Quadratic reciprocity via generalized Fibonacci numbers?

Solution 1:

The following paper seems to answer your question: P. T. Young, "Quadratic reciprocity via Lucas sequences", Fibonacci Quart. 33 (1995), no. 1, 78–81.

Here's its MathSciNet Review by A. Grytczuk:

Let $\{\gamma_n\}^\infty_{n=0}$ be a given Lucas sequence defined by $\gamma_0=0$, $\gamma_1=1$, $\gamma_{n+1}=\lambda \gamma_n+\mu \gamma_{n-1}$, $n\geq 1$, $\lambda, \mu\in{\bf Z}$, and let $q$ be an odd prime such that $D=(\frac{-1}q)q=\lambda^2+4\mu$. Then the author proves that there is a unique formal power series $\Phi$ with integer coefficients and constant term zero such that (1) $\sum^\infty_{n=1}\gamma_n\Phi^n(t)/n=\sum^\infty_{n=1}(\frac nq)t^n/n$ holds, where $(\frac nq)$ is the Legendre symbol.
From this result follows the Gauss law of quadratic reciprocity in the following form: (2) $(\frac pq)=(\frac Dp)$, where $p$, $q$ are distinct odd primes and $D=(\frac{-1}q) q=\lambda^2+4\mu$.

Here's the direct link to the paper.