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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 byIt 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 $a\ne 1$. Its norm interpretation isFor $N=2$ the quadratic Hilbert symbol is symmetric and sign-valued. It is also characterized by the existence of a nonzero solution of $z^2=a{x}^2+b{y}^2$. 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 givesThe 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: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 $u\equiv 1(\mathrm{mod}8)$, 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 givesFor an odd unit putInspection of the residues $1,3,5,7$ gives $[u]=[-1{]}^{ϵ(u)}[5{]}^{\omega (u)}$. It remains to compute six basis pairings. Explicit field norms giveindeed $2=1^2+1^2$, $5=1^2+2^2$, $2=2^2-2\cdot 1^2$, and $5=5^2-5\cdot 2^2$.

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 giveThe two boxed formulas determine the quadratic Hilbert symbol on all nonzero elements, including every rational prime $p$.