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Tentative conjecture

Let x, y and z be complex quadratic algebraic integers (a and b not equal to 0) then x^2 + y^2 not equal to z^2.

 paulunderwood 2018-06-12 10:58

What are a and b? What is wrong with any pythagorian triple such as 3,4 and 5? Are you trying to say all real and imaginary parts are non-zero?

Tentative conjecture

[QUOTE=paulunderwood;489674]What are a and b? What is wrong with any pythagorian triple such as 3,4 and 5? Are you trying to say all real and imaginary parts are non-zero?[/QUOTE]

a and b are the coefficients of the real and imaginary parts.Pythagorean triplets
exist only when b = 0. When x, y and z are complex quadratic algebraic integers x^2 + y^2 is not equal to z^2. Trust my point is clear.

 Dr Sardonicus 2018-06-12 14:08

Finding counterexamples is easy-peasy...

I^2 = -1

(7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2

 axn 2018-06-12 14:12

[QUOTE=Dr Sardonicus;489693]I^2 = -1

(7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2[/QUOTE]

 CRGreathouse 2018-06-12 15:41

Yes.

 Dr Sardonicus 2018-06-12 16:35

Yes.

7 - 6*I has minimum polynomial (x - 7)^2 + 36 or x^2 - 14*x + 85

6 - 2*I has minimum polynomial (x - 6)^2 + 4 or x^2 - 12*x + 40

-9 + 6*I has minimum polynomial (x + 9)^2 + 36 or x^2 + 18*x + 117

 axn 2018-06-12 16:58

[QUOTE=CRGreathouse;489697]Yes.[/QUOTE]

[QUOTE=Dr Sardonicus;489701]Yes.

7 - 6*I has minimum polynomial (x - 7)^2 + 36 or x^2 - 14*x + 85

6 - 2*I has minimum polynomial (x - 6)^2 + 4 or x^2 - 12*x + 40

-9 + 6*I has minimum polynomial (x + 9)^2 + 36 or x^2 + 18*x + 117[/QUOTE]

Thanks!

 Dr Sardonicus 2018-06-15 19:36

Algebraic formulas are algebraic formulas...

Substituting Gaussian integers z[sub]1[/sub] and z[sub]2[/sub] into the usual parametric formulas for Pythagorean triples,

(A, B, C) = (z[sub]1[/sub][sup]2[/sup] - z[sub]2[/sub][sup]2[/sup], 2*z[sub]1[/sub]*z[sub]2[/sub], z[sub]1[/sub][sup]2[/sup] + z[sub]2[/sub][sup]2[/sup])

We assume that z[sub]1[/sub] and z[sub]2[/sub] are nonzero. We obtain primitive triples if gcd(z[sub]1[/sub], z[sub]2[/sub]) = 1 and gcd(z[sub]1[/sub] + z[sub]2[/sub], 2) = 1. The latter condition rules out z[sub]1[/sub] and z[sub]2[/sub] being complex-conjugate.

We obviously obtain thinly disguised versions of rational-integer triples when one of z[sub]1[/sub] and z[sub]2[/sub] is real, and the other is pure imaginary.

Obviously A, B, and C are real when z[sub]1[/sub] and z[sub]2[/sub] are rational integers.

Clearly B is real when z[sub]2[/sub] is a real multiple of conj(z[sub]1[/sub]).

Also, B/C is real when z[sub]2[/sub]/z[sub]1[/sub] is real, or |z[sub]1[/sub]| = |z[sub]2[/sub]|.

A/C is only real when z[sub]2[/sub]/z[sub]1[/sub] is real.

The nontrivial primitive solutions with A, B, C all complex having the smallest coefficients appear to be

z[sub]1[/sub] = 1, z[sub]2[/sub] = 1 + I: A = 1 - 2*I, B = 2 + 2*I, C = 1 + 2*I

and variants.

Tentative conjecture

[QUOTE=Dr Sardonicus;489693]I^2 = -1

(7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2[/QUOTE]

Geometric interpretation like Pythagoras theorem pl?

Tentative conjecture

[QUOTE=Dr Sardonicus;489693]I^2 = -1

(7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2[/QUOTE]

Hypothesis: Fermat's last conjecture extended to include triples in which each variable is a quadratic algebraic integer. Q: Does Andrew Wiles's proof cover this case?

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