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Use the strong form of Hensel lemma: for $f\in {\mathbb{Z}}_p[T]$ and $a\in {\mathbb{Z}}_p$, ifthere is a unique polynomial root $b$ in the ball $v_p(b-a)>v_p(f^{\prime }(a))$. In particular, a simple polynomial root modulo $p$ lifts uniquely in its residue class. Newton iteration gives $v_p(b-a)=v_p(f(a))-v_p(f^{\prime }(a))$ when $f(a)\ne 0$.

For odd $p\notin \{5,7,13\}$, the setseach have $(p+1)/2$ elements, so they intersect. This gives a zero modulo $p$ with $z=1$. The derivative with respect to $z$ is $26$, nonzero modulo $p$, so Hensel's lemma lifts $z$ while holding $x,y$ fixed. This is the elementary proof of isotropy of nondegenerate ternary quadratic forms over finite fields followed by a simple-root lift.

At $p=5$, the residue vector $(0,1,1)$ is a zero, and the derivative $14y$ is nonzero. At $p=13$, use $(3,1,0)$, for which the value is 52 and the derivative $10x=30$ is nonzero modulo 13. Again a single-variable Hensel lift suffices.

At $p=2$, hold $x=2,z=1$ and use $f(Y)=7Y^2+33$. At $Y=1$, $f(1)=40$ has valuation 3, while $f^{\prime }(1)=14$ has valuation 1. Since $3>2$, the strong Hensel inequality gives a polynomial root $y\in {\mathbb{Z}}_2$, indeed $y\equiv 1(\mathrm{mod}4)$. This yields a nonzero solution.

At $p=7$, suppose a nonzero solution exists and scale it so that all coordinates are in ${\mathbb{Z}}_7$ and at least one is a unit. Reduction modulo 7 gives $z^2=5x^2$. Since 5 is not a square modulo 7, both $x,z$ are divisible by 7. In the original equation their terms are then divisible by 49, so $7y^2$ is divisible by 49 as well, forcing $y$ divisible by 7. This contradicts the normalization. Thus local isotropy of the five seven thirteen form gives

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