20180612, 04:40  #1 
May 2004
2^{2}×79 Posts 
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.

20180612, 10:58  #2 
Sep 2002
Database er0rr
2×1,811 Posts 
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?
Last fiddled with by paulunderwood on 20180612 at 11:14 
20180612, 13:20  #3  
May 2004
2^{2}·79 Posts 
Tentative conjecture
Quote:
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. 

20180612, 14:08  #4 
Feb 2017
Nowhere
10567_{8} Posts 
Finding counterexamples is easypeasy...
I^2 = 1
(7  6*I)^2 + (6  2*I)^2 = (9 + 6*I)^2 
20180612, 14:12  #5 
Jun 2003
1337_{16} Posts 
Are these quadratic integers?

20180612, 15:41  #6  
Aug 2006
5985_{10} Posts 
Quote:


20180612, 16:35  #7  
Feb 2017
Nowhere
10567_{8} Posts 
Quote:
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 

20180612, 16:58  #8 
Jun 2003
4,919 Posts 

20180615, 19:36  #9 
Feb 2017
Nowhere
1000101110111_{2} Posts 
Algebraic formulas are algebraic formulas...
Substituting Gaussian integers z_{1} and z_{2} into the usual parametric formulas for Pythagorean triples,
(A, B, C) = (z_{1}^{2}  z_{2}^{2}, 2*z_{1}*z_{2}, z_{1}^{2} + z_{2}^{2}) We assume that z_{1} and z_{2} are nonzero. We obtain primitive triples if gcd(z_{1}, z_{2}) = 1 and gcd(z_{1} + z_{2}, 2) = 1. The latter condition rules out z_{1} and z_{2} being complexconjugate. We obviously obtain thinly disguised versions of rationalinteger triples when one of z_{1} and z_{2} is real, and the other is pure imaginary. Obviously A, B, and C are real when z_{1} and z_{2} are rational integers. Clearly B is real when z_{2} is a real multiple of conj(z_{1}). Also, B/C is real when z_{2}/z_{1} is real, or z_{1} = z_{2}. A/C is only real when z_{2}/z_{1} is real. The nontrivial primitive solutions with A, B, C all complex having the smallest coefficients appear to be z_{1} = 1, z_{2} = 1 + I: A = 1  2*I, B = 2 + 2*I, C = 1 + 2*I and variants. 
20180713, 10:36  #10 
May 2004
316_{10} Posts 
Tentative conjecture

20180722, 05:38  #11 
May 2004
2^{2}·79 Posts 
Tentative conjecture

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