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Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Let $\chi\colon (\mathbb{Z}/D\mathbb{Z})^\times\rightarrow \mathbb{Z}^\times = \{-1, 1\}$ be the map defined in this question. Let $n$ be an integer. We denote by $[n]$, the image of $n$ by the canonical map $\mathbb{Z} \rightarrow \mathbb{Z}/D\mathbb{Z}$. Computing values of $\chi$ at $[-1]$ for several $D$s, it seems that $\chi([-1]) = 1$ if $D \gt 0$ and $\chi([-1]) = -1$ if $D \lt 0$.

Is this true? If yes, how can we prove it?

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1 Answers1

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We will freely use the properties of the Jacobi symbol as stated in this question In particular, we have the folowing:

If $m \equiv 1$ (mod $4$), $\left(\frac{m}{n}\right) = \left(\frac{n}{m}\right)$.

If $m \equiv 3$ (mod $4$) and $n \equiv 3$ (mod $4$), $\left(\frac{m}{n}\right) = -\left(\frac{n}{m}\right)$.

If $m \equiv 1$ (mod $4$), $\left(\frac{-1}{m}\right) = 1$.

If $m \equiv 3$ (mod $4$), $\left(\frac{-1}{m}\right) = -1$.

Case 1 $D \gt 0$ and $D \equiv 0$ (mod $4$)

$\chi([-1]) = \chi([D - 1]) = \left(\frac{D}{D-1}\right) = \left(\frac{D-1 +1 }{D-1}\right) = \left(\frac{1}{D-1}\right) = 1$

Case 2 $D \gt 0$ and $D \equiv 1$ (mod $4$)

$\chi([-1]) = \chi([2D - 1]) = \left(\frac{D}{2D-1}\right) = \left(\frac{2D-1}{D}\right) = \left(\frac{-1}{D}\right) = 1$

Case 3 $D \lt 0$ and $D \equiv 0$ (mod $4$)

We first note that $-D - 1 \equiv -1$ (mod $4$).

$\chi([-1]) = \chi([-D - 1]) = \left(\frac{D}{-D-1}\right) = \left(\frac{-(-D-1) -1 }{-D-1}\right) = \left(\frac{-1}{-D-1}\right) = -1$

Case 4 $D \lt 0$ and $D \equiv 1$ (mod $4$)

We first note that $-2D -1 \equiv 1$ (mod $4$).

$\chi([-1]) = \chi([-2D - 1]) = \left(\frac{D}{-2D-1}\right) = \left(\frac{-1}{-2D-1}\right)\left(\frac{-D}{-2D-1}\right) = \left(\frac{-2D - 1}{-D}\right) = \left(\frac{-1}{-D}\right) = -1$.

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