My Math Forum Pseudo-Diophantine Nuisance

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 August 17th, 2007, 07:58 AM #1 Newbie   Joined: Jul 2007 Posts: 7 Thanks: 0 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
 August 17th, 2007, 09:25 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms You can transform this from an equation over Gaussian integers to two over (rational) integers and just use standard methods to solve, I suppose.
 August 30th, 2007, 08:03 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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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