1.
Let $n$ be a positive odd integer.
Let $a, b$ be integers such that $a \equiv b$ (mod $n$).
Then
$$\left(\frac{a}{n}\right) = \left(\frac{b}{n}\right)$$
Proof:
Let $n = \prod p$ be the prime decomposition of $n$ with every prime factor repeated according to its multiplicity.
Since $a \equiv b$ (mod $p$) for every prime factor of $n$, $\left(\frac{a}{p}\right) = \left(\frac{b}{p}\right)$.
Hence $\prod \left(\frac{a}{p}\right) = \prod \left(\frac{b}{p}\right)$.
Hence $\prod \left(\frac{a}{n}\right) = \prod \left(\frac{b}{n}\right)$.
2.
Let $a$ be integers.
Let $m, n$ be positive odd integers.
Then
$$\left(\frac{a}{mn}\right) = \left(\frac{a}{m}\right)\left(\frac{a}{n}\right)$$
Proof:
Clear from the definition of Jacobi symbol.
3.
Let $a, b$ be integers.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{ab}{n}\right) = \left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$$
Proof:
Let $n = \prod p$ be the prime decomposition of $n$ as in the proof of $1.$
$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$ for every prime factor $p$ of $n$.
Hence $\prod \left(\frac{ab}{p}\right) = \prod \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$.
Hence $\left(\frac{ab}{n}\right) = \left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$.
Lemma 1
Let $a, b$ be odd integers.
Then $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Proof:
Since $a - 1$ and $b - 1$ are even,
$(a - 1)(b - 1) \equiv 0$ (mod $4$).
Hence $ab - a - b + 1 \equiv 0$ (mod $4$).
Hence $ab - 1 \equiv (a - 1) + (b - 1)$ (mod $4$).
Hence $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Lemma 2
Let $a, b$ be odd integers.
Then $(a^2b^2 - 1)/8 \equiv (a^2 - 1)/8 + (b^2 - 1)/8$ (mod $2$).
Proof:
$a^2 - 1 \equiv 0$ (mod $4$).
$b^2 - 1 \equiv 0$ (mod $4$).
Hence
$(a^2 - 1)(b^2 - 1) \equiv 0$ (mod $16$).
Hence
$a^2b^2 - a^2 - b^2 + 1 \equiv 0$ (mod $16$).
Hence $a^2b^2 - 1 \equiv (a^2 - 1) + (b^2 - 1)$ (mod $16$).
Hence
$(a^2b^2 - 1)/8 \equiv (a^2 - 1)/8 + (b^2 - 1)/8$ (mod $2$).
Lemma 3
Let $a, b, c$ be positive odd integers.
Suppose
$\left(\frac{a}{c}\right) \left(\frac{c}{a}\right) = (-1)^{\frac{a-1}{2}\frac{c-1}{2}}$.
$\left(\frac{b}{c}\right) \left(\frac{c}{b}\right) = (-1)^{\frac{b-1}{2}\frac{c-1}{2}}$.
Then
$\left(\frac{ab}{c}\right) \left(\frac{c}{ab}\right) = (-1)^{\frac{ab-1}{2}\frac{c-1}{2}}$.
Proof:
$\left(\frac{ab}{c}\right) \left(\frac{c}{ab}\right) =\left(\frac{a}{c}\right) \left(\frac{c}{a}\right)\left(\frac{b}{c}\right) \left(\frac{c}{b}\right) = (-1)^{\frac{a-1}{2}\frac{c-1}{2} + \frac{b-1}{2}\frac{c-1}{2}}
= (-1)^{(\frac{a-1}{2} + \frac{b-1}{2})\frac{c-1}{2}}
= (-1)^{\frac{ab-1}{2}\frac{c-1}{2}}$.
The first equality follows from $2.$ and $3.$
The last equality follows from Lemma 1.
4.
Let $m, n$ be positive odd integers such that gcd$(m,n) = 1$.
Then
$$\left(\frac{m}{n}\right) \left(\frac{n}{m}\right) = (-1)^{\frac{m-1}{2}\frac{n-1}{2}}$$.
Proof:
This follows immediately from Quadratic Reciprocity Theorem using Legendre symbol and Lemma 3.
5.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$$.
Proof:
This follows immediately from the first supplementary law of quadratic reciprocity using Legendre symbol and $2.$ and Lemma 1.
6.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{2}{n}\right) = (-1)^{\frac{n^2-1}{8}}$$.
Proof:
This follows immediately from the second supplementary law of quadratic reciprocity using Legendre symbol and $2.$ and Lemma 2.
