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.

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 nonzero?

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 nonzero?[/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. 
Finding counterexamples is easypeasy...
I^2 = 1
(7  6*I)^2 + (6  2*I)^2 = (9 + 6*I)^2 
[QUOTE=Dr Sardonicus;489693]I^2 = 1
(7  6*I)^2 + (6  2*I)^2 = (9 + 6*I)^2[/QUOTE] Are these [URL="https://en.wikipedia.org/wiki/Quadratic_integer"]quadratic integers[/URL]? 
[QUOTE=axn;489694]Are these [URL="https://en.wikipedia.org/wiki/Quadratic_integer"]quadratic integers[/URL]?[/QUOTE]
Yes. 
[QUOTE=axn;489694]Are these [URL="https://en.wikipedia.org/wiki/Quadratic_integer"]quadratic integers[/URL]?[/QUOTE]
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=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! 
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 complexconjugate. We obviously obtain thinly disguised versions of rationalinteger 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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