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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 $1,3,5,7$.