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Let $q(\mathbf{x}) = q(x_1, \cdots, x_n)$ be a quadratic form with integer coefficients. For $n \geq 3$, is there a reasonable theory for the set of integer solutions to the equation

$$\displaystyle q(\mathbf{x}) = 1?$$

When $n = 2$ this is the well-known theory of Pell equations. Indeed, it is known that $q(\mathbf{x}) = 1$ has a solution in integers if and only if $q$ is $\text{GL}_2(\mathbb{Z})$-equivalent to the principal form $x^2 - dy^2$ for some $d \geq 1$, and if $d$ is not a square, then there will be infinitely many solutions generates by a single fundamental solution.

Obviously, the same type of structure cannot exist for $n \geq 3$ since $q$ is in general geometrically irreducible, unlike the case when $n = 2$. Nevertheless, it would be nice to be able to describe the set of integer solutions in some reasonable way.

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    $\begingroup$ Do you have any sort of criteria for what would constitute a reasonable decription? For example, would an emptiness vs. non-emptiness, or finite vs. infinite, classification be reasonable, or too weak? (I don't know the answers even to these special cases, just am curious about what sort of answer is sought.) $\endgroup$
    – LSpice
    Dec 8 at 18:44
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    $\begingroup$ @LSpice certainly Hasse's principle on rational solutions would be too weak (i.e., that there is a rational solution if and only if the equation has a local solution in every completion of $\mathbb{Q}$). I am mostly thinking about the case when there are infinitely many solutions, and what can be said about the set of solutions. $\endgroup$
    – Stanley Yao Xiao
    Dec 8 at 18:53
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    $\begingroup$ I am sure you are aware of this, but the solutions of $q(x_1,\dotsc,x_n)=1$ form finitely many orbits under the group of automorphs of $q$, and a solution exists if and only if $q$ is properly equivalent to a form $x_1^2+r(x_2,\dotsc,x_n)$. $\endgroup$
    – GH from MO
    Dec 8 at 21:03
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    $\begingroup$ See my answer to this question mathoverflow.net/questions/142938 about general quadratic equations. $\endgroup$
    – Bogdan Grechuk
    Dec 8 at 23:54
  • $\begingroup$ For a nice example of equation of your type, see mathoverflow.net/questions/432091 $\endgroup$
    – Bogdan Grechuk
    Dec 8 at 23:55

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