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Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$.

We say $D = b^2 – 4ac$ is the discriminant of $F$.

It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $ax^2 + bxy + cy^2$ is primitive.

Let $m$ be an integer.

If $m = ax^2 + bxy + cy^2$ has a solution in $\mathbb{Z}^2$, we say $m$ is represented by $F$.

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

Let $\Phi_1,\dots,\Phi_{\mu}$ be the system of genus characters of discriminant $D$[see this question].

Let $n$ be an integer.

We denote by $[n]$ the image of $n$ by the canonical homomorphism $\mathbb{Z} \rightarrow \mathbb{Z}/D\mathbb{Z}$.

Let $(\mathbb{Z}/D\mathbb{Z})^\times$ be the group of invertible elements of the ring $\mathbb{Z}/D\mathbb{Z}$.

we define a homomorphism $\Phi\colon(\mathbb{Z}/D\mathbb{Z})^\times \rightarrow (\mathbb{Z}^\times)^\mu$ by

$\Phi([n]) = (\Phi_1(n),\dots,\Phi_{\mu}(n))$.

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We would like to determine the kernel and the image of $\Phi$.

To do this, we need the following definition.

**Definition**

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

If $D \equiv 0$ (mod $4$), $x^2 – \frac{D}{4}y^2$ is a primitive form of discriminant $D$.

If $D \equiv 1$ (mod $4$), $x^2 + xy + \frac{1 – D}{4}y^2$ is a primitive form of discriminant $D$.

We call $x^2 – \frac{D}{4}y^2$ or $x^2 + xy + \frac{1 – D}{4}y^2$ the principal form of discriminant $D$.

**My question**

Is the following proposition true?

If yes, how do we prove it?

**Proposition**

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

Let $H = \{ [m] \in (\mathbb{Z}/D\mathbb{Z})^\times$; $m$ is represented by the principal form of discriminant $D\}$.

Then $\Phi$ is surjective and $H$ = Ker($\Phi$).

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It is true, the proof is easier for case that $D\equiv 1 \mod D$. There it is a easy congurence of the Chinese reminder theorem. In the case $-4n$ the proof is similar but requires case distinction. (see Cox, Primes of the form $x^2+ny^2$, Lemma 3.17)

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