precisely.  It is the scratching of the surface.
 It is a nice
relationship of the sort that can be the starting
point of some
research but no more.  An experienced mathematician
is familiar with
work of incredibly greater depth and with what has
been learned about
this problem and similar problems.  Thus he is not
going to be wowed
by a technically simple and quite unoriginalll,
meaning that this kind

Yeah I suggest to you that math people HATE simple,
which is where I
think you finally got something right.

More complexity allows pedantic behavior like you
displayed in your
reply though I wonder if the solution question
knocked any of that
proud arrogance off of you?



My analysis is that mathematicians crave complexity
because difficult
explanations can fuel more research and keep more
jobs, for
mathematicians.

I call it white collar welfare.

i partially agree , but must not that complexity must grow by time to avoid repetition of old math.

i.e. sometimes complexity is necc for progress.



of manipulation has been used gazillions of times
before and so the
method itself is not anything new.  Not to say that
there is anything
wrong with using old methods to solve new problems.
 You just get more
credit when you use original methods.

Like using a technique that I call tautological
spaces where I use
expressions like

x+y+vz = 0(mod x+y+vz)?

i have used 0 mod ' something ' quite often myself.

i like it , though i dont know if you use it in the same manner as i do.

it sure seems underestimated doing 0 mod ' something ' , ill grant you that.



So that I can have an additional degree of freedom so
that v is a free
variable?

And subtract equations to be analyzed from identities
in that way so
that I can probe them by adjusting my free variable?

Yet you dismiss my research as trivial and
pontificate about all this
research I'm supposed to read done by old geezers
long dead who don't
have work that interests me.

I think people like you are more like antique
collectors than
discoverers.

It's not the knowledge you crave but the SOCIAL stuff
around the
knowledge, like trying to look smart before an
audience by babbling
about the obsolete work of dead people who used to do
math.

thats trolling ...




James Harris

regards

tommy1729
.

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