= Solution
\b[<Herbrand quotient> and local norm indices.] Let $G=\langle\sigma\rangle$ have order $n$. For an additive $G$-module $M$, put $D=\sigma-1$ and $N=1+\sigma+\cdots+\sigma^{n-1}$. The relevant <Tate cohomology of a cyclic group> groups are
$$
\widehat H^0(G,M)=M^G/NM,\qquad \widehat H^{-1}(G,M)=\ker N/DM.
$$
If both groups are finite, their size ratio is the <Herbrand quotient>
$$
h_G(M)=\frac{\#\widehat H^0(G,M)}{\#\widehat H^{-1}(G,M)}.
$$
Multiplicative modules use products for $N$ and $D(m)=\sigma(m)/m$. A finite module has quotient one: $\#\ker N=\#M/\#NM$, while $\#DM=\#M/\#M^G$, and the two cohomology orders are equal. The six-term periodic cohomology sequence shows $h(B)=h(A)h(C)$ for a short exact sequence $0\to A\to B\to C\to0$ when these groups are finite. Thus the quotient is unchanged by finite-index changes of lattices. For the trivial $G$-module $\mathbb Z$, the zeroth group is $\mathbb Z/n\mathbb Z$ and the negative first group is zero, so $h_G(\mathbb Z)=n$.
Now let $L/K$ be a cyclic extension of <p-adic fields> of degree $n$. For a sufficiently deep <principal unit> subgroup $U_L^r=1+\mathfrak m_L^r$, the <p-adic logarithm> is a $G$-equivariant isomorphism with the additive <p-adic lattice> $\mathfrak m_L^r$; it suffices to take $r>v_L(p)/(p-1)$. The <normal basis theorem> makes $L$ a regular $K[G]$-module. Consequently $\mathfrak m_L^r$, as a $\mathbb Z_p[G]$-lattice, is commensurable with a direct sum of $[K:\mathbb Q_p]$ copies of $\mathbb Z_p[G]$. These regular lattices have zero Tate groups: its invariants are the multiples of the sum of the basis elements, every such element is a norm, and vectors with coefficient sum zero are images of $D$. Commensurability and the finite-module calculation give $h_G(U_L^r)=1$. Since $U_L/U_L^r$ is finite, $h_G(U_L)=1$. These comparisons also establish finiteness of the Tate groups concerned.
The valuation exact sequence $1\to U_L\to L^\times\to\mathbb Z\to0$ now gives $h_G(L^\times)=n$. <Hilbert theorem 90> makes $\widehat H^{-1}(G,L^\times)$ trivial. One can prove the cyclic statement directly: for $a$ of norm one set $A_0=1$, $A_i=a\sigma(a)\cdots\sigma^{i-1}(a)$ and choose $c$ for which $b=\sum_i A_i\sigma^i(c)\ne0$. Such a $c$ exists by linear independence of distinct field automorphisms. Then $\sigma(b)=b/a$, so $a=b/\sigma(b)$ is a coboundary. Therefore \b[the local cyclic norm index is]
$$
\boxed{[K^\times:N_{L/K}L^\times]=[L:K]=n.}
$$
This is the central use of the <Herbrand quotient>: it calculates a norm index without first constructing the local reciprocity map.
Let $e,f$ be the <ramification index> and <residue degree>. With normalized integer valuations, $v_K(Nx)=f v_L(x)$. A norm is a unit exactly when its preimage is a unit, and the norm valuations fill $f\mathbb Z$. Hence there is an exact sequence
$$
1\to\mathcal O_K^\times/N\mathcal O_L^\times\to K^\times/NL^\times\to\mathbb Z/f\mathbb Z\to0.
$$
\b[The unit norm index is $\boxed{[\mathcal O_K^\times:N\mathcal O_L^\times]=e}$.] In an unramified extension all units are norms and the obstruction is the valuation modulo $n$; in a totally ramified cyclic extension the entire index comes from units. The norm subgroup is open: on deep <principal units>, logarithm carries the norm to the <field trace>, and the trace of a full <p-adic lattice> contains a sufficiently deep <p-adic lattice> in $K$.
\b[<Hilbert norm residue symbol> and the local-to-global principle.] More generally, for a local field $F$ containing $\mu_m$, fix the local reciprocity map $\operatorname{rec}_F$ with uniformizers acting as arithmetic Frobenius on unramified extensions. The <Hilbert norm residue symbol> is
$$
(a,b)_{F,m}=\frac{\operatorname{rec}_F(b)(a^{1/m})}{a^{1/m}}\in\mu_m.
$$
This is independent of the chosen root, is bilinear, and has value one exactly when $b$ is a norm from $F(a^{1/m})$. These are standard consequences of <Local Artin reciprocity>. The case $m=2$, for which the values are signs, is the one directly governing <quadratic forms>. The <quadratic Hilbert symbol> at a place $v$ is defined for $a,b\in K_v^\times$ by
$$
(a,b)_v=\begin{cases}1,&b\in N_{K_v(\sqrt a)/K_v}(K_v(\sqrt a)^\times),\\-1,&\text{otherwise},\end{cases}
$$
with value one for all $b$ when $a$ is a square. Equivalently it is one precisely when $z^2=ax^2+by^2$ has a nonzero solution over $K_v$. For nonsquare $a$, a solution has $y\ne0$ and gives $b=(z/y)^2-a(x/y)^2$; the converse follows from the same norm identity. When $a$ is square the conic is already isotropic. Symmetry follows from this conic criterion. The local cyclic norm index gives a norm subgroup of index two; its sign character is multiplicative in $b$, and symmetry gives multiplicativity in $a$. Thus the symbol is a nondegenerate bilinear pairing on the <square-class group of a field>, since every nonsquare $a$ gives a nontrivial norm character. Also $(a,-a)_v=1$, because $N(\sqrt a)=-a$, and $(a,a)_v=(a,-1)_v$.
For an odd-residue-characteristic <p-adic field> with residue size $q$, write $a=\pi^r u$, $b=\pi^s w$, and let $\chi$ be the quadratic character of the residue units. Then
$$
\boxed{(a,b)=(-1)^{rs(q-1)/2}\chi(\overline u)^s\chi(\overline w)^r.}
$$
The unramified <quadratic extension> has every unit as a norm and only even norm valuations; in a ramified <quadratic extension> the norm of a unit has square residue. The preceding unit norm index is two, so the square-residue condition in the ramified case is also sufficient. These facts, together with $N(\sqrt\pi)=-\pi$, determine the formula on the generators $u,\pi$ of the <square-class group>. At a real place the symbol is negative exactly when both arguments are negative, and at a complex place it is always one. For $\mathbb Q_2$, with $u,w$ odd units, the dyadic formula is
$$
(a,b)_2=(-1)^{\frac{u-1}{2}\frac{w-1}{2}+r\frac{w^2-1}{8}+s\frac{u^2-1}{8}}.
$$
Only residue classes modulo eight and the parities of $r,s$ enter; other dyadic fields retain the norm definition.
The <Hilbert reciprocity law> states $\prod_v(a,b)_v=1$ for $a,b\in K^\times$. Only finitely many factors can be nontrivial: outside the places above two, the Archimedean places, and the finite places where $a$ or $b$ is not a unit, the odd-residue formula gives one. For $K=\mathbb Q$ this product formula is a formulation of <quadratic reciprocity>, including its supplementary laws. It forces local norm obstructions to occur with compatible parity. It is a necessary compatibility law, not by itself a substitute for the following local-to-global theorem.
The <Hasse-Minkowski theorem> states that a nondegenerate <quadratic form> over a <number field> has a nonzero isotropic vector if and only if it does over every completion. Equivalently, two nondegenerate <quadratic forms> are isometric globally if and only if they are isometric at every place. The forward directions are immediate; the reverse directions are the substantive global theorem. Over non-Archimedean completions, local isometry classes are determined by dimension, determinant square class and the <Hasse invariant of a quadratic form>
$$
\epsilon_v(\langle a_1,\ldots,a_m\rangle)=\prod_{i<j}(a_i,a_j)_v.
$$
At real places one uses signature, and at complex places dimension suffices. Reciprocity gives $\prod_v\epsilon_v(q)=1$ for a globally diagonalized form. These invariants make the theorem practically usable: local square classes and norm characters replace an unrestricted search for rational solutions. In particular every <quadratic form> of dimension at least five over a <p-adic field> is isotropic; for such a form over a <number field> the only isotropy obstructions are definite signatures at real places.
As a concrete application, take nonsquare $a\in K^\times$. Then $b$ is a global norm from $K(\sqrt a)$ if and only if $(a,b)_v=1$ at every place. The local conditions make the ternary form $\langle1,-a,-b\rangle$ isotropic everywhere. <Hasse-Minkowski theorem> gives a global solution of $z^2-a x^2=b y^2$; since $a$ is nonsquare, $y$ cannot be zero, and division by $y$ gives the global norm. For example $-1$ fails to be a norm from $\mathbb Q(i)$ already at the real place. \b[Thus the <quadratic Hilbert symbol> detects local norm solvability, while the <Hasse-Minkowski theorem> turns solvability at all places into a global quadratic solution.] The reciprocity and local-to-global theorems in this essay are stated as standard results; the norm interpretation, bilinearity and application are derived above.
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