Still there is more as it indicates a route to finding a general solution for all 2 variable Diophantine equations using what I now call quadratic chains, which are infinite chains of related Diophantine equations.

To derive the full theory I will use

c

_{1}x

^{2}+ c

_{2}xy + c

_{3}y

^{2}= c

_{4}z

^{2}+ c

_{5}zx + c

_{6}zy

with z=1, so I have

((c

_{2}- 2c

_{1})

^{2}+ 4c

_{1}(c

_{2}- c

_{1}- c

_{3}))v

^{2}+ (2(c

_{2}- 2c

_{1})(c

_{6}- c

_{5}) + 4c

_{5}(c

_{2}- c

_{1}- c

_{3}))v + (c

_{6}- c

_{5})

^{2}- 4c

_{4}(c

_{2}- c

_{1}- c

_{3}) = n

^{2}mod p

for some n, where p is any prime coprime to z for a given solution, when

v = -(x+y)z

^{-1}mod p,

like before but because z=1, I can immediately substitute and generalize to all primes as I did in my previous post with Pell's Equation to get

((c

_{2}- 2c

_{1})

^{2}+ 4c

_{1}(c

_{2}- c

_{1}- c

_{3}))(x+y)

^{2}- (2(c

_{2}- 2c

_{1})(c

_{6}- c

_{5}) + 4c

_{5}(c

_{2}- c

_{1}- c

_{3}))(x+y) + (c

_{6}- c

_{5})

^{2}- 4c

_{4}(c

_{2}- c

_{1}- c

_{3}) = S

^{2}

for some integer S, and to simplify doing the next calculations let

A = (c

_{2}- 2c

_{1})

^{2}+ 4c

_{1}(c

_{2}- c

_{1}- c

_{3})

B = 2(c

_{2}- 2c

_{1})(c

_{6}- c

_{5}) + 4c

_{5}(c

_{2}- c

_{1}- c

_{3})

and

C = (c

_{6}- c

_{5})

^{2}- 4c

_{4}(c

_{2}- c

_{1}- c

_{3})

so I have

A(x+y)

^{2}- B(x+y) + C = S

^{2}

and it's immediately clear that I just have another quadratic Diophantine equation!

Now manipulating to complete the square gives

(A(x+y) - B/2)

^{2}+ AC - (B/2)

^{2}= AS

^{2}

which is

(2A(x+y) - B)

^{2}+ 4AC - B

^{2}= 4AS

^{2}

and I have then the new quadratic Diophantine:

(2A(x+y) - B)

^{2}- 4AS

^{2}= B

^{2}- 4AC

and have the stunning result that every quadratic Diophantine in two variables is connected to a quadratic Diophantine of the form:

u

^{2}- Dv

^{2}= F

and I get an existence condition as

(2A(x+y) - B)

^{2}= 4AS

^{2}mod (B

^{2}- 4AC)

so I have that A must be a quadratic residue modulo (B

^{2}- 4AC) if A is coprime to B.

Also I have that if solutions exist and A is negative, then there are only a finite number of solutions.

But further

(2A(x+y) - B)

^{2}- 4AS

^{2}= B

^{2}- 4AC

is another Diophantine equation so I can use it to get yet another equation of the same form.

And if one has rational solutions the next must have solutions, all the way out to infinity if the series goes on forever.

Next I note that if B does not equal 0 using the starting equation, with the next member of the chain you have that

c

_{1}=1, c

_{2}=0, c

_{5}= 0, and c

_{6}= 0, so I have

A = - 4c

_{3}, B=0, C = 4c

_{4}(1 + c

_{3})

so I have

(2c

_{3}(x+y))

^{2}+ c

_{3}S

^{2}= 4c

_{3}c

_{4}(1 + c

_{3})

which is

4c

_{3}(x+y)

^{2}+ S

^{2}= 4c

_{4}(1 + c

_{3})

which is

(S/2)

^{2}+ c

_{3}(x+y)

^{2}= c

_{4}(1 + c

_{3})

so for the next iteration I'd have

4c

_{3}(S/2+x+y)

^{2}+ T

^{2}= 4c

_{4}(1 + c

_{3})

^{2}

and I can again re-group and divide off 4 from both sides to have

(T/2)

^{2}+ c

_{3}(S/2+x+y)

^{2}= c

_{4}(1 + c

_{3})

^{2}

and see the emergence of a pattern where c

_{3}remains the same throughout.

Intriguingly some prior number theory can now be explained as in general after the initial equation, following equations in the chain look like

u

^{2}+ c

_{3}v

^{2}= c

_{4}(1 + c

_{3})

^{j}

where j can be 0 or a natural number

So with c

_{3}=-2, you just have

u

^{2}- 2v

^{2}= c

_{4}(-1)

^{j}

and the remarkable result that you have only two primary forms.

The only simpler case is with c

_{3}=-1, when you just get

u

^{2}= v

^{2}.

So if A is negative and B

^{2}- 4AC is 1, -1, 2 or -2--where the quadratic residue requirement is of no use--I can simply go down the chain until it applies and may have a solution.

However, there has to be an additional constraint along with the initial quadratic residue constraint, as consider

x

^{2}- 10y

^{2}= 3

as it has no solutions; although 10 mod 3 = 1, so 10 is a quadratic residue modulo 3.

Going back to the general case

u

^{2}+ c

_{3}v

^{2}= c

_{4}(1 + c

_{3})

^{j}

I have also that -c

_{3}must be a quadratic residue modulo c

_{4}(1+c

_{3}), but I also importantly have that

c

_{4}(1 + c

_{3})

^{j}

must be a quadratic residue modulo c

_{3}as long as c

_{4}does not have any square factors so that dual requirement provides the full existence conditions, and the dual residue requirement can be a means to a solution.

Author's note: Square factors are a big deal which have to be addressed more fully than I first hoped. If x^{2}+ Dy^{2}= n^{2}then -D may not be a quadratic residue modulo n^{2}, but dividing n^{2}from both sides gives

(x/n)^{2}+ D(y/n)^{2}= 1

which has any solutions to u^{2}+ Dv^{2}= 1, so x=nu, and y=nv.

In general with u^{2}+ Dv^{2}= F, if F = Gn^{2}, and -D is not a quadratic residue modulo F, then check modulo G. 10/19/08

For instance, if v

^{2}=1 mod c

_{4}(1 + c

_{3})

^{j}from my little congruence result:

With c

_{3}coprime to c

_{4}(1 + c

_{3}), if u

^{2}= r

_{1}mod c

_{3}and u

^{2}= r

_{2}mod c

_{4}(1 + c

_{3})

^{j}, you can find u

^{2}mod c

_{3}c

_{4}(1 + c

_{3})

^{j}with

u

^{2}= r

_{1}+ kc

_{3}mod c

_{3}c

_{4}(1 + c

_{3})

^{j}

where k = (r

_{2}- r

_{1})c

_{3}

^{-1}mod c

_{4}(1 + c

_{3})

^{j}

where

r

_{1}= c

_{4}(1 + c

_{3})

^{j}mod c

_{3}and

r

_{2}= -c

_{3}mod c

_{4}(1 + c

_{3})

^{j}

and you find k, such that r

_{1}+ kc

_{3}is a perfect square, where you increment up from j=0.

Picking v may seem suspect but if there are an infinity of solutions then unless particular residues are excluded for some unknown reason there would be solutions for any residue, but if there is a problem finding solutions with v

^{2}=1 mod c

_{4}(1 + c

_{3})

^{j}, you can search with other values using

r

_{2}= -c

_{3}v

^{2}mod c

_{4}(1 + c

_{3})

^{j}.

So if no residues are excluded that is the general theory for finding solutions.

Not bad for a theorem just recently discovered, and it shows the incredible power of what I call tautological spaces.

They have the power to simplify.

James Harris