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Old 2018-06-12, 04:40   #1
devarajkandadai
 
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Default 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.
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Old 2018-06-12, 10:58   #2
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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
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Old 2018-06-12, 13:20   #3
devarajkandadai
 
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Default Tentative conjecture

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Originally Posted by paulunderwood View Post
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.
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Old 2018-06-12, 14:08   #4
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Default Finding counterexamples is easy-peasy...

I^2 = -1

(7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2
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Old 2018-06-12, 14:12   #5
axn
 
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Quote:
Originally Posted by Dr Sardonicus View Post
I^2 = -1

(7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2
Are these quadratic integers?
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Old 2018-06-12, 15:41   #6
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Quote:
Originally Posted by axn View Post
Yes.
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Old 2018-06-12, 16:35   #7
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Quote:
Originally Posted by axn View Post
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
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Old 2018-06-12, 16:58   #8
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Quote:
Originally Posted by CRGreathouse View Post
Yes.
Quote:
Originally Posted by Dr Sardonicus View Post
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!
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Old 2018-06-15, 19:36   #9
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Default 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.
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Old 2018-07-13, 10:36   #10
devarajkandadai
 
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Default Tentative conjecture

Quote:
Originally Posted by Dr Sardonicus View Post
I^2 = -1

(7 - 6*I)^2 + (6 - 2*I)^2 = (-9 + 6*I)^2
Geometric interpretation like Pythagoras theorem pl?
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Old 2018-07-22, 05:38   #11
devarajkandadai
 
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Default Tentative conjecture

Quote:
Originally Posted by Dr Sardonicus View Post
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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