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I would like to find all integer triples (x,y,z) such that: $\prod_{\theta}(x + y \theta + z \theta^2)=1$, where $\theta$ runs through the solutions to the cubic $x^3 + x^2 - 2x - 1=0$.

In his book "Diophantine equations" (p. 111-12) Mordell gives the equation

$w^n = \prod_{\theta}(x + y \theta + z \theta^2)$

where $\theta=\theta_1, \theta_2, \theta_3$ are the solutions to a cubic equation with integer coefficients.

Mordell's partial solution is,

$x + y\theta_1 + z \theta_1^2 = (p + q\theta_1 + r\theta_1^2)^n$,

$x + y\theta_2 + z \theta_2^2 = (p + q\theta_2 + r\theta_2^2)^n$,

$x + y\theta_3 + z \theta_3^2 = (p + q\theta_3 + r\theta_3^2)^n$,

$w = \prod_{\theta}(p + q \theta + r \theta^2) $

where $p,q,r$ are arbitrary integers and $n$ runs through the integers.''

He continues to say, " the general solution depends upon the theory of algebraic numbers and is connected with the units in an algebraic number field".

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  • $\begingroup$ So: you want to find all triples $(x, y, z)$ of integers such that $\prod_{\theta} (x + y\theta + z \theta^2) = 1$, where $\theta$ runs through the solutions to your cubic? (I think Mordell is trying to do something else much harder -- for him w is not part of the question; he's solving for the quadruple $(x, y, z, w)$.) $\endgroup$
    – David Loeffler
    Commented Jun 7, 2011 at 14:25
  • $\begingroup$ David, thank you for your reply. Indeed your restatement of the problem is correct. It may be less difficult than the problem he tries to solve. I looked at Mordell's particular solution as only a particular solution to my problem. $\endgroup$
    – Tom Hunt
    Commented Jun 7, 2011 at 15:10
  • $\begingroup$ In your example $1,\theta,\theta^2$ generate the maximal order, so the equation is simply the norm equation whose solutions are exactly the units of the field generated by $\theta$. The field is totally real, hence has unit rank 2, and the group of units is generated by: {$-1,\theta+1,\theta^2-1$}. So the solutions are those such that $p+q\theta+r\theta^2=\pm (\theta+1)^a(\theta^2-1)^b$, for any integers $a,b$. (all computations were done using Sage) $\endgroup$
    – Dror Speiser
    Commented Jun 7, 2011 at 15:29
  • $\begingroup$ Dror: You have to restrict to norm 1 units, cf. my answer below, which crossed over with yours. $\endgroup$
    – David Loeffler
    Commented Jun 7, 2011 at 15:45
  • $\begingroup$ Oh yes! Thanks David. I've looked at too many quadratic imaginary fields lately that I forgot that units with negative norm can even exist... $\endgroup$
    – Dror Speiser
    Commented Jun 7, 2011 at 18:39

1 Answer 1

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This isn't really a research-level question, and hence belongs more on math.SE than here, but here goes anyway...

Let $K$ be the number field $\mathbb{Q}(\theta)$ where $\theta$ is a root of your cubic $f$. Then the ring of integers $\mathcal{O}_K$ of $K$ is $\mathbb{Z}(\theta)$, and we are reduced to looking for elements $\omega \in \mathcal{O}_K^\times$ such that $\operatorname{Norm}_{K/\mathbb{Q}}(\omega) = 1$. It takes about 0.01 seconds for my computer to tell me that $\mathcal{O}_K^\times$ is isomorphic to $\mathbb{Z}^2 \times \mathbb{Z}/2$, and the index 2 subgroup of units of norm 1 is isomorphic to $\mathbb{Z}^2$, with independent generators $\theta-1$ and $-\theta-1$.

So the general solution is $$ x + y \theta + z \theta^2 = (\theta - 1)^a (-\theta-1)^b$$ for any $a, b \in \mathbb{Z}$.

(If you're unfamiliar with the theory of algebraic number fields I'm using here, Stewart and Tall's book is a good reference.)

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  • $\begingroup$ David, thank you for the response, I will refer to Stewart and Tall. $\endgroup$
    – Tom Hunt
    Commented Jun 7, 2011 at 16:08
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    $\begingroup$ It's not research-level for a number theorist, but any other type of mathematician could come across such a problem, and would likely have no idea how to solve it, and would have difficulty making sense of our literature. $\endgroup$
    – Kevin O'Bryant
    Commented Jun 12, 2011 at 16:14

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