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When regarding a ternary quadratic form $Q(x,y,z)$, is a classic question to consider which integers $n$ can be represented by $Q$. It is also classic to wonder how "well distributed" are the lattice points $(x,y,z)\in\mathbb{Z}^3$ that satisfies $Q(x,y,z)=n$ when $n\rightarrow\infty$ along a certain sequence.

When $Q$ is just the modulus square, I have found a lot of literature but I am interested in more general $Q$. On chapter 11 of Topics in Classical Automorphic Forms by Henryk Iwaniec, we find conditions to $n$ so that this condition holds. However, Iwaniec said that the result is due to Duke and that he asked a few less on $n$.

I have tried to read Duke papers:

Representation of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids

And

On ternary quadratic forms,

But I couldn't find the conditions he imposed over $n$ so that we have equidistribution. I am interested in asking the less possible to $n$ and knowing the difference between the hypothesis of Duke and those of Iwaniec.

Thank you to everyone.

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I recommend that you study the paper by Duke and Schulze-Pillot, Representation of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids (Inventiones, 1990).

Theorem 1 shows that if the number of representations $r(n,Q)$ exceeds $n^{1/2-1/176}$, then the representations become equidistributed on the relevant ellipsoid as $n\to\infty$. So the question is how to guarantee that $r(n,Q)$ is large. By Theorem 3 (see also the subsequent Corollary), it suffices that $n$ is primitively represented by some form in the spinor genus of $Q$. By Theorem 2, if we exclude finitely many explicitly computable square classes for $n$, the condition simplifies to: $n$ is primitively represented by some form in the genus of $Q$. The latter is equivalent to: $n$ is primitively represented by $Q$ modulo $4\det(a_{ij})$, where $(a_{ij})\in\mathrm{M}_3(\mathbb{Z})$ is the symmetric matrix such that $$Q(x_1,x_2,x_3)=\frac{1}{2}\sum_{i,j}a_{ij}x_ix_j.$$ It remains to determine the exceptional square classes. Regarding that, see the proof of Theorem 2 and the references given. In particular, if $n$ is coprime to $\det(a_{ij})$, then the only exceptional square class is the set of squares (see Footnote 5 in arXiv:1402.1332).

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    $\begingroup$ That is exactly what I was looking for. I needed some help to follow Duke and Schulze-Pillot paper. Thank you! $\endgroup$
    – MathqA
    Commented Jun 8, 2022 at 5:41
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    $\begingroup$ I needed to come back to this topic and a question about your answer just hit me. Could you please justify the statement "The latter is equivalent to: $n$ is primitively represented by $Q$ modulo $4det(a_{ij})$. All the references I have consulted seem to need a congruence with $8det(a_{ij})^2$. What I am understanding wrong? $\endgroup$
    – MathqA
    Commented Oct 5, 2022 at 8:38
  • $\begingroup$ @MathqA I think you are right. In my response, I secretly assumed that $n$ was coprime to $\det(a_{ij})$. In that case, I believe that modulo $4\det(a_{ij})$ is sufficient. See Hilfssatz 13 in Siegel: Über die analytische Theorie der quadratischen Formen, Ann. of Math. 36 (1935), 527-606. Note that when applying this theorem, $b=0$ for $p>2$, and $b=1$ for $p=2$ (because $p\mid S$ implies $p\nmid T$ by our coprimality assumption). $\endgroup$
    – GH from MO
    Commented Oct 5, 2022 at 14:03

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