I think by far the simplest easiest to remember elementary proof of QR is due to Rousseau (On the quadratic reciprocity law). All it uses is the Chinese remainder theorem and Euler's formula $a^{(p-1)/2}\equiv (\frac{a}{p}) \mod p$. The mathscinet review does a very good job of outlining the proof. I'll try to explain how I remember it here (but the lack of formatting is really rough for this argument).
Here's the outline. Consider $(\mathbb{Z}/p)^\times \times (\mathbb{Z}/q)^\times = (\mathbb{Z}/pq)^\times$. We want to split that group in "half", that is consider a subset such that exactly one of $x$ and $-x$ is in it. There are three obvious ways to do that. For each of these we take the product of all the elements in that "half." The resulting three numbers are equal up to an overall sign. Calculating that sign on the $(\mathbb{Z}/p)^\times$ part and the $(\mathbb{Z}/q)^\times$ part give you the two sides of QR.
In more detail. First let me describe the three "obvious" halves:
- Take the first half of $(\mathbb{Z}/p)^\times$ and all of the other factor.
- Take all of the first factor and the first half of $(\mathbb{Z}/q)^\times$.
- Take the first half of $(\mathbb{Z}/pq)^\times$.
The three products are then (letting $P = (p-1)/2$ and $Q=(q-1)/2$):
- $(P!^{q-1}, (q-1)!^P)$.
- $((p-1)!^Q, Q!^{p-1})$.
- $\left(\dfrac{(p-1)!^Q P!}{q^P P!},\dfrac{(q-1)!^P Q!}{p^Q Q!}\right)$.
All of these are equal to each other up to overall signs. Looking at the second component it's clear that the sign relating 1 and 3 is $\left(\frac{p}{q}\right)$. Similarly, the sign relating 2 and 3 is $\left(\frac{q}{p}\right)$. So the sign relating 1 and 2 is $\left(\frac{p}{q}\right) \left(\frac{q}{p}\right)$. But to get from 1 to 2 we just changed the signs of $\frac{p-1}{2} \frac{q-1}{2}$ elements. QED
Hi, Pete. There are a few observations related to this, not widely known although basic, and that includes your colleague. First, Conway gives a quick proof on page 142 of The Sensual Quadratic Form, including over the rationals.
Next, also Conway, the form (five variables) that he and Schneeberger found that represents all the numbers from 1 to 289, fails to represent 290, then represents 291 and on forever, he initially called Methusaleh. It is just a binary added to a ternary that represents the numbers from 1 to 28 consecutive, discriminant 29. However, for ternaries that is not the record. The form he called Little Methusaleh, discriminant 31, represents 1 to 30 consecutive. The theorem is in this material, as the conditions for a positive ternary to represent, say, 1,2,3,5, places strong restrictions on a partly reduced form. Kap wrote this sort of argument up several times, including a repeat in the unpublished 1996 Classification. It is quite easy. OK, Little Methusaleh and your result over the integers are proved on page 81 of The Sensual Quadratic Form
Finally, a positive form is anisotropic at the "prime" infinity. In Cassels Rational Quadratic Forms he shows global relations on the Hilbert Norm Residue symbol that show that any ternary is anisotropic at an even number of primes. So a positive ternary is anisotropic at an odd number of finite primes. Taken with the observation above that at least one number below 31 is missed, and a positive ternary fails to integrally represent an infinite number of positive integers.
I will look up some of my tables and fill things in. Note that some of this is discussed in an early article by William Duke, 1997 Notices, but he mistyped the form with discriminant 29.
Let's see, Conway and Schneeberger probably had an acceptable proof of the 15 Theorem scattered about, but it never got put together. Bhargava was looking for diversions from his own dissertation, Conway mentioned this in passing. Bhargava showed the fundamental result that one of these forms must have a regular ternary as a sub-form, thus the project became a careful inspection of my paper with Kap on all possible regular ternaries. Also, correspondence between Kap and Bhargava first revealed some important errors in Magma relating to calculating the spinor genus, and hilarity ensued.
EDIT: thinking about the history question, it is quite possible that this result was never written down as a separate proposition, by Gauss, Legendre, etc. The reason I suggest this is the great weight placed on positive ternary forms missing certain "progressions," in the language of Jones, Dickson, other early books. So, in Jones, chapter 8, we read "Thus there will be a finite number of arithmetical progressions of this type" of numbers not represented by any form in the genus under consideration. Not much motivation for proving that a form misses at least one number if you are going to quickly show that it misses an entire arithmetic progression.
EDIT TOOO: note that Conway replaces the prime usually called $\infty$ by the prime $-1.$
No definite ternary form is universal
However, a simple argument shows that
any definite ternary form must fail to
represent infinitely many integers,
even over the rationals. For if a
ternary form $f$ of determinant $d$
represents anything in the $p$-adic
squareclass of $-d$ over $\mathbf
> Q_p,$ then it must be $p$-adically
equivalent to $[ -d,a,b]$ where the
"quotient form" $[a,b]$ has
determinant $-1,$ and so $p$-adically,
$f$ must be the isotropic form $[
> -d,1,-1].$
But a positive definite form fails to
represent $-1,$ and so it is not
$p$-adically isotropic for $p=-1.$ By
the global relation, there must be
another $p$ for which it is not
$p$-adically isotropic, and so it
also fails to represent all numbers in
the $p$-adic square-class of $-d$ for
this $p$ too!
The Three Squares Theorem illustrates
this nicely--the form $[ 1,1,1]$ fails
to represent $-1$ both $-1$-adically
and $2$-adically. In the Third
Lecture, we showed that The Little
Methusaleh Form $$ x^2 + 2 y^2 + y z
> + 4 z^2 $$ fails to represent 31. We now see that since it fails to
represent the $-1$-adic class of its
determinant $-31/4$ (i.e., the
negative numbers), it must also fail
to represent the infinitely many
positive integers in the $31$-adic
squareclass of $-31/4.$
Best Answer
$\def\FF{\mathbb{F}}$I'm just guessing, but I would have thought it was the following: Hilbert reciprocity for function fields can be deduced from Weil reciprocity. Weil reciprocity is the following statement: Let $X$ be a complete curve over an algebraically closed field $k$. For any point $x \in X$ and nonzero meromorphic functions $f$ and $g$, define $(f,g)_x = (-1)^{(\mathrm{ord}_x f)(\mathrm{ord}_x g)}(f^{\mathrm{ord}_x g}/g^{\mathrm{ord}_x f})(x)$. Then $\prod_{x \in X} (f,g)_x=1$. See here and here for the connection.
Now, over $\mathbb{C}$, we can prove Weil reciprocity as follows: Choose a path $\delta$ connecting $0$ to $\infty$ in $\mathbb{CP}^1$ and avoiding the critical values of $f$. For simplicity, let us assume $f$ has simple zeroes and poles $\zeta^{\pm}_1$, $\zeta^{\pm}_2$, ..., $\zeta^{\pm}_n$. Set $\gamma = f^{-1}(\delta)$. Then $\gamma$ is the union of $\deg(f)$ closed line segments. After reordering, we may assume $\zeta^+_i$ is joined to $\zeta^-_i$, say by $\gamma_i$.
We can define $\log(f)$ on $X \setminus \gamma$, by composing $f$ with a branch of $\log$ on $\mathbb{CP}^1 \setminus \delta$. The differential form $\omega:= \tfrac{1}{2 \pi i} \log(f) \tfrac{dg}{g}$ therefore makes sense on $X \setminus (\gamma \cup g^{-1}(\{ 0,\infty \}))$. If we integrate $\omega$ on little contours around the zeroes and poles of $g$, we get $\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x))$.
On the other hand, if we integrate around a tubular neighborhood of $\gamma_i$, we pick up $\int_{\gamma_i} \tfrac{dg}{g} = \log(g(\zeta^{+}_i) - \log(g(\zeta^-_i))$ for some branch of $\log$. Summing on $i$, this is $\sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$
The sum of the contours around the zeroes of $f$ is homologous to the sum over the neighborhoods of the $\gamma_i$, so we deduce $$\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x)) = \sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$$ and exponentiating gives the result.