Is the Hilbert symbol non-degenerate over $\mathbb{Q}$? For $a,b\in\mathbb{Q}^{\times}$ let $(a,b)$ be the Hilbert symbol of $a$ and $b$, i.e. the function equal to $1$ if $ax^2+by^2=1$ has a solution $(x,y)\in\mathbb{Q}$ and equal to $-1$ if not.
Does there exists $a\in\mathbb{Q}^{\times}$ not a square, such that $(a,b)=1$ for all $b\in\mathbb{Q}^{\times}$?
I think that local class field theory tells us that this is impossible over p-adic fields.
 A: Note that for $p$ a prime or $\infty$ (using the convention $\Bbb Q_{\infty}=\Bbb R$), the natural map
$$\Bbb Q^\times/(\Bbb Q^\times)^2\to \Bbb Q_p^\times/(\Bbb Q_p^\times)^2$$ is surjective, which means that any $b \in \Bbb Q_p^\times$ may be written as $b=uv^2$ with $u \in \Bbb Q^\times, v \in \Bbb Q_p^\times$.
Assume that there is some $a \in \Bbb Q^\times$ with $(a,b)=1$ for all $b \in \Bbb Q$.
We denote the local Hilbert symbol by $(x,y)_p$. Let $b \in \Bbb Q_p^\times$, then we can write $b=uv^2$ as above. Then we get by our assumption and bimultiplicativity of the Hilbert symbol $$(a,b)_p=(a,u)_p(a,v^2)_p=1$$ since $(a,u)=1$ implies $(a,u)_p=1$.
Thus $a$ is a square in $\Bbb Q_p$.
By the Grunwald-Wang theorem, if an element $a \in \Bbb Q$ is square in $\Bbb Q_p$ for all $p$, it is also a square in $\Bbb Q$.
A: meanwhile, here is page 43 from Cassels, Rational Quadratic Forms, online preview.   For each row $b \neq 1$ there are   values of $a$ that result in $-1.$   For odd primes, $r$   stands for a nonresidue, and $\varepsilon = (-1|p) \; . \;$
You can see the preview at  PREVIEW

