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 2018-06-12, 04:40 #1 devarajkandadai     May 2004 22×79 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.
 2018-06-12, 10:58 #2 paulunderwood     Sep 2002 Database er0rr 2×3×691 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 non-zero? Last fiddled with by paulunderwood on 2018-06-12 at 11:14
2018-06-12, 13:20   #3

May 2004

22×79 Posts
Tentative conjecture

Quote:
 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.

 2018-06-12, 14:08 #4 Dr Sardonicus     Feb 2017 Nowhere 3×1,931 Posts Finding counterexamples is easy-peasy... I^2 = -1 (7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2
2018-06-12, 14:12   #5
axn

Jun 2003

3·1,789 Posts

Quote:
 Originally Posted by Dr Sardonicus I^2 = -1 (7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2

2018-06-12, 15:41   #6
CRGreathouse

Aug 2006

3×1,993 Posts

Quote:
 Originally Posted by axn Are these quadratic integers?
Yes.

2018-06-12, 16:35   #7
Dr Sardonicus

Feb 2017
Nowhere

10110101000012 Posts

Quote:
 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

2018-06-12, 16:58   #8
axn

Jun 2003

3×1,789 Posts

Quote:
 Originally Posted by CRGreathouse Yes.
Quote:
 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!

 2018-06-15, 19:36 #9 Dr Sardonicus     Feb 2017 Nowhere 3·1,931 Posts Algebraic formulas are algebraic formulas... 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.
2018-07-13, 10:36   #10

May 2004

22×79 Posts
Tentative conjecture

Quote:
 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?

2018-07-22, 05:38   #11

May 2004

22·79 Posts
Tentative conjecture

Quote:
 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?

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