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Friday, September 26, 2008

Solvability and Diophantine Quadratic Chains

Deciding to take my newly discovered Quadratic Diophantine Theorem for a spin against "Pell's Equation" turned out to be a good idea as besides letting me validate that I had derived the theorem correctly, it also showed me that the result I had didn't simply lead in a BFC--Big Freaking Circle.

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

c1x2 + c2xy + c3y2 = c4z2 + c5zx + c6zy

with z=1, so I have

((c2 - 2c1)2 + 4c1(c2 - c1 - c3))v2 + (2(c2 - 2c1)(c6 - c5) + 4c5(c2 - c1 - c3))v + (c6 - c5)2 - 4c4(c2 - c1 - c3) = n2 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

((c2 - 2c1)2 + 4c1(c2 - c1 - c3))(x+y)2 - (2(c2 - 2c1)(c6 - c5) + 4c5(c2 - c1 - c3))(x+y) + (c6 - c5)2 - 4c4(c2 - c1 - c3) = S2

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

A = (c2 - 2c1)2 + 4c1(c2 - c1 - c3)

B = 2(c2 - 2c1)(c6 - c5) + 4c5(c2 - c1 - c3)

and

C = (c6 - c5)2 - 4c4(c2 - c1 - c3)

so I have

A(x+y)2 - B(x+y) + C = S2

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 = AS2

which is

(2A(x+y) - B)2 + 4AC - B2 = 4AS2

and I have then the new quadratic Diophantine:

(2A(x+y) - B)2 - 4AS2 = B2 - 4AC

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

u2 - Dv2 = F

and I get an existence condition as

(2A(x+y) - B)2 = 4AS2 mod (B2 - 4AC)

so I have that A must be a quadratic residue modulo (B2 - 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 - 4AS2 = B2 - 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

c1=1, c2=0, c5 = 0, and c6 = 0, so I have

A = - 4c3, B=0, C = 4c4(1 + c3)

so I have

(2c3(x+y))2 + c3S2 = 4c3c4(1 + c3)

which is

4c3(x+y)2 + S2 = 4c4(1 + c3)

which is

(S/2)2 + c3(x+y)2 = c4(1 + c3)

so for the next iteration I'd have

4c3(S/2+x+y)2 + T2 = 4c4(1 + c3)2

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

(T/2)2 + c3(S/2+x+y)2 = c4(1 + c3)2

and see the emergence of a pattern where c3 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

u2 + c3v2 = c4(1 + c3)j

where j can be 0 or a natural number

So with c3=-2, you just have

u2 - 2v2 = c4(-1)j

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

The only simpler case is with c3=-1, when you just get

u2 = v2.

So if A is negative and B2 - 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

x2 - 10y2 = 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

u2 + c3v2 = c4(1 + c3)j

I have also that -c3 must be a quadratic residue modulo c4(1+c3), but I also importantly have that

c4(1 + c3)j

must be a quadratic residue modulo c3 as long as c4 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 x2 + Dy2 = n2 then -D may not be a quadratic residue modulo n2, but dividing n2 from both sides gives

(x/n)2 + D(y/n)2 = 1

which has any solutions to u2 + Dv2 = 1, so x=nu, and y=nv.

In general with u2 + Dv2 = F, if F = Gn2, and -D is not a quadratic residue modulo F, then check modulo G. 10/19/08


For instance, if v2=1 mod c4(1 + c3)j from my little congruence result:

With c3 coprime to c4(1 + c3), if u2 = r1 mod c3 and u2 = r2 mod c4(1 + c3)j, you can find u2 mod c3c4(1 + c3)j with

u2 = r1 + kc3 mod c3c4(1 + c3)j

where k = (r2 - r1)c3-1 mod c4(1 + c3)j

where

r1 = c4(1 + c3)j mod c3 and

r2 = -c3 mod c4(1 + c3)j

and you find k, such that r1 + kc3 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 v2=1 mod c4(1 + c3)j, you can search with other values using

r2 = -c3v2 mod c4(1 + c3)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
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