Prove that $L_{6k} \equiv 2$ (mod $4$)

Question

Here is my reasoning so far

$L_{6k} =F_{6k-1} + F_{6k+1}$

I have proved that any $F_n$ with n a multiple of 3 is even

i.e. $F_{3n}$ is even and so is $F_{6n}$, it follows that $F_{6k-1}$ and $F_{6k+1}$ are odd and an odd $+$ and odd $=$ an even.

Therefore, $L_{6k} = $ an even number

the following is where I am stuck

I have this conjecture

an even number $\equiv 2$ (mod $4$)

Answer 1

First we prove that $F_{6n-3}\equiv2\%4,$ where $\%$ is the modulo operation, by induction. For $n=0$ we have $F_{-3}=2\equiv2\%4.$ Assume that $F_{6n-3}\equiv2\%4,$ then \begin{align} F_{6n+3}&=F_{6n+2}+F_{6n+1}\\&=2F_{6n+1}+F_{6n}\\&=3F_{6k}+2F_{6n-1}\\&=5F_{6n-1}+3F_{6n-2}\\&=8F_{6n-2}+5F_{6n-3}\\&\equiv5F_{6n-3}\%4\\&\equiv10\%4\\&\equiv2\%4, \end{align} hence $F_{6n-3}\equiv2\%4$ for all $n\in\mathbb{N}.$ Now we prove that $L_{6n}\equiv2\%4$ for all $n\in\mathbb{N}.$ \begin{align} L_{6n}&=F_{6n+1}+F_{6n-1}\\&=F_{6n}+2F_{6n-1}\\&=3F_{6n-1}+F_{6n-2}\\&=4F_{6n-2}+3F_{6n-3}\\&\equiv3F_{6n-3}\%4\\&\equiv6\%4\\&\equiv2\%4. \end{align}