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The general theorem is: for all odd, distinct primes $p, q$, the following holds: $$\left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}$$

I've discovered the following proof for the case $q=3$: Consider the Möbius transformation $f(x) = \frac{1}{1-x}$, defined on $F_{p} \cup {\infty}$. It is a bijection of order 3: $f^{(3)} = Id$.

Now we'll count the number of fixed points of $f$, modulo 3:

1) We can calculate the number of solutions to $f(x) = x$: it is equivalent to $(2x-1)^2 = -3$. Since $p \neq 2,3$, the number of solutions is $\left( \frac{-3}{p} \right) + 1$ (if $-3$ is a non-square, there's no solution. Else, there are 2 distinct solutions, corresponding to 2 distinct roots of $-3$).

2) We know the structure of $f$ as a permutation: only 3-cycles or fixed points. Thus, number of fixed points is just $|F_{p} \cup {\infty}| \mod 3$, or: $p+1 \mod 3$.

Combining the 2 results yields $p = \left( \frac{-3}{p} \right) \mod 3$. Exploiting Euler's criterion gives $\left( \frac{p}{3} \right) = p^{\frac{3-1}{2}} = p \mod 3$, and using $\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}{2}}$, we get: $$\left( \frac{3}{p} \right) \left( \frac{p}{3} \right) = (-1)^{\frac{p-1}{2}\frac{3-1}{2}} \mod 3$$ and equality in $\mathbb{Z}$ follows.

My questions:

  • Can this idea be generalized, with other functions $f$?
  • Is there a list\article of proofs to special cases of the theorem?
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Not a duplicate, but related:… – David Speyer Nov 4 '11 at 0:24
Very nice proof! – Alon Amit Nov 4 '11 at 0:44
Nice indeed! But it's not immediately clear how to generalize this: it's easy to find 3-torsion in $PSL_2(\mathbb Z)\cong\mathbb Z_2\ast\mathbb Z_3$ but not, say, 5-torsion... – Grigory M Jan 4 '14 at 14:44
up vote 4 down vote accepted

As for your second question, a (partial) list of articles dealing with the quadratic character of small primes can be found here.

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Dear franz: I don't have the privilege of knowing you personally, but of course I know of you. Welcome to MSE! – Pierre-Yves Gaillard Jan 21 '12 at 16:00

Well, at least it can be extended to a proof of the case $q=5$. $\def\lf#1#2{\left(\dfrac{#1}{#2}\right)}$

0. $\lf5p=1\iff\exists\phi\in\mathbb F_p:\phi^2+\phi-1=0$.

1. $\lf5p=1\implies\lf p5=1$.

$\sigma\colon x\mapsto\dfrac1{\phi-x}$ is a Möbius transformation of $P^1(\mathbb F_p)$ of order 5 which has either 0 or 2 fixed points. So $p\pm1$ is divisible by 5, or equivalently $\lf p5=1$.

2. $\lf p5=1\implies\lf5p=1$.

$\lf p5=1$ means that $p^2+1$ is not divisible by 5, so the action of $\sigma$ on $P^1(\mathbb F_{p^2})$ has fixed points. A fixed point of $\sigma$ is a solution of the equation $x^2-\phi x+1=0$, so $\phi=x+x^{-1}$ — in particular, $\phi\in\mathbb F_p$.

// It, of course, would be nice to have a different, more geometric proof of (2)…

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