It suffices to prove the following lemmas.
Lemma 1 Let $D \gt 0$ be a positive integer such that $D \equiv 1$ (mod $4$). Let $a$ and $b$ be positive odd integers such that $a \equiv b$ (mod $D$). Then $\left(\frac{D}{a}\right) = \left(\frac{D}{b}\right)$
Proof: Since $a \equiv b$ (mod $D$), if gcd$(a, D) \ne 1$, then gcd$(b, D) \ne 1$. Hence we can assume that gcd$(a, D) = 1$ and gcd$(b, D) = 1$.
By 4. of this question,
$\left(\frac{D}{a}\right) \left(\frac{a}{D}\right) = (-1)^{\frac{D-1}{2}\frac{a-1}{2}}$.
$\left(\frac{D}{b}\right) \left(\frac{b}{D}\right) = (-1)^{\frac{D-1}{2}\frac{b-1}{2}}$.
Since $D \equiv 1$ (mod $4$), $(D - 1)/2 \equiv 0$ (mod $2$). Hence
$\left(\frac{D}{a}\right) = \left(\frac{a}{D}\right)$.
$\left(\frac{D}{b}\right) = \left(\frac{b}{D}\right)$.
Hence $\left(\frac{D}{a}\right) = \left(\frac{D}{b}\right)$ by 1. of this question
Lemma 2 Let $D \lt 0$ be a negative integer such that $D \equiv 1$ (mod $4$). Let $a$ and $b$ be positive odd integers such that $a \equiv b$ (mod $D$). Then $\left(\frac{D}{a}\right) = \left(\frac{D}{b}\right)$
Proof: Since $a \equiv b$ (mod $D$), if gcd$(a, D) \ne 1$, then gcd$(b, D) \ne 1$. Hence we can assume that gcd$(a, D) = 1$ and gcd$(b, D) = 1$.
By 3. of this question, $\left(\frac{D}{a}\right) = \left(\frac{-1}{a}\right)\left(\frac{-D}{a}\right)$.
By 5. of this question, $\left(\frac{-1}{a}\right) = (-1)^{\frac{a-1}{2}}$.
By 4. of this question, $\left(\frac{-D}{a}\right) \left(\frac{a}{-D}\right) = (-1)^{\frac{-D-1}{2}\frac{a-1}{2}}$.
Since $-D \equiv 3$ (mod $4$), $(-D - 1)/2 \equiv 1$ (mod $2$).
Hence $\left(\frac{-D}{a}\right) \left(\frac{a}{-D}\right) = (-1)^{\frac{a-1}{2}}$.
Hence $\left(\frac{-D}{a}\right) = (-1)^{\frac{a-1}{2}} \left(\frac{a}{-D}\right)$.
Hence, $\left(\frac{D}{a}\right) = \left(\frac{-1}{a}\right)\left(\frac{-D}{a}\right) = \left(\frac{a}{-D}\right)$.
Similarly $\left(\frac{D}{b}\right) = \left(\frac{-1}{b}\right)\left(\frac{-D}{b}\right) = \left(\frac{b}{-D}\right)$.
Since $a \equiv b$ (mod $D$), $\left(\frac{a}{-D}\right) = \left(\frac{b}{-D}\right)$ by 1. of this question. Hence $\left(\frac{D}{a}\right) = \left(\frac{D}{b}\right)$
Lemma 3 Let $D \gt 0$ be a positive integer such that $D \equiv 0$ (mod $4$). There exists an integer $\alpha \ge 2$ such that $D = (2^\alpha)m$, where $m$ is a positive odd integer. Suppose $\alpha$ is even. Let $a$ and $b$ be positive odd integers such that $a \equiv b$ (mod $D$). Then $\left(\frac{D}{a}\right) = \left(\frac{D}{b}\right)$
Proof: Since $a \equiv b$ (mod $D$), if gcd$(a, D) \ne 1$, then gcd$(b, D) \ne 1$. Hence we can assume that gcd$(a, D) = 1$ and gcd$(b, D) = 1$.
By 3. of this question, $\left(\frac{D}{a}\right) = \left(\frac{2}{a}\right)^\alpha\left(\frac{m}{a}\right)$.
Since $\alpha$ is even,
$\left(\frac{D}{a}\right) = \left(\frac{m}{a}\right)$.
$\left(\frac{D}{b}\right) = \left(\frac{m}{b}\right)$.
Hence it suffices to prove that $\left(\frac{m}{a}\right) = \left(\frac{m}{b}\right)$.
By 4. of this question,
$\left(\frac{m}{a}\right) \left(\frac{a}{m}\right) = (-1)^{\frac{m-1}{2}\frac{a-1}{2}}$.
$\left(\frac{m}{b}\right) \left(\frac{b}{m}\right) = (-1)^{\frac{m-1}{2}\frac{b-1}{2}}$.
Case 1: $m \equiv 1$ (mod $4$)
Since $(m - 1)/2 \equiv 0$ (mod $2$),
$\left(\frac{m}{a}\right) = \left(\frac{a}{m}\right)$.
$\left(\frac{m}{b}\right) = \left(\frac{b}{m}\right)$.
Since $a \equiv b$ (mod $D$), $a \equiv b$ (mod $m$),
Hence $\left(\frac{a}{m}\right) = \left(\frac{b}{m}\right)$ by 1. of this question
Hence $\left(\frac{m}{a}\right) = \left(\frac{m}{b}\right)$.
Case 2: $m \equiv 3$ (mod $4$)
Since $(m - 1)/2 \equiv 1$ (mod $2$),
$\left(\frac{m}{a}\right) \left(\frac{a}{m}\right) = (-1)^{\frac{a-1}{2}}$.
$\left(\frac{m}{b}\right) \left(\frac{b}{m}\right) = (-1)^{\frac{b-1}{2}}$.
Since $D \equiv 0$ (mod $4$) and $a \equiv b$ (mod $D$), $a \equiv b$ (mod $4$).
Hence ${\frac{a-1}{2}} \equiv (-1)^{\frac{b-1}{2}}$ (mod $2$).
Hence $(-1)^{\frac{a-1}{2}} = (-1)^{\frac{b-1}{2}}$.
Since $a \equiv b$ (mod $D$), $a \equiv b$ (mod $m$).
Hence $\left(\frac{a}{m}\right) = \left(\frac{b}{m}\right)$ by 1. of this question.
Hence $\left(\frac{m}{a}\right) = \left(\frac{m}{b}\right)$.
Lemma 4 Let $D \gt 0$ be a positive integer such that $D \equiv 0$ (mod $4$). There exists an integer $\alpha \ge 2$ such that $D = (2^\alpha)m$, where $m$ is a positive odd integer. Suppose $\alpha$ is odd. Let $a$ and $b$ be positive odd integers such that $a \equiv b$ (mod $D$). Then $\left(\frac{D}{a}\right) = \left(\frac{D}{b}\right)$
Proof: Since $a \equiv b$ (mod $D$), if gcd$(a, D) \ne 1$, then gcd$(b, D) \ne 1$. Hence we can assume that gcd$(a, D) = 1$ and gcd$(b, D) = 1$.
By 3. of this question, $\left(\frac{D}{a}\right) = \left(\frac{2}{a}\right)^\alpha\left(\frac{m}{a}\right)$.
Since $\alpha$ is odd,
$\left(\frac{D}{a}\right) = \left(\frac{2}{a}\right)\left(\frac{m}{a}\right)$.
$\left(\frac{D}{b}\right) = \left(\frac{2}{b}\right)\left(\frac{m}{b}\right)$.
By 6. of this question,
$\left(\frac{2}{a}\right) = (-1)^{\frac{a^2-1}{8}}$.
$\left(\frac{2}{b}\right) = (-1)^{\frac{b^2-1}{8}}$.
Since $\alpha \ge 3$ and $a \equiv b$ (mod $D$), $a \equiv b$ (mod $8$).
Hence $\left(\frac{2}{a}\right) = \left(\frac{2}{b}\right)$.
Hence it suffices to prove that $\left(\frac{m}{a}\right) = \left(\frac{m}{b}\right)$.
By 4. of this question,
$\left(\frac{m}{a}\right) = (-1)^{\frac{m-1}{2}\frac{a-1}{2}}\left(\frac{a}{m}\right)$.
$\left(\frac{m}{b}\right) = (-1)^{\frac{m-1}{2}\frac{b-1}{2}}\left(\frac{b}{m}\right)$.
Case 1: $m \equiv 1$ (mod $4$)
Since $(m - 1)/2 \equiv 0$ (mod $2$),
$\left(\frac{m}{a}\right) = \left(\frac{a}{m}\right)$.
$\left(\frac{m}{b}\right) = \left(\frac{b}{m}\right)$.
Since $a \equiv b$ (mod $D$), $a \equiv b$ (mod $m$).
Hence $\left(\frac{a}{m}\right) = \left(\frac{b}{m}\right)$ by 1. of this question
Hence $\left(\frac{m}{a}\right) = \left(\frac{m}{b}\right)$.
Case 2: $m \equiv 3$ (mod $4$)
Since $(m - 1)/2 \equiv 1$ (mod $2$),
$\left(\frac{m}{a}\right) = (-1)^{\frac{a-1}{2}}\left(\frac{a}{m}\right)$.
$\left(\frac{m}{b}\right) = (-1)^{\frac{b-1}{2}}\left(\frac{b}{m}\right)$.
Since $a \equiv b$ (mod $D$), $a \equiv b$ (mod $m$).
Hence $\left(\frac{a}{m}\right) = \left(\frac{b}{m}\right)$ by 1. of this question
Since $D \equiv 0$ (mod $4$) and $a \equiv b$ (mod $D$), $a \equiv b$ (mod $4$).
Hence $(-1)^{\frac{a-1}{2}} \equiv (-1)^{\frac{b-1}{2}}$ (mod $2$).
Hence $(-1)^{\frac{a-1}{2}} = (-1)^{\frac{b-1}{2}}$.
Hence $\left(\frac{m}{a}\right) = \left(\frac{m}{b}\right)$.
Lemma 5 Let $D \lt 0$ be a negative integer such that $D \equiv 0$ (mod $4$). Let $a$ and $b$ be positive odd integers such that $a \equiv b$ (mod $D$). Then $\left(\frac{D}{a}\right) = \left(\frac{D}{b}\right)$
Proof: Since $a \equiv b$ (mod $D$), if gcd$(a, D) \ne 1$, then gcd$(b, D) \ne 1$. Hence we can assume that gcd$(a, D) = 1$ and gcd$(b, D) = 1$.
There exists an integer $\alpha \ge 2$ such that $D = -(2^\alpha)m$, where $m$ is a positive odd integer.
By 3. of this question,
$\left(\frac{D}{a}\right) = \left(\frac{-1}{a}\right)\left(\frac{-D}{a}\right)$.
$\left(\frac{D}{b}\right) = \left(\frac{-1}{b}\right)\left(\frac{-D}{b}\right)$.
Since $-D \equiv 0$ (mod $4$), $\left(\frac{-D}{a}\right) = \left(\frac{-D}{b}\right)$ by Lemma 3 and Lemma 4. Hence it suffices to prove $\left(\frac{-1}{a}\right) = \left(\frac{-1}{b}\right)$.
By 5. of this question,
$\left(\frac{-1}{a}\right) = (-1)^{\frac{a-1}{2}}$.
$\left(\frac{-1}{b}\right) = (-1)^{\frac{b-1}{2}}$.
Since $D \equiv 0$ (mod $4$) and $a \equiv b$ (mod $D$), $a \equiv b$ (mod $4$).
Hence ${\frac{a-1}{2}} \equiv (-1)^{\frac{b-1}{2}}$ (mod $2$).
Hence $\left(\frac{-1}{a}\right) = \left(\frac{-1}{b}\right)$.