If $c$ is not a square in $\mathbb{Q}_p$, then there exists a rational witness $b \in \mathbb{Q}^\times$ such that the $p$-adic Hilbert symbol $(b, c)_p = -1$.
Solvability of the Hilbert equation $b x^2 + c y^2 = 1$ over $\mathbb{Q}$ implies its solvability over $\mathbb{Q}_p$, obtained by base change of the rational solution.
If the rational Hilbert symbol $(b, c)_\mathbb{Q} = 1$, then the $p$-adic Hilbert symbol $(b, c)_p = 1$ for every prime $p$.
For any field $F$ with $2 \neq 0$ and any unit $x$, the Hilbert symbol $(x, -x) = 1$, witnessed by the identity solution $x \cdot ((x^{-1}+1)/2)^2 + (-x) \cdot ((x^{-1}-1)/2)^2 = 1$.
The Hilbert equation $(a c^2) x^2 + b y^2 = 1$ is solvable iff $a x^2 + b y^2 = 1$ is, since multiplying $a$ by a square does not change the square class.
The Hilbert equation $a x^2 + (b c^2) y^2 = 1$ is solvable iff $a x^2 + b y^2 = 1$ is,
by symmetry from IsSolvable_sq_mul_left.
The Hilbert symbol is invariant under scaling the first argument by a square: $(a c^2, b) = (a, b)$.
The Hilbert symbol is invariant under scaling the second argument by a square: $(a, b c^2) = (a, b)$.
The "division" identity for the Hilbert symbol over a field with bilinear Hilbert symbol: $(a/t, b/t) = (a, b) \cdot (t, -ab)$. A key step in relating binary form representation to the Hilbert symbol.
Representation criterion via Hilbert symbols. Over a field with bilinear Hilbert symbol, the binary form $a x^2 + b y^2$ represents $t \in F^\times$ if and only if $(a, b) = (t, -ab)$ in $\{1, -1\}$.
If the binary form $a x^2 + b y^2$ represents $-c$, then the ternary form $a x^2 + b y^2 + c z^2$ represents zero (with $z = 1$).
If the ternary form $a x^2 + b y^2 + c z^2$ represents zero, then the binary form $a x^2 + b y^2$ represents $-c$: either dehomogenize by dividing by $z$ if $z \neq 0$, or use the "represents-all" property of an isotropic binary form.
The ternary form $a x^2 + b y^2 + c z^2$ represents zero iff the binary form $a x^2 + b y^2$ represents $-c$.
Corollary 11.4. The ternary form $a x^2 + b y^2 + c z^2$ represents zero over $F$ iff $(a, b) = (-c, -(ab))$ in $\{1, -1\}$.
Corollary 11.2. A diagonal quadratic form $\sum_{i < n} w_i x_i^2$ with $n > 2$ variables and nonzero rational coefficients represents zero in $\mathbb{Q}_p$ for all but finitely many primes $p$. The exceptional set consists of $\{2\}$ together with primes dividing some coefficient.
The Hilbert symbol at the archimedean place: $(a, b)_\infty$ equals the real Hilbert symbol of $a, b$ regarded as units of $\mathbb{R}^\times$.
Instances For
The global Hilbert product $\prod_v (a, b)_v$, taken over all places of $\mathbb{Q}$: the archimedean place $\infty$ and all finite primes $p$.
Instances For
Hilbert reciprocity (product formula). For any $a, b \in \mathbb{Q}^\times$, the global product $\prod_v (a, b)_v = 1$.
For any $a, b \in \mathbb{Q}^\times$, the local Hilbert symbol $(a, b)_p$ is equal to $1$ for all but finitely many primes $p$.
The global Hilbert product can be rewritten using the hilbertAt notation indexed by
places: $\prod_v (a, b)_v = (a, b)_\infty \cdot \prod_p (a, b)_p$.
A consequence of Hilbert reciprocity. If two pairs $(a, b)$ and $(c, d)$ have matching local Hilbert symbols at all places except possibly one place $v_0$, then they also match at $v_0$. This follows from the product formula $\prod_v (a, b)_v = 1$.
Corollary 11.5. If the local equality $(a, b)_v = (-c, -(ab))_v$ (the ternary representation-of-zero criterion) holds at every place except possibly $v_0$, then it also holds at $v_0$. Applied to deduce global representation of zero from local conditions.
The element $2 \in \mathbb{Q}$ is invertible (needed for converting between quadratic forms and their associated symmetric matrices over $\mathbb{Q}$).
A symmetric matrix $M$ over $\mathbb{Q}$ defines a quadratic form $x \mapsto x^T M x$, and represents zero $p$-adically if there exists a nonzero vector $x \in \mathbb{Q}_p^n$ with $x^T M x = 0$.
Instances For
Diagonalization of nondegenerate quadratic forms over $\mathbb{Q}$. Any nondegenerate quadratic form $Q$ over $\mathbb{Q}$ in $n$ variables is equivalent (via a linear change of coordinates) to a diagonal weighted sum of squares $\sum w_i x_i^2$ with all $w_i \neq 0$.
Transport of $p$-adic isotropy along an equivalence: if $Q$ is rationally equivalent to the diagonal form $\sum w_i x_i^2$ and the latter represents zero $p$-adically, then the matrix $Q.\mathrm{toMatrix}'$ of $Q$ also represents zero $p$-adically.