Some time ago, I asked a [question](https://mathoverflow.net/questions/424067/duke-and-schulze-pillot-condition-for-equidistribution) 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](https://arxiv.org/pdf/1402.1332.pdf) , 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).



Primitive representation of integers by some form on the genus of a quadratic form
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).


