How to prove Fibonacci recurrence holds mod p? Let $$J_n \equiv c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n \Big) \ \text{(mod  p)},$$ with $c$ and $c^{-1}$ integers such that $c^2 \equiv 5 \ \text{(mod  p)}$ and $cc^{-1} \equiv 1 \ \text{(mod  p)}$. And $c$ is an odd integer. 
It is easy to check that $J_1 \equiv J_2 \equiv 1 \ \text{(mod  p)}$. However, I can't prove that $J_{n} \equiv J_{n-1} + J_{n-2} \  \text{(mod  p)}$.
My attempt: $$J_{n-1} + J_{n-2} = c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^{n-2} - \Big( \dfrac{1-c}{2} \Big)^{n-2}\Big) \Big( \dfrac{6+2c}{4} \Big) + k_3 p = c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n\Big) + c^{-1} \Big( \dfrac{1+c}{2} \Big)^{n-2} \dfrac{k_1p}{4} - c^{-1} \Big( \dfrac{1-c}{2} \Big)^{n-2} \Big(1+ \dfrac{k_2p}{4} \Big) + k_3 p$$ which after a long try still I can't reduce it to the desired result because of factor 4 in the denominator. 
PS I always keep  $k_ip$'s to avoid mistakes working in mod p.
Please help! 
Edit. If $\dfrac{c^{-1}}{2^2} ((1+c)^{n-2} - (1-c)^{n-2})$ were an integer for any $n$ then we are done! But that also I couldn't prove. 

Write $a=\frac12(1+c)$ and $b=\frac12(1-c)$. Then $a+b=1$ and
$$ab=\frac14(1-c^2)\equiv\frac14(1-5)\equiv-1\pmod p.$$
Therefore
$$a^2-a-1\equiv a^2-(a+b)a+ab\equiv0\pmod p$$
and
$$b^2-b-1\equiv b^2-(a+b)b+ab\equiv0\pmod p.$$
Then
$$c(J_n-J_{n-1}-J_{n-2})\equiv a^n-b^n- a^{n-1}+b^{n-1}-a^{n-2}+b^{n-2}
\equiv(a^2-a-1)a^{n-2}+(b^2-b-1)b^{n-2}\equiv0\pmod p.
$$
It follows that $J_n\equiv J_{n-1}+J_{n-2}$ (mod $p$).
