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I'm trying to generate the set of solutions of a specific diophantine equation over Z[i]. The equation is the following:

$$ z_1^2 + z_2^2 + z_1*z_2 + 39 = 0$$

with $ z_1$ (resp $z_2$) such that $\exists a,b \in \mathbb{Z} , z_1$ (resp $z_2$) $= a + ib $

Is there a specific tehnique to deal with Gaussian Integers in Diophantine equations ?

Regards

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    $\begingroup$ The equation asks in effect to write $-39$ as a norm $N_{K/{\bf Q}(i)}(z)$, where $K = Q(i,\sqrt{-3})$ is the 12th cyclotomic field and $z = z_1 - e^{2\pi i /3} z_2$. The relative unit group has rank $1$, so this equation behaves like a Pell's equation, with finitely many families of solutions each obtained from a linear recurrence of degree 2 with constant coefficients. Some small solutions are $(2i,5i)$, $(3+4i,-3+4i)$, $(6+5i,-6+5i)$. $\endgroup$
    – Noam D. Elkies
    Commented Jun 4, 2012 at 14:30
  • $\begingroup$ Thanks for your comment ! I've been able to generate those small solutions. But I'd like to generate all primitive solutions and to find this linear reccurence relation/ $\endgroup$
    – Michel
    Commented Jun 4, 2012 at 14:39
  • $\begingroup$ For starters, if $z_1,z_2 = a \pm b i$ then we get $3a^2 - b^2 = -39$, which actually is a Pell equation (with non-unit discriminant). The resulting recurrence must already give some positive fraction of the solutions, and then a few linear transformations (probably coming from roots of unity, like $(z_1,z_2) \mapsto (z_1,-z_1-z_2)$) should provide the rest. What do you need this for? $\endgroup$
    – Noam D. Elkies
    Commented Jun 4, 2012 at 19:41
  • $\begingroup$ Thanks. Just to clarify: Is the equation $3*b^2 - a^2 = 39$ ? With this Pell equation, I only generate conjugate solutions, and I'm not sure to understand what you mean with "a few linear transform". How can I be sure to have all the families ? $\endgroup$
    – Michel
    Commented Jun 4, 2012 at 21:51
  • $\begingroup$ Sorry: $3a^2-b^2 = -39$ is right, but that equation has no solution, so we must multiply $z_1,z_2$ by $i$ to get $ia \pm b$, and then it's $3a^2-b^2=39$ which is the equation you gave. At the point the linear transformation $(z_1,z_2) \mapsto (z_1,-z_1-z_2)$ and its iterate converts conjugate solutions to two other families, which together with $ia \pm b$ seems to give half the solutions; probably another linear transformation gives the others. Again I ask: what's the source of this problem? $\endgroup$
    – Noam D. Elkies
    Commented Jun 4, 2012 at 22:11

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