This is a question in the same spirit than this one, trying to prove algebraic number theoretic statements from zeta functions. I want to prove the Hasse-Minkowski principle for quadratic forms in two variables. Let $q(x, y) = qx^2+by^2$.
The Hasse-Minkowski principle is equivalent to the square theorem: $a\in K^2$ iff $a\in K_v^2$ for all place $v$ of the base field $K$.
Let $L = K(\sqrt{a})$ at extension of $K$ of degree 1 or 2. If it is everywhere isomorphic to $K_v \oplus K_v$, the associated zeta function would have a double pole and hence $L$ would not be a field (which is not).
Why is $L\simeq K_v \oplus K_v$ equivalent to $a \in K_v^2$?
The minimal polynomial of $a^{1/2}$ is $x^2-a\in K[x]$.
$a$ is not a square in $K_v$ iff $x^2-a\in K_v[x]$ is irreducible iff $$K_v\otimes_K K(a^{1/2})\cong K_v\otimes_K K[x]/(x^2-a)\cong K_v[x]/(x^2-a)$$ is a field.
If $a=b^2\in K_v$ is a square then $x^2-a=(x-b)(x+b)$ and $$K_v\otimes_K K(a^{1/2})\cong K_v[x]/((x-b)(x+b))\cong K_v[x]/(x-b) \times K_v[x]/(x+b) \cong K_v\times K_v$$
(isomorphisms as rings and $K_v$ algebras)