TITLE: How can one deduce quadratic reciprocity from Hilbert reciprocity?
QUESTION [8 upvotes]: Hilbert reciprocity says the following:
Define $(a,b)_p$ to be $1$ if there is a non-trivial solution in $\mathbb{Q}_p$ to $z^2=ax^2+by^2$, and $-1$ if there isn't. Then $\prod_p (a,b)_p =1$, where the product runs also over the infinite prime (and where $\mathbb{Q}_{\infty}$ is $\mathbb{R}$).
Quadratic reciprocity says:
Define $\left( \frac{p}{q}\right)$ to be $1$ if $p$ is a square in $\mathbb{F}_q$, and $-1$ otherwise. Then for every two different odd primes $p$ and $q$, $\left( \frac{p}{q}\right)\left( \frac{q}{p}\right)=(-1)^{\frac{(p-1)(q-1)}{4}}$.
I have heard that people think of Hilbert reciprocity as a generalization of quadratic reciprocity. How does one deduce it as a consequence?

REPLY [5 votes]: I happened to post on this (one hour before the question!) in another answer:
Conceptual Proof of Quadratic Reciprocity
The proof is quite easy.
For places $r$ different from $p, q, 2$, and $\infty$, the local Hilbert symbol is $+1$ because the quadratic form $x^2  -py^2 - qz^2$ has rank higher than 2.  More combinatorially, let $z=1$ and observe that the sets of values of $x^2$ and of $py^2$ have cardinality more than $|r|/2$ so that their sum has to cover all possibilities mod $r$.  Because the degree $2$ is less than $r$, Hensel lifting shows that an $r$-adic solution can be found from a mod $r$ solution.
The values of the local symbol at $p$ and $q$ are, by definition, the Legendre symbols.  (The existence of a solution mod $p$ or $q$ is the information recorded in a Legendre symbol, and Hensel lifting of the modular square roots to p-adic square roots can always be done for odd primes.)
The value of the the local symbol at $\infty$ is $+1$ because the form is not positive definite or negative definite.  Or write down a particular real solution such as $y=z=1, x = \sqrt{p+q}$.
This leaves the value of the local symbol at $2$. The condition for $py^2 + qz^2$ to be a square 2-adically is that it be a square mod $8$, i.e., congruent to $0,1$, or $4$, and a primitive nonzero solution has $y$ or $z$ odd.  Case analysis is not difficult:

The symbol is +1 if $p$ or $q$ is $1 \mod 8$. If $p=1$ there is a solution with $y=1$ and $z=0$.
The symbol is +1 if $p$ or $q$ is $5 \mod 8$. If $p=5$ there is a solution with $y=1$ and $z=2$.
The symbol is -1 if $p$ and $q$ are both $-1 \mod 4$.  Both $y$ and $z$ must be even in this case.

This pattern of signs is the same as the $(-1)^{(p-1)(q-1)/4}$ in the quadratic reciprocity formula.  
Finally, the product of all the local symbols is 1:
$$(1) (\frac{p}{q}) (\frac{q}{p}) (-1)^{(p-1)(q-1)/4} = 1$$