The system of genus characters determined by a binary quadratic form

Question

Let $f = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$. Let $D = b^2 -4ac$ be its discriminant. It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). We suppose $D$ is not a square integer. 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 $p$ be an odd prime divisor of $D$. By this question, $\left(\frac{m}{p}\right)$ does not depend on the choice of $m$. So it is natural to ask what can be said for the prime 2 if $D \equiv 0$ (mod $4$).

We define a map $\psi_1:\mathbb{Z} \rightarrow \mathbb{C}$ as follows.

If $r$ is even, $\psi_1(r) = 0$.

If $r$ is odd, $\psi_1(r) = (-1)^{(r-1)/2}$.

We define a map $\psi_2:\mathbb{Z} \rightarrow \mathbb{C}$ as follows.

If $r$ is even, $\psi_2(r) = 0$.

If $r$ is odd, $\psi_2(r) = (-1)^{(r^2 - 1)/8}$.

By this question, $\{1, \psi_1, \psi_2, \psi_1\psi_2\}$ are the set of Dirichlet characters modulo $8$.

Let $p$ be an odd prime number. We define a map $\chi_p:\mathbb{Z} \rightarrow \mathbb{C}$ by $\chi_p(m) = \left(\frac{m}{p}\right)$. We call $\chi_p$ the quadratic residue character modulo $p$.

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Let $\{p_1, p_2, . . . , p_r\}$ be the set of odd prime divisors of $D$. We allocate a finite sequence of elements of $\{\chi_{p_1},\dots,\chi_{p_r},\psi_1, \psi_2, \psi_1\psi_2\}$ to each of the following cases as follows.

1) $D ≡ 1$ (mod $4$): $\chi_{p_1},\dots,\chi_{p_r}$

2) $D ≡ 0$ (mod $4$) and $D/4\equiv 0$ (mod $8$): $\chi_{p_1},\dots,\chi_{p_r}, \psi_1, \psi_2$

3) $D ≡ 0$ (mod $4$) and $D/4\equiv 1, 5$ (mod $8$): $\chi_{p_1},\dots,\chi_{p_r}$

4) $D ≡ 0$ (mod $4$) and $D/4\equiv 2$ (mod $8$): $\chi_{p_1},\dots,\chi_{p_r}, \psi_2$

5) $D ≡ 0$ (mod $4$) and $D/4\equiv 3, 4, 7$ (mod $8$): $\chi_{p_1},\dots,\chi_{p_r}, \psi_1$

6) $D ≡ 0$ (mod $4$) and $D/4\equiv 6$ (mod $8$): $\chi_{p_1},\dots,\chi_{p_r}, \psi_1\psi_2$

We call each of these sequences the system of genus characters of discriminant $D$.

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 $\Phi_1,\dots,\Phi_{\mu}$ be the system of genus characters of discriminant $D$. Let $f = ax^2 + bxy + cy^2$ be a primitive binary quadratic form of discriminant $D$. If $D < 0$, we assume $a > 0$. By this question, there exists an integer $m$ which is represented by $f$ and gcd($m, D) = 1$. Then $\Phi_1(m),\dots,\Phi_{\mu}(m)$ do not depend on the choice of $m$.

Remark The notion of the system of genus characters is due to Gauss(Disquisitiones Arithmeticae, art.230).

A related question

Answer 1

If $D \equiv 1$ (mod $4$), the assertion of the proposition is clear from this question. So we may assume $D \equiv 0$ (mod $4$). Let $m, k$ be odd integers which can be represented by $f$. By this question, we have the following results.

1) If $D/4 \equiv 0$ (mod $8$), $mk \equiv 1$ (mod $8$).

2) If $D/4 \equiv 1, 5$ (mod $8$), $mk \equiv 1, 3, 5, 7$ (mod $8$).

3) If $D/4 \equiv 2$ (mod $8$), $mk \equiv 1, 7$ (mod $8$).

4) If $D/4 \equiv 3, 4, 7$ (mod $8$), $mk \equiv 1, 5$ (mod $8$).

5) If $D/4 \equiv 6$ (mod 8), $mk \equiv 1, 3$ (mod $8$).

We would like to describe these results using $\psi_1$ and $\psi_2$. We note that

$\psi_1(１) = 1$, $\psi_1(3) = -1$, $\psi_1(5) = 1$, $\psi_1(7) = -1$.

$\psi_2(１) = 1$, $\psi_2(3) = -1$, $\psi_2(5) = -1$, $\psi_2(7) = 1$.

Hence we get:

a) $mk \equiv 1$ (mod $8$) if and only if $\psi_1(mk) = \psi_2(mk) = 1$.

b) $mk \equiv 1, 7$ (mod $8$) if and only if $\psi_2(mk) = 1$.

c) $mk \equiv 1, 5$ (mod $8$) if and only if $\psi_1(mk) = 1$.

d) $mk \equiv 1, 3$ (mod $8$) if and only if $\psi_1(mk)\psi_2(mk) = 1$.

The assertion of the proposition follows immediately.

Answer 2

I think for some special cases, it is written formula. At least not hurt.

Though it is necessary to bring the decisions some pretty simple solutions:

the equation: $aX^2+bXY+cY^2=f$

If the root of the whole: $\sqrt{\frac{f}{a+b+c}}$

Then use the solution of Pell's equation: $p^2-(b^2-4ac)s^2=1$

Solutions can be written:

$Y=((4a+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$

$X=(-(4c+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$

If a root: $\sqrt{fa}$ then the solutions are of the form:

$Y=4ps\sqrt{fa}$

$X=(-2bps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a}}$

Although it should be mentioned, and the equation: $aX^2-qY^2=f$

If the root of the whole: $\sqrt{\frac{f}{a-q}}$

Using equation Pell: $p^2-aqs^2=1$ solutions can be written:

$Y=(2aps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$

$X=(2qps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$

And for that decision have to find double formula.

$Y_2=Y+2as(qsY-pX)$

$X_2=X+2p(qsY-pX)$

Well it is possible for such unpretentious.

For a private quadratic form:

$Y^2=aX^2+bX+1$

Using solutions of Pell's equation: $p^2-as^2=1$

Solutions can be expressed through them is quite simple.

$Y=p^2+bps+as^2$

$X=2ps+bs^2$

$p,s$ - can be any character.