Tentative conjecture

[devarajkandadai]

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]

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?

[devarajkandadai]

Originally Posted by paulunderwood

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?

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]

I^2 = -1

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

[axn]

Originally Posted by Dr Sardonicus

I^2 = -1

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

Are these quadratic integers?

[CRGreathouse]

Originally Posted by axn

Are these quadratic integers?

Yes.

[Dr Sardonicus]

Originally Posted by axn

Are these quadratic integers?

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]

Originally Posted by CRGreathouse

Yes.

Originally Posted by Dr Sardonicus

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

Thanks!

[Dr Sardonicus]

Substituting Gaussian integers z₁ and z₂ into the usual parametric formulas for Pythagorean triples,

(A, B, C) = (z₁² - z₂², 2*z₁*z₂, z₁² + z₂²)

We assume that z₁ and z₂ are nonzero. We obtain primitive triples if gcd(z₁, z₂) = 1 and gcd(z₁ + z₂, 2) = 1. The latter condition rules out z₁ and z₂ being complex-conjugate.

We obviously obtain thinly disguised versions of rational-integer triples when one of z₁ and z₂ is real, and the other is pure imaginary.

Obviously A, B, and C are real when z₁ and z₂ are rational integers.

Clearly B is real when z₂ is a real multiple of conj(z₁).

Also, B/C is real when z₂/z₁ is real, or |z₁| = |z₂|.

A/C is only real when z₂/z₁ is real.

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

z₁ = 1, z₂ = 1 + I: A = 1 - 2*I, B = 2 + 2*I, C = 1 + 2*I

and variants.

[devarajkandadai]

Originally Posted by Dr Sardonicus

I^2 = -1

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

Geometric interpretation like Pythagoras theorem pl?

[devarajkandadai]

Originally Posted by Dr Sardonicus

I^2 = -1

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

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?