Three Variable Quadratic Reduction PDF
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JSH
Feb 4, 2018, 7:54:00 PM2/4/18
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Three variable quadratic reduction
Have long talked my method for reducing binary quadratic Diophantine equations which of course is two variable, but actually simplified from a three variable analysis. Was considering:
c1x2 + c2xy + c3y2 = c4z2 + c5zx + c6zy
where the c's are constants. And found what I call my Quadratic Diophantine Theorem which I talk in this post from September 2008.
And figured out:
((c2 - 2c1)2 + 4c1(c2 - c1 - c3))(x+y)2 - (2(c2 - 2c1)(c6 - c5) + 4c5(c2 - c1 - c3))(x+y)z + [(c6 - c5)2 - 4c4(c2 - c1 - c3)]z2 = m2
where m is some integer, if integer solutions exist for x, y and z. And can see that m has an explicit solution as a function of x, y and z. But is SO hard to calculate to something simpler than given squared one. Kind of thing for which math software is good for those who know how to use, and wish to solve for it. I never bother.
Way too complicated looking so introduced some more variables to collect things.
A = (c2 - 2c1)2 + 4c1(c2 - c1 - c3), B = (c2 - 2c1)(c6 - c5) + 2c5(c2 - c1 - c3)
and
C = (c6 - c5)2 - 4c4(c2 - c1 - c3)
where neither A nor B is zero for what follows.
Then have: A(x+y)2 - 2B(x+y)z + Cz2 = m2
Which is easier to handle. And multiplying through by A, moving some things around, and completing the square:
A2(x+y)2 - 2BA(x+y)z + B2z2 - Am2 = B2z2 - CAz2
Which is easily enough:
[A(x+y) - Bz]2 - Am2 = (B2 - AC)z2
And that is simple enough. Not far at all from what I already had. Where also for certain situations would NOT complete the square, but work with the prior form.
Shows yet another type of general reduced form: u2 - Dv2 = Fw2
Maybe should come up with another cool name for iterator am sure can get from it. Maybe later.
As a first example let c1 = 1, c2 = 1, c3 = 1, c4 = 1, c5 = 1, c6 = 1, so:
x2 + xy + y2 = z2 + zx + zy
Which gives:
A = -3, B = -2, and C = 4
[-3(x+y) + 2z]2 + 3m2 = 16z2
Where now have two unknowns m and z determining existence of a rational solution. But by inspection has an infinite number of rational solutions with m = 0. Still shouldn't just assume that then x and y always will be integers. Easy enough to check:
-3(x+y) + 2z = +/- 4z, so: -3(x+y) = 2z or -6z
So there will be a set of integer solutions for every nonzero integer z.
Well I DO like easy. So yeah, very straightforward from my original research.
Primary point of this post was to just get a look at result without setting z = 1, which did before just to focus on the simpler case.
So now have the three variable general reduced form for the quadratic case.
James Harris
c1x2 + c2xy + c3y2 = c4z2 + c5zx + c6zy
where the c's are constants. And found what I call my Quadratic Diophantine Theorem which I talk in this post from September 2008.
And figured out:
((c2 - 2c1)2 + 4c1(c2 - c1 - c3))(x+y)2 - (2(c2 - 2c1)(c6 - c5) + 4c5(c2 - c1 - c3))(x+y)z + [(c6 - c5)2 - 4c4(c2 - c1 - c3)]z2 = m2
where m is some integer, if integer solutions exist for x, y and z. And can see that m has an explicit solution as a function of x, y and z. But is SO hard to calculate to something simpler than given squared one. Kind of thing for which math software is good for those who know how to use, and wish to solve for it. I never bother.
Way too complicated looking so introduced some more variables to collect things.
A = (c2 - 2c1)2 + 4c1(c2 - c1 - c3), B = (c2 - 2c1)(c6 - c5) + 2c5(c2 - c1 - c3)
and
C = (c6 - c5)2 - 4c4(c2 - c1 - c3)
where neither A nor B is zero for what follows.
Then have: A(x+y)2 - 2B(x+y)z + Cz2 = m2
Which is easier to handle. And multiplying through by A, moving some things around, and completing the square:
A2(x+y)2 - 2BA(x+y)z + B2z2 - Am2 = B2z2 - CAz2
Which is easily enough:
[A(x+y) - Bz]2 - Am2 = (B2 - AC)z2
And that is simple enough. Not far at all from what I already had. Where also for certain situations would NOT complete the square, but work with the prior form.
Shows yet another type of general reduced form: u2 - Dv2 = Fw2
Maybe should come up with another cool name for iterator am sure can get from it. Maybe later.
As a first example let c1 = 1, c2 = 1, c3 = 1, c4 = 1, c5 = 1, c6 = 1, so:
x2 + xy + y2 = z2 + zx + zy
Which gives:
A = -3, B = -2, and C = 4
[-3(x+y) + 2z]2 + 3m2 = 16z2
Where now have two unknowns m and z determining existence of a rational solution. But by inspection has an infinite number of rational solutions with m = 0. Still shouldn't just assume that then x and y always will be integers. Easy enough to check:
-3(x+y) + 2z = +/- 4z, so: -3(x+y) = 2z or -6z
So there will be a set of integer solutions for every nonzero integer z.
Well I DO like easy. So yeah, very straightforward from my original research.
Primary point of this post was to just get a look at result without setting z = 1, which did before just to focus on the simpler case.
So now have the three variable general reduced form for the quadratic case.
James Harris
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