Re: JSH: Binary quadratic Diophantines

On Sep 20, 12:19 pm, Enrico <ungerne...@xxxxxxx> wrote:
On Sep 19, 7:01 pm, MichaelW <ms...@xxxxxxxxxx> wrote:



On Sep 20, 8:04 am, JSH <jst...@xxxxxxxxx> wrote:

On Sep 18, 11:53 pm, MichaelW <ms...@xxxxxxxxxx> wrote:

On Fri, 18 Sep 2009 18:00:29 -0700, JSH wrote:
On Sep 17, 8:10 pm, MichaelW <ms...@xxxxxxxxxx> wrote:
On Sep 17, 10:35 am, JSH <jst...@xxxxxxxxx> wrote:

The s is actually I think x-y, but I usually just give it as s, as
it's still just another unknown.

The value of s varies depending upon the original equation and is of
the form ax+by+c where (a,b,c) are some integers. For example taking a

Thanks for that as it explains why I left it as s.  It has been a while
since I derived the expression, and I'd forgotten.

reverse Pell

x^2 - 5y^2 = -1                   (1)

gives A=20, B=0 and C=16 which leads to (after some dividing through of
gcd)

5u^2 + 4 = v^2                    (2)

where u = (x+y) and v = s = (x+5y), so in this case (a,b,c) =(1,5,0).

If we were solving these equations we can therefore say that for some
solution (u,v) of (2) then y=(v-u)/4 and x=(5u-v)/4 is a solution of
(1).

One solution of (2) is (-1,3) giving x=-2 and y=1 which does indeed
solve (1).
Another solution of (2) is (0,2) giving x=-1/2 and y=1/2 which are not
integer solutions (although they do solve the equation).

That's the x=-y case.

Yeah, to be sure you get integers additional mathematical expertise is
required.

If I remember correctly though, there is a way to do it.

A further solution of (2) is (21,47) giving x=29/2 and y=13/2 which are
also not integer solutions.

The full expression is:

(2A(x+y) - B)^2 - 4As^2 = B^2 - 4AC

(where A, B and C are given in my original post)

so 2A is a potential denominator.

So you can consider solutions mod 2A, so as I said, there is modular
arithmetic to figure out the integer case.

Regards, Michael W.

The quick thing is finding out if rational solutions exist.

Finding out if integer solutions exist takes just a little bit more
expertise in handling the various residues.

Actually to be a complete and general solution the question is *how many*
integer solutions exist.

And you can answer that with modular arithmetic.  I know you've said
you're a computer programmer and not a mathematician, can you at least
consider the possibility that what you believe is a major issue, is
not?

I don't remember why I didn't handle that fully last year when I derived
the thing, but am not terribly interested now as I think it's fairly
easy mathematics.

I have stated that the algorithm as presented is not a complete and
general means of finding all integer solutions to a binary quadratic
Diophantine equation. Does you previous statement mean that you agree?

I said that's fairly easy mathematics.

Clearly you believe differently.

The main thing is that the identity works.  And it IS new, against
something you said earlier in this thread.

You are making me feel like I am trying to convince someone that their
baby is ugly. Okay, how about this. The first step of your algorithm

So are you saying it doesn't work or it's not new?

Looks below like your claim is it's not new.  I address that issue
below.

I'm not the one trying to cast doubt here, you are.  I'm showing
physics people some easy algebra that they can very quickly test to
see that it does what is claimed.

To me that's simple: facts forward.

You are disparaging the content, but anyone who checks can see that it
works, so why am I supposed to be the one making the major effort
here?

takes the general form and translates it into a form that in terms of
analytic geometry is centred at (0,0) and symmetrical about the x and y
axes. Now I remember learning this technique in high school so *in
general* I was saying that it is not new. Can you agree on that at least?

And in general doing algebra is solving for an unknown variable like
x.

In physics there is the difference between describing what you wish to
do, and actually showing how you do it, which is in other areas as
well.

Can you GIVE the mathematics which does what you said above?

Reality test.

I've given a method in my original post that starts this thread.

And hey, I deleted out the rest of your post and in part of it you
whine about what you see as personality attacks, but you keep making
these false statements, and sorry, but calling you on b.s. is not a
personality attack.

Just tell the truth.  Mathematics that is as simple as what I show in
my original post is kind of hard to make false statements about, if
you hadn't noticed.

If I'm wrong, give the mathematics which backs up your claim, and I
quote:

"Now I remember learning this technique in high school..."

Give the math.  I've already given my math.  And I emphasize readers
can CHECK me on that simply by reading my original post.

James Harris

James,

You should be ashamed of your dishonesty. I have clearly stated
several times that the first part that translates to the simplified
form is not the problem but the second part which claims to take the
simplified form and find solutions. I have even run the first part and
and demonstrated that it works. I have never said any different. In
addition I gave two links where analytic geometry is used to transform
ellipses and hyperbolas to achieve the same effect that you have. You
did little but glance at them and say that you could not see the
similarity.

You did the same thing to Tim Smith last year. He also noted the two
part nature of the algorithm and asked questions about the second part
and you carried on as if the first part was the part in question.

If you cannot make the second part work then just say so. I don't
think you can and the fact that you keep coming up with these
distractions only reinforces the belief.

You want me to tell the truth? Here is the truth. If I have
x^2-5y^2=-4 then your algorithm will fail to find all solutions. No
one can use your algorithm to find solutions. Not you, not me, no one.

This is what you are going to do now. You are going to make some sad
excuse why you won't take up the challenge. You will appeal to the
readers to agree with you. No one will. You will think that you have
answered me, but you won't. There is a high chance you will make a
long post entirely missing the point and ignoring the challenge
entirely and carry on about how this is all some conspiracy. But what
you won't do is face reality and actually do the hard work of checking
your algorithm. If this is the sort of response you are planning then
I suggest you don't bother as I have heard it all before.

Regards, Michael W.- Hide quoted text -

- Show quoted text -

======================================================

You want me to tell the truth? Here is the truth. If I have
x^2-5y^2=-4 then your algorithm will fail to find all solutions. No
one can use your algorithm to find solutions. Not you, not me, no one.

That's strange - I ran x^2-5y^2=-4 thru my spead*** implimentation
of
James' method and it works fine.

I suspect that the equations have gotten changed over time.
My equations are:

c1*x^2 + c2*x*y + c3*y^2 + c4 +c5*x + c6*y = 0

A = (c2 - 2c1)^2 + 4c1(c2 - c1 - c3)
B = 2(c2 - 2c1)(c6 - c5) + 4c5(c2 - c1 - c3)
C = (c6 - c5)^2 + 4c4(c2 - c1 - c3).

S^2 = A*(x + y)^2 + B*(x+y) + C     where S is any square

The square result gives you the value of x+y, I had to try values to
get
x, y pairs that satisfiy c1*x^2 + c2*x*y + c3*y^2 + c4 +c5*x + c6*y =
0

(I checked the Alpern site first to see if solutions existed, copied
the
first few and the general form.)

JSH's method worked fine up to x=29, y=13 and
also gave no false indications of solution existing for
values of x+y. I tried some more solutions generated
by the general form on the Alpern page - worked fine.

Check these against the equations in the first post of this thread ...
they look different.

                                                 Enrico

That's the first part, which as I say I never disputed. I get

S^2 = 20*(x+y)^2+64

Did this match your spread***? The solution (x,y) values do indeed
work, which matches my own coding.

I have even said that the way it is done here is actually a pretty
good representation of a known technique of approaching this type of
problem. Of course in this particular case the original equation is
just as easy to solve using a searching method as the derived one but
when you have the co-efficients for xy, x and y non-zero then the
first step of the algorithm is valid and useful.

I am running from the algorithm as published on the blog and contained
in a file in the Google group. This algorithm claims that having
obtained the new equation there is a process using modular arithmetic
and cyclically generating new equations that allows the extraction of
all solutions. Over a number of threads I have simply asked that the
author answer some questions about how the second part works since as
presented it is unclear. He keeps focusing on the first part; if I
don't agree that this is some new and brilliant technique then nothing
else I have to say seems to matter. I offered a "let's agree to
disagree" compromise but this just got more abuse in return.

Thanks for the response. Did your work on the algorithm deal with the
second step?

Regards, Michael W.
.

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