Quadratic reciprocity via generalized Fibonacci numbers?

This is a pet idea of mine which I thought I'd share. Fix a prime q congruent to 1mod4 and define a sequence Fn by F0=0,F1=1, and
Fn+2=Fn+1+q−14Fn.

Then Fn=αn−βnα−β where α,β are the two roots of f(x)=x2−x−q−14. When q=5 we recover the ordinary Fibonacci numbers. The discriminant of f(x) is q, so it splits modp if and only if q is a quadratic residue modp.
If (qp)=−1, then the Frobenius morphism x↦xp swaps α and β (working over Fp), hence Fp≡−1modp. And if (qp)=1, then the Frobenius morphism fixes α and β, hence Fp≡1modp. In other words,
Fp≡(qp)modp.
Quadratic reciprocity in this case is equivalent to the statement that
Fp≡(pq)modp.
Question: Does anyone have any ideas about how to prove this directly, thereby proving quadratic reciprocity in the case that q≡1mod4?
My pet approach is to think of Fp as counting the number of ways to tile a row of length p−1 by tiles of size 1 and 2, where there is one type of tile of size 1 and q−14 types of tiles of size 2. The problem is that I don't see, say, an obvious action of the cyclic group Z/pZ on this set. Any ideas?