Hilbert symbol

Remarks

The value only depends on the classes of $a$ and $b$ modulo squares. Equivalently, $(a, b)_K = 1$ exactly when $a$ is a norm from the field $K(\sqrt{b})$. The symbol is symmetric and bimultiplicative, and $(a, 1 - a)_K = 1$ for $a \neq 1$. Example: $(-1, -1)_\R = -1$, because $z^2 = -x^2 - y^2$ forces $x = y = z = 0$ over $\R$. Example: $(2, 7)_\R = 1$, because $(x, y, z) = (1, 1, 3)$ solves $z^2 = 2x^2 + 7y^2$. Non-example: the solution $(0, 0, 0)$ must be excluded; it solves every such equation, so allowing it would make every symbol equal to $1$. For $K = \Q_p$ with $p$ odd, write $a = p^\alpha u$ and $b = p^\beta v$ with units $u, v \in \Z_p^\times$. Then $(a, b)_p = (-1)^{\alpha \beta (p - 1)/2} \left(\frac{u}{p}\right)^\beta \left(\frac{v}{p}\right)^\alpha$, with Legendre symbols of the residues of $u$ and $v$.