How to show that : $4(-1)^nL_n^2+L_{4n}-L_n^4=2$

Question

How can we prove that: $$4(-1)^nL_n^2+L_{4n}-L_n^4=2$$

Where $L_n$ is Lucas number

We got $L_n=\phi^n+(-\phi)^{-n}$

$4(-1)^nL_n^2=8(-1)^n\phi^{2n}+8$

$L_{4n}=\phi^{4n}+(-\phi)^{-4n}$

$L_n^4=4\phi^{4n}+4(-1)^n\phi^{2n}+4$

$$8(-1)^n\phi^{2n}+8+\phi^{4n}+(\phi)^{-4n}-4\phi^{4n}-4(-1)^n\phi^{2n}-4=2$$

$$3\phi^{4n}-4(-1)^n\phi^{2n}-\phi^{-4n}=2$$

Answer 1

$4(-1)^nL_n^2=8(-1)^n\phi^{2n}+8$

$L_n^4=4\phi^{4n}+4(-1)^n\phi^{2n}+4$

These are incorrect.

We have $$\begin{align}4(-1)^nL_n^2&=4(-1)^n(\phi^n+(-\phi)^{-n})^2 \\\\&=4(-1)^n(\phi^{2n}+2\phi^n(-\phi)^{-n}+(-\phi)^{-2n}) \\\\&=4(-1)^n(\phi^{2n}+2(-1)^n+\phi^{-2n}) \\\\&=4(-1)^n\phi^{2n}+8+4(-1)^n\phi^{-2n} \\\\&=4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+8\end{align}$$

and

$$\begin{align}L_n^4&=(L_n^2)^2 \\\\&=(\phi^{2n}+2(-1)^n+\phi^{-2n})^2 \\\\&=\phi^{4n}+4+\phi^{-4n}+4(-1)^n\phi^{2n}+2+4(-1)^n\phi^{-2n} \\\\&=\phi^{4n}+\phi^{-4n}+4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+6\end{align}$$

It follows from these that $$\begin{align}&4(-1)^nL_n^2+L_{4n}-L_n^4 \\\\&=4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+8+(\phi^{4n}+\phi^{-4n}) \\&\qquad\qquad -(\phi^{4n}+\phi^{-4n}+4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+6) \\\\&=8-6 \\\\&=2\end{align}$$