Let $p$ be an odd prime and $L_n$ be the $n$th Lucas number. Can anyone prove this? $$\frac{L_1}{1}+\frac{L_3}{3}+\frac{L_5}{5}+\cdots+\frac{L_{p-2}}{p-2}\neq0\pmod{p}$$ Please help me!
I am thinking about the period of Fibonacci sequence. The purpose is to prove that (the period mod $p$) $\neq$ (the period mod $p^2$). It is known that the period mod $p$ divides $p-1$ or $2p+2\ (p\neq5)$. So, I tried to prove $F^{p^2-1}\neq I\ \pmod{p^2}$, where $F$ is the Fibonacci matrix, then I got this inequality.
If $p\neq5$, this question is equivalent to prove that $$\frac{\phi}{1}+\frac{\phi^2}{2}+\frac{\phi^3}{3}+\cdots+\frac{\phi^{p-1}}{p-1}\neq0\pmod{p}.$$ Here, $\phi$ is a root of $x^2-x-1$. See this page. I have showed that $$\frac{\phi}{1}+\frac{\phi^2}{2}+\frac{\phi^3}{3}+\cdots+\frac{\phi^{p-1}}{p-1}\equiv\sum_{k=1}^{\frac{p-1}{2}} \frac{{(-1)}^k}{2k} {{2k}\choose{k}}\pmod{p}.$$
