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Over a finite field of odd cardinality $q$, a diagonal nondegenerate ternary quadratic form $A{x}^2+B{y}^2+C{z}^2$ has a zero with $z=1$: the sets $\{A{x}^2\}$ and $\{-C-B{y}^2\}$ both have $(q+1)/2$ elements and must intersect. This gives a nonzero nonsingular zero. Over a local integer ring with these coefficients units, the Hensel lemma lifts it to an isotropic vector in the fraction field.

This ternary quadratic form has a nonzero zero over ${\mathbb{Q}}_p$ exactly when $p\ne 7$. Away from $2,5,7,13$, the finite-field square-set intersection proof supplies a nonsingular residue zero, and Hensel lemma lifts it. At 5 use $(0,1,1)$ modulo 5; at 13 use $(3,1,0)$. At 2 hold $x=2,z=1$ and apply the strong form of Hensel lemma to $7Y^2+33$ at $Y=1$: the function and derivative have valuations 3 and 1. At 7, the equation modulo 7 forces $x=z=0$ because 5 is not a square; reduction after dividing by 7 then forces $y=0$. Any normalized integral nonzero solution would therefore have all coordinates divisible by 7, a contradiction.

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

By Fermat's little theorem,Thus the sum of $g$ over ${\mathbb{F}}_p^3$ counts, modulo $p$, the triples on which the ternary quadratic form $f$ vanishes.

Expand $f^{p-1}$ with the multinomial theorem. A term indexed by $r+s+t=p-1$ contributes a constant multiple ofAt least one of $r,s,t$ is strictly less than $(p-1)/2$. Its corresponding exponent is therefore in $\{0,1,\dots ,p-2\}$, so part (b), including its $k=0$ case, makes that power sum zero modulo $p$. Every term in the expansion consequently vanishes after summation. The constant term contributes $p^3\equiv 0$, and therefore

The zero triple is one solution of $f=0$. If it were the only solution, the displayed sum would be congruent to $1$ rather than $0$. Hence there is another triple $(x,y,z)$, not all divisible by $p$, satisfying