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Let $(L,q)$ be an even Lorentzian lattice of signature $(1,l-1)$, i.e., $q(\lambda) \in \Bbb Z$ for all $\lambda \in L$ and the (non-degenerate) quadratic form $q$ is of type $(1,l-1)$ (the type/signature shouldn't matter here, I think...). I am interested in counting the number of solutions $x \in L/8L$ modulo $a \in (\Bbb Z / 8 \Bbb Z)^*$, i.e., what is $$N(a) := \# \{ x \in L/8L : q(x) \equiv a \pmod 8 \}$$

We could rewrite this as $N(a) = \# \{ x \in \Bbb Z / 8 \Bbb Z : q(x) \equiv a \pmod 8 \}$ again using $q(x)$ for the quadratic form $\frac 1 2 x^t S x$, $S$ the Gram matrix, for vectors $x \in \Bbb Z / 8 \Bbb Z$.

A trivial bound would be $8^l$ assuming that all vectors are solutions. One can verify, that all even vectors are no solutions, i.e., vectors $x$ for which there is a vector $x' \in (\Bbb Z / 8 \Bbb Z)^l$ such that $x = 2 x'$. Because then $q(x) = 4 q(x') \in \{0,4\}$ but we assumed that $a \in (\Bbb Z / 8 \Bbb Z)^*$ so $x$ cannot be a solution. This gives a slightly better bound of $8^l - 4^l$. But I was hoping for a bound like $4 \cdot 8^{l-1} = 8^l /2$, i.e., that only at most half of the vectors are solutions. My intuition tells me that this should be possible, assuming that $a$ is a unit.

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