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Some time ago, I asked a question about equidistribution on a paper of Duke and Schulze-Pillot that was usefully answered.

However, on the answer there was a statement that was unimportant for me back then but catch my attention now. One of the steps of the answer was to use the following equivalence property:

Let $Q$ be a ternary quadratic form and $n$ a non-negative integer. "$n$ is primitively represented by some form in the genus of $Q$" 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 $$Q(x_1,x_2,x_3)=\frac{1}{2}\sum_{i,j}a_{ij}x_ix_j."$$

I believe that this statement is true but have failed to found a reference for it, although I have read some other papers that use the same or similar results (e.g. this one , page 5, line 10). I wonder if somebody could give me a reference or a simple proof for this property in case it's true. Otherwise, I would really appreciate a similar congruence property that's correct (if exists).

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  • $\begingroup$ See my page zakuski.math.utsa.edu/~kap for a start $\endgroup$
    – Will Jagy
    Commented Sep 12, 2022 at 15:02
  • $\begingroup$ Thank you! By searching in your page I have found other references where this result is used, e.g. the coment after the theorem [1] of this Duke article of 1997 (zakuski.math.utsa.edu/~kap/Duke_1997.pdf). But again I am not able to prove why this is true (why we only need the congruence with a single value depending on the determinant?) or a reference where stated explicitly. $\endgroup$
    – MathqA
    Commented Sep 13, 2022 at 15:20
  • $\begingroup$ In that case, I suggest Burton W. Jones, The Arithmetic Theory of Quadratic Forms, especially chapter 8. Rationally represented is Hasse-Minkowski. Then, Jones was the first to show that every integer rationally represented by a form is integrally represented by something in its genus. .... It is also Theorem 1.3, page 129, chapter 9, of Rational Quadratic Forms by Cassels. $\endgroup$
    – Will Jagy
    Commented Sep 13, 2022 at 15:40
  • $\begingroup$ Again in Cassels, Theorem 5.1 on page 143. A cautionary note on page 168, example 23, primitive spinor exceptions. google.com/books/edition/Rational_Quadratic_Forms/… $\endgroup$
    – Will Jagy
    Commented Sep 13, 2022 at 16:15

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The quoted text was written by me. As I corrected myself recently in a comment below the original post: 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).

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  • $\begingroup$ Yes, thank you. You clarification in the comment was totally helpful. By the time I posted this question I didn't know it. $\endgroup$
    – MathqA
    Commented Oct 11, 2022 at 9:52

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