Special Cases of Quadratic Reciprocity and Counting Fixed Points

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

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

// Note that in $\mathbb R$ one can take $\phi=2\cos(2\pi/5)$. So in the next step we'll use something like 'rotation by $2\pi/5$' (unfortunately $\sin(2\pi/5)\notin\mathbb Q(\sqrt5)$ so we'll have to use a slightly different matrix).

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 (its matrix has two 5th roots of unity as eigenvalues) 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 $P^1(\mathbb F_p(\phi)=\mathbb F_{p^2})$ has fixed points. A fixed point of $\sigma$ is a solution of the equation $x^2-\phi x+1=0$, so $\phi=x+x^{-1}\in\mathbb F_p$.

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