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August 17th, 2007, 07:58 AM   #1
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Pseudo-Diophantine Nuisance

I have the equation

a^2+a^-2+b^2+b^-2+c^2+c^-2=0

and would need just a "nice" solution.
E.g. a=1 b=i c=sqr(i) but that is a) too special and
b) actually contains a fourth root.
a=i b=3 c=(sqrt(82)*i - sqrt(46)*i)/6 or something
like that is the best I can come up with - can someone
find a solution in Gaussian integers, maybe?

THX, Hauke
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August 17th, 2007, 09:25 AM   #2
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You can transform this from an equation over Gaussian integers to two over (rational) integers and just use standard methods to solve, I suppose.
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August 30th, 2007, 08:03 AM   #3
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The solutions are the solutions of the equations

(d^2-e^2)*((f^2-g^2)*(k^4-6*j^2*k^2+j^4+1)-2*f*g*(4*j^3*k-4*j*k^3))-2*d*e*(2*f*g*(k^4-6*j^2*k^2+j^4+1)+(f^2-g^2)*(4*j^3*k-4*j*k^3))+(d^2-e^2)*((g^4-6*f^2*g^2+f^4+1)*(j^2-k^2)-2*(4*f^3*g-4*f*g^3)*j*k)-2*d*e*((4*f^3*g-4*f*g^3)*(j^2-k^2)+2*(g^4-6*f^2*g^2+f^4+1)*j*k)+(e^4-6*d^2*e^2+d^4+1)*((f^2-g^2)*(j^2-k^2)-4*f*g*j*k)-(4*d^3*e-4*d*e^3)*(2*f*g*(j^2-k^2)+2*(f^2-g^2)*j*k) = 0 and 2*d*e*((f^2-g^2)*(k^4-6*j^2*k^2+j^4+1)-2*f*g*(4*j^3*k-4*j*k^3))+(d^2-e^2)*(2*f*g*(k^4-6*j^2*k^2+j^4+1)+(f^2-g^2)*(4*j^3*k-4*j*k^3))+2*d*e*((g^4-6*f^2*g^2+f^4+1)*(j^2-k^2)-2*(4*f^3*g-4*f*g^3)*j*k)+(d^2-e^2)*((4*f^3*g-4*f*g^3)*(j^2-k^2)+2*(g^4-6*f^2*g^2+f^4+1)*j*k)+(4*d^3*e-4*d*e^3)*((f^2-g^2)*(j^2-k^2)-4*f*g*j*k)+(e^4-6*d^2*e^2+d^4+1)*(2*f*g*(j^2-k^2)+2*(f^2-g^2)*j*k) = 0

where

a = d+ei
b = f + gi
c = j + ki

But six variables, two equations... you'll have a lot of solutions.
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