= Solution
For a <local field> $F$ containing the Nth <roots of unity>, fix the <Local Artin map> with arithmetic <Frobenius> convention. The <Hilbert norm residue symbol> can be defined by
$$
(a,b)_{F,N}=\frac{\operatorname{rec}_F(b)(a^{1/N})}{a^{1/N}}\in\mu_N.
$$
It is independent of the chosen root, continuous and bimultiplicative, factors through $F^\times/(F^\times)^N$ in each variable, and gives a nondegenerate pairing on that finite <group>. It satisfies $(a,b)_{F,N}=(b,a)_{F,N}^{-1}$ and $(a,1-a)_{F,N}=1$ for $a\ne1$. Its <norm> interpretation is
$$
(a,b)_{F,N}=1\iff b\in N_{F(a^{1/N})/F}F(a^{1/N})^\times.
$$
For $N=2$ the <quadratic Hilbert symbol> is symmetric and sign-valued. It is also characterized by the existence of a nonzero solution of $z^2=ax^2+by^2$. Moreover $(a,-a)_F=1$, and therefore $(a,a)_F=(a,-1)_F$. We now compute both the <square-class groups> and the pairing explicitly.
For odd $p$, write $a=p^\alpha u$ with $u\in\mathbb Z_p^\times$. An element is a square exactly when $\alpha$ is even and the residue of $u$ is a square in $\mathbb F_p^\times$. Necessity follows by reduction; sufficiency follows from the simple-root form of <Hensel lemma> applied to $X^2-u$, since $2x$ is then a unit. If $\eta$ is any unit with nonsquare residue, this gives
$$
\boxed{\mathbb Q_p^\times/(\mathbb Q_p^\times)^2\cong C_2\times C_2,\qquad\text{representatives }1,\eta,p,\eta p\quad(p\text{ odd}).}
$$
The nonsquare unit defines an <unramified> quadratic extension. Its <field norms> include every unit and have even <valuation>, as follows from the local norm calculation below. Thus $(\eta,u)_p=1$ for every unit, but $(\eta,p)_p=-1$. Finally $(p,p)_p=(p,-1)_p=(-1/p)$, by symmetry and the preceding unit calculation. Bimultiplicativity now proves <explicit quadratic Hilbert symbols over the p-adic numbers>:
$$
\boxed{(p^\alpha u,p^\beta v)_p=(-1)^{\alpha\beta(p-1)/2}\left(\frac up\right)^\beta\left(\frac vp\right)^\alpha\quad(p\text{ odd}).}
$$
Here the factors are <Legendre symbols> of the nonzero residues. The formula works for arbitrary integer <valuations> $\alpha,\beta$, since only their parity matters.
At $p=2$, the square of every odd integer is one modulo eight. Conversely, for an odd $u\equiv1\pmod8$, apply the <strong form of Hensel lemma> to $X^2-u$ at $1$: the error has <valuation> at least three, while the <derivative> has <valuation> one. Thus the <unit square classes of the p-adic integers> are precisely their four residues modulo eight. Combining this with parity of the <valuation> gives
$$
\boxed{\mathbb Q_2^\times/(\mathbb Q_2^\times)^2\cong C_2^3,\qquad\text{basis }[-1],[2],[5],\quad\text{representatives }\pm1,\pm2,\pm5,\pm10.}
$$
For an odd unit put
$$
\epsilon(u)=\frac{u-1}{2}\pmod2,\qquad\omega(u)=\frac{u^2-1}{8}\pmod2.
$$
Inspection of the residues $1,3,5,7$ gives $[u]=[-1]^{\epsilon(u)}[5]^{\omega(u)}$. It remains to compute six basis pairings. Explicit <field norms> give
$$
(-1,2)_2=(-1,5)_2=(2,2)_2=(5,5)_2=1:
$$
indeed $2=1^2+1^2$, $5=1^2+2^2$, $2=2^2-2\cdot1^2$, and $5=5^2-5\cdot2^2$.
In contrast $(-1,-1)_2=-1$. To justify the obstruction even for rational 2-adic coordinates, write $x=2^t x_0$, $y=2^t y_0$ with $x_0,y_0$ integral and at least one odd. Then $v_2(x_0^2+y_0^2)$ is zero if exactly one is odd, and one if both are odd. A sum of squares equal to the unit $-1$ consequently forces $t=0$ and exactly one coordinate odd. Its residue modulo four would then be one, whereas $-1$ is three.
Similarly $(2,5)_2=-1$. If $x^2-2y^2$ is a unit, the <valuations> of the two terms have opposite parity, so cannot cancel. It follows that $x$ is an odd integral element and $y$ is integral. Modulo eight its <norm> is then $1$ or $7$, never $5$. These signs, symmetry, and bimultiplicativity give
$$
\boxed{(2^\alpha u,2^\beta v)_2=(-1)^{\epsilon(u)\epsilon(v)+\alpha\omega(v)+\beta\omega(u)}.}
$$
The two boxed formulas determine the <quadratic Hilbert symbol> on all nonzero elements, including every rational prime $p$.
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