[proofplan]
We prove the three identities directly from the standard interpretation of the Hilbert symbol. Symmetry and invariance under multiplying the second argument by a square follow from the defining quadratic equation for the Hilbert symbol. Multiplicativity in the second argument follows from the norm criterion for the Hilbert symbol: for fixed $a$, the symbol detects whether an element lies in the norm subgroup of the quadratic algebra $\mathbb{Q}_v(\sqrt a)$. This norm subgroup is the kernel of the corresponding quadratic character, so the character is multiplicative.
[/proofplan]
[step:Fix the local field and recall the defining criterion]
Let
\begin{align*}
F := \mathbb{Q}_v
\end{align*}
be the local field attached to the place $v$. For $a,b \in F^\times$, the Hilbert symbol is defined by
\begin{align*}
(a,b)_v =
\begin{cases}
1, & \text{if there exists } (x,y,z) \in F^3 \setminus \{(0,0,0)\} \text{ with } z^2 = ax^2 + by^2,\\
-1, & \text{otherwise.}
\end{cases}
\end{align*}
We also use the standard norm criterion for the Hilbert symbol: for fixed $a \in F^\times$, if $K_a$ denotes the quadratic $F$-algebra
\begin{align*}
K_a := F[t]/(t^2-a),
\end{align*}
and if
\begin{align*}
N_{K_a/F}: K_a^\times \to F^\times
\end{align*}
denotes its norm map, then for every $b \in F^\times$,
\begin{align*}
(a,b)_v = 1 \iff b \in N_{K_a/F}(K_a^\times).
\end{align*}
Equivalently, the map
\begin{align*}
\chi_a: F^\times &\to \{-1,1\}\\
b &\mapsto (a,b)_v
\end{align*}
is the quadratic character with kernel $N_{K_a/F}(K_a^\times)$, with the convention that $\chi_a$ is the trivial character when $a \in (F^\times)^2$.
(citing a result not yet in the wiki: Norm Criterion for the Hilbert Symbol)
[/step]
[step:Swap the two coefficients in the defining quadratic equation]
Let $a,b \in F^\times$. By definition, $(a,b)_v = 1$ exactly when there exists $(x,y,z) \in F^3 \setminus \{(0,0,0)\}$ such that
\begin{align*}
z^2 = ax^2 + by^2.
\end{align*}
The map
\begin{align*}
F^3 &\to F^3\\
(x,y,z) &\mapsto (y,x,z)
\end{align*}
is a bijection from the set of nonzero triples satisfying $z^2 = ax^2 + by^2$ to the set of nonzero triples satisfying
\begin{align*}
z^2 = bx^2 + ay^2.
\end{align*}
Therefore the first equation has a nonzero solution if and only if the second equation has a nonzero solution. Hence
\begin{align*}
(a,b)_v = (b,a)_v.
\end{align*}
[/step]
[step:Absorb a square factor by rescaling one variable]
Let $a,b,c \in F^\times$. By definition, $(a,bc^2)_v = 1$ exactly when there exists $(x,y,z) \in F^3 \setminus \{(0,0,0)\}$ such that
\begin{align*}
z^2 = ax^2 + bc^2y^2.
\end{align*}
Define the invertible change of variables
\begin{align*}
T_c: F^3 &\to F^3\\
(x,y,z) &\mapsto (x,cy,z).
\end{align*}
Since $c \in F^\times$, the map $T_c$ is a bijection of $F^3$ preserving the property of being nonzero. Under this substitution, the equation becomes
\begin{align*}
z^2 = ax^2 + b(cy)^2.
\end{align*}
Thus the equation defining $(a,bc^2)_v = 1$ has a nonzero solution if and only if the equation defining $(a,b)_v = 1$ has a nonzero solution. Therefore
\begin{align*}
(a,bc^2)_v = (a,b)_v.
\end{align*}
[/step]
[step:Use the norm character to multiply the second argument]
Fix $a \in F^\times$. Define
\begin{align*}
\chi_a: F^\times &\to \{-1,1\}\\
b &\mapsto (a,b)_v.
\end{align*}
By the norm criterion recalled above, $\chi_a$ is a group homomorphism from $F^\times$ to $\{-1,1\}$. Hence, for all $b_1,b_2 \in F^\times$,
\begin{align*}
(a,b_1b_2)_v
= \chi_a(b_1b_2)
= \chi_a(b_1)\chi_a(b_2)
= (a,b_1)_v(a,b_2)_v.
\end{align*}
This proves multiplicativity in the second variable.
[/step]
[step:Collect the three identities]
The previous steps prove, for arbitrary $a,b,b_1,b_2,c \in F^\times = \mathbb{Q}_v^\times$,
\begin{align*}
(a,b)_v &= (b,a)_v,\\
(a,bc^2)_v &= (a,b)_v,\\
(a,b_1b_2)_v &= (a,b_1)_v(a,b_2)_v.
\end{align*}
Since $v$ was an arbitrary place of $\mathbb{Q}$, the identities hold for every place $v$ of $\mathbb{Q}$.
[/step]
