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For every positive integer $n$,Under Local Artin reciprocity, the cyclotomic character identifies the quotient with $(\mathbb{Z}/2^n\mathbb{Z}{)}^{\times }$: a uniformizer acts trivially on the totally ramified extension, while a 2-adic unit acts through its residue class modulo $2^n$.

For an odd integer $x=2a+1$, $x^2=1+4a(a+1)\equiv 1(\mathrm{mod}8)$. The same conclusion holds for every 2-adic unit by reduction modulo eight. Thus an odd square must be congruent to one modulo eight.

Conversely let $u\in {\mathbb{Z}}_2^{\times }$ with $u\equiv 1(\mathrm{mod}8)$. For digit lifting for square roots of 2-adic units, start with $x_3=1$. If $x_n$ is odd and $x_n^2\equiv u(\mathrm{mod}{2}^n)$, considerFor $n\ge 3$,since the remaining term is divisible by $2^{2n-2}$ and $2n-2\ge n+1$. The odd multiplier $x_n$ lets the choice of ${\epsilon }_n$ correct exactly the next binary digit. Thus $x_{n+1}^2\equiv u(\mathrm{mod}{2}^{n+1})$. The corrections have valuations tending to infinity, so completeness gives $x=\lim x_n\in {\mathbb{Z}}_2^{\times }$ with $x^2=u$.

The unit $3$ is congruent to one modulo two but is not a square, and $5$ is congruent to one modulo four but is not a square, by the necessary modulo-eight condition. Therefore

Every nonzero P-adic number in ${\mathbb{Q}}_2$ is uniquely $2^{m}u$ with $m\in \mathbb{Z}$ and $u$ a 2-adic unit. A square has even valuation, and the unit square criterion shows that the unit class is determined by its odd residue modulo eight. The four residues form $C_2\times C_2$, generated by $-1$ and $5$; the valuation parity supplies the third factor, generated by $2$. Hence the square-class group of the 2-adic numbers isIts eight classes are represented by $1,-1,5,-5,2,-2,10,-10$. They are distinct: the first four have even valuation and distinct unit residues modulo eight, while multiplication by $2$ gives odd valuation and the same four distinct unit residues.

In characteristic different from two, completing the square in a quadratic minimal polynomial makes every quadratic extension $K(\sqrt{d})$ for a nonsquare $d\in K$. If $K(\sqrt{d})=K(\sqrt{e})$, write $\sqrt{e}=a+b\sqrt{d}$ and square. The coefficient $2ab$ must vanish. Since $e$ is nonsquare, $b\ne 0$, so $a=0$ and $e/d=b^2$. Conversely a square ratio plainly gives the same field. Thus the seven quadratic extensions of the 2-adic numbers are exactly

The unique unramified extension in this list is ${\mathbb{Q}}_2(\sqrt{5})$. To verify this, $\alpha =(1+\sqrt{5})/2$ is integral and satisfies $X^2-X-1=0$. Its reduction $X^2+X+1$ is irreducible over ${\mathbb{F}}_2$, so its residue field contains ${\mathbb{F}}_4$. The standard degree formula for finite extensions of complete discretely valued fields with perfect residue field is $[L:K]=e(L/K)f(L/K)$. Here the field degree is two, so the residue degree is two and the ramification index is one.

Conversely an unramified quadratic extension has residue field ${\mathbb{F}}_4$. A root there of $X^2+X+1$ lifts by Hensel lemma to a root of $X^2-X-1$, whose derivative is a unit. For this lift $\beta$, $(2\beta -1{)}^2=5$, so the extension contains ${\mathbb{Q}}_2(\sqrt{5})$ and must equal it. Every other field in the list is therefore ramified.

For the requested ramification groups, choose $L={\mathbb{Q}}_2(\sqrt{2})$ and $\pi =\sqrt{2}$. We use uniqueness of the extension of the valuation to a finite extension, together with the degree formula just stated. Since $v_2(\pi )=1/2$, the extension has ramification index two and residue degree one. Normalize $v_L(\pi )=1$, giving $v_L(2)=2$. Every element is $a+b\pi$ with $a,b\in {\mathbb{Q}}_2$, and the two summands have valuations of opposite parity:Thus $a+b\pi$ is integral exactly when $a,b\in {\mathbb{Z}}_2$, proving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$ directly.

The nonidentity automorphism $\sigma$ negates $\pi$. For any integral $a+b\pi$,of valuation at least three, with equality when $b$ is a unit. By the definition $G_i=\{\sigma :v_L(\sigma x-x)\ge i+1\text{ for every }x\in {\mathcal{O}}_L\}$, the ramification in the quadratic extension generated by the square root of two is thereforeThe lower ramification break is two; the valuation used here is the integral normalization on $L$.

First reduce the polynomial modulo seven. At residues $0,1,\dots ,6$, its values are respectively $4,2,6,1,0,2,6$. Its only residue root is therefore $4$, and $f^{\prime }(4)=45\equiv 3(\mathrm{mod}7)$ is a unit. By Hensel lemma, this residue root lifts uniquely to the 7-adic integers; any other root would have to reduce to the same residue root. Thus there is exactly one root in ${\mathbb{Z}}_7$, congruent to $4$ modulo seven.

Modulo five the values at $0,1,\dots ,4$ are $4,2,1,2,1$, so there is no residue root and hence no root in ${\mathbb{Z}}_5$. Modulo three the only possible residue root is $-1$. But writing $x=-1+3t$ givesThus that residue root cannot lift even modulo nine, and there is no root in ${\mathbb{Z}}_3$. A repeated residue root alone would not establish this obstruction; the second congruence is needed.

Modulo two both residues are roots. The even residue is simple because $f^{\prime }(0)=-3$ is a 2-adic unit, so Hensel lemma gives exactly one even root. For any odd 2-adic integer, $x^3\equiv x(\mathrm{mod}4)$, whence $f(x)\equiv -2x\equiv 2(\mathrm{mod}4)$. No odd root can exist. Consequently the full answer is

The Hasse-Minkowski theorem says that a nondegenerate quadratic form over a number field has a nonzero isotropic vector over that field if and only if it has one over every completion. In particular, over $\mathbb{Q}$ one tests $\mathbb{R}$ and every ${\mathbb{Q}}_{\ell }$. For the present integral form, a rational isotropic vector can be multiplied by a common denominator to give an integer one. This is the Hasse-Minkowski principle.

Write $P=p$, $R=p+12$ and $S=p+24$. All three primes are odd and exceed three: $p=2$ or $3$ would make $p+12$ composite. The second supplementary law for quadratic reciprocity gives

Necessity at $P$. Scale any nonzero vector over ${\mathbb{Q}}_P$ so that its coordinates lie in ${\mathbb{Z}}_P$ and at least one is a unit. Reduction of its equation modulo $P$ gives $y^2=2z^2$, since twelve is invertible modulo $P$. If the Legendre symbol $(2/P)$ is $-1$, this forces $P\mid y,z$. The original equation then forces $P\mid x$, because its other two terms are divisible by $P^2$. This contradicts the normalization. Thus

Sufficiency at the coefficient primes. Suppose $(2/P)=1$. A nonzero solution modulo $P$ is $(x,y,z)=(0,\sqrt{2},1)$. Its $y$-derivative is nonzero modulo $P$, so fixing the other two coordinates and applying the Hensel lemma lifts it to ${\mathbb{Q}}_P$.

Since $S\equiv P(\mathrm{mod}8)$, the same Legendre symbol criterion gives $(2/S)=1$. Modulo $S$, the equation reduces to $y^2=2x^2$, and the vector $(1,\sqrt{2},0)$ is a simple zero in the $y$ coordinate. It lifts to ${\mathbb{Q}}_S$. Modulo $R$, the coefficients of $x^2$ and $z^2$ are respectively $-12$ and $12$, so $(1,0,1)$ is a zero with nonzero $x$-derivative. It lifts to ${\mathbb{Q}}_R$.

The remaining odd primes. If $\ell$ is odd and divides none of $PRS$, the reduced ternary quadratic form is nondegenerate. An elementary finite-field argument finds a zero with $z=1$: if its coefficients are $A,B,C$, the setsboth have $(\ell +1)/2$ elements, so intersect. The resulting vector has nonzero $z$-derivative $2C$, and the Hensel lemma lifts it. This isotropy of nondegenerate ternary quadratic forms over finite fields verifies every remaining odd completion without an extra congruence condition.

The dyadic and real places. Setting $y=1,z=2$ reduces the dyadic problem toThe right side is a 2-adic unit congruent to $-3-4=1$ modulo eight. The unit square classes of the p-adic integers computed above show that it is a square in ${\mathbb{Q}}_2$. This also explains the printed hint: its integer can be taken as $m=-(p+21)/2$. Over $\mathbb{R}$, the positive and negative coefficients make the form indefinite; for instance $x=0,z=1,y=\sqrt{S/R}$ gives a nonzero zero.

All completions now have a nonzero isotropic vector. The Hasse-Minkowski theorem, followed by clearing denominators, proves

A strong lifting lemma. Let $R$ be a complete discrete valuation ring, with fraction field $F$ and normalized discrete valuation $v$. A useful strong form of Hensel lemma says that for $f\in R[T]$ and $a\in R$ satisfyingthere is a root $b\in R$ withIf $f(a)=0$, take $b=a$. Otherwise the hypothesis ensures that $f^{\prime }(a)\ne 0$. The root is unique among $b$ satisfying $v(b-a)>v(f^{\prime }(a))$. In particular, a simple root of the reduction of $f$ lifts uniquely with its specified residue.

Here is a proof by Newton iteration over a valued field. Put $a_0=a$ andWrite $A_j=v(f(a_j))$ and $B=v(f^{\prime }(a_0))$. Taylor expansion over $R$ shows, inductively, that the increment has valuation $A_j-B>B$, that $v(f^{\prime }(a_{j+1}))=B$, and thatIndeed, the constant and linear terms of $f(a_{j+1})$ cancel, leaving terms of valuation at least twice that of the increment. The change in $f^{\prime }$ has valuation greater than $B$, so its valuation stays $B$. Thus all iterates lie in $R$ and form a Cauchy sequence. Completeness and continuity give a root $b$. The later increments have strictly larger valuations than the first, proving the displayed distance formula. For uniqueness, if $b,c$ are in the stated ball, Taylor expansion factorsThe second factor is nonzero, so two roots must coincide. This proves Hensel's lemma in the form needed here.

Odd residue characteristic. For odd $p$, a p-adic unit $u$ is a square exactly when its reduction is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction. For sufficiency, a nonzero residue root of $T^2-u$ has derivative $2T$ invertible modulo $p$, so the Hensel lemma lifts it. The multiplicative group of a finite field is cyclic, and its subgroup of squares has index two. ThereforeRepresentatives are $1$ and any unit whose reduction is a quadratic nonresidue.

Residue characteristic two. An odd integer has square congruent to one modulo eight, so a square 2-adic unit must lie in $1+8{\mathbb{Z}}_2$. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $f(T)=T^2-u$ and $a=1$. Then $v_2(f(1))\ge 3>2v_2(f^{\prime }(1))=2$, and the strong form of Hensel lemma produces a square root. HenceThe four unit square classes of the p-adic integers have representatives $1,3,5,7$.

Let $L_n={\mathbb{Q}}_2({\xi }_{2^n})$. The extension is a cyclotomic extension of a p-adic field and is totally ramified. Under Local Artin reciprocity, the action on ${\xi }_{2^n}$ is the reduction of a 2-adic unit modulo $2^n$, up to the harmless inverse determined by the Frobenius convention. A uniformizer acts trivially because $L_n/{\mathbb{Q}}_2$ has no unramified part. The kernel is thereforeBy the norm subgroup of a local field extension, this kernel is exactly the required norm group. Hence, for every $n>0$,For $n=1$, the factor $1+2{\mathbb{Z}}_2$ is all of ${\mathbb{Z}}_2^{\times }$, consistently with ${\mathbb{Q}}_2({\xi }_2)={\mathbb{Q}}_2$.

Put $Y=X^2$. Modulo two, $Y^2+9Y-2$ has the two simple roots zero and one, so Hensel lemma lifts them to roots $y_0,y_1\in {\mathbb{Z}}_2$, with $v_2(y_0)=1$ and $y_1\equiv 1(\mathrm{mod}8)$. The 2-adic unit criterion makes $y_1$ a square, so $X^2-y_1$ splits into two linear factors. The polynomial $X^2-y_0$ is Eisenstein and remains irreducible. Thus $X^4+9X^2-2$ has three irreducible factors, of degrees $1,1,2$, as in factorization of X4 plus 9X2 minus 2 over the 2-adic numbers.

Every $x\in {\mathbb{Q}}_2^{\times }$ has a unique form $2^{n}u$ with $n\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_2^{\times }$. Its square class first records $n\mathrm{mod}2$. An odd 2-adic unit is a square precisely when it is congruent to $1$ modulo $8$: necessity follows by squaring an odd integer, and sufficiency follows from Hensel lemma applied in its standard $2$-adic square-root form.

The odd residues $1,3,5,7$ modulo $8$ therefore give four unit square classes. Together with valuation parity this yieldsFor example, the classes of $-1$, $2$, and $5$ form a basis of the square-class group of the 2-adic numbers.