A condition for an odd integer to be properly represented by a primitive binary quadratic form

Question

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 a non-square integer and gcd$(a, b, c) = 1$, we say $f$ 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 $F$. If $m = ax^2 + bxy + cy^2$ has a solution $(s, t)$ such that gcd$(s, t) = 1$, we say $m$ is properly represented by $f$.

Suppose $m$ is properly represented by $ax^2 + bxy + cy^2$. Then $D$ is a quadratic residue modulo $m$ by this question.

Now I would like to ask if the converse holds. Namely, is the following proposition correct? 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 $m$ be an odd integer such that gcd$(m, D) = 1$. Suppose $D$ is a quadratic residue modulo $m$. Then there exists a primitive form $f = ax^2 + bxy + cy^2$ of discriminant $D$ and $m$ is properly represented by $f$.

Answer 1

Since $D$ is a quadratic residue modulo $m$, there exists an integer $b$ such that $D \equiv b^2 \pmod m$. Since $m$ is odd, $b + m$ is odd when $b$ is even. Similarly $b + m$ is even when $b$ is odd. Hence we may assume that $D$ and $b$ have the same parity. Then

If $D \equiv 0 \pmod 4$, $D - b^2 \equiv 0 \pmod 4$.

If $D \equiv 1 \pmod 4$, $D - b^2 \equiv 0 \pmod 4$.

In either case, $D \equiv b^2 \pmod {4m}$. Hence there exists an integer $c$ such that $b^2 - D = 4mc$. Then the discriminant of the form $f = mx^2 + bxy + cy^2$ is $D$. Let $d = \gcd (m, b, c)$. Since $m$ and $D = b^2 - 4mc$ are both divisible by $d$, $d$ must be $1$. Hence $f$ is primitive. Since $m = f(1, 0)$, $m$ is properly represented by $f$.