Fibonacci numbers and perfect squares

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Can this be simplified any further?:

$\phi^{2n}$ + $\psi^{2n}$$-2(-1)^n$ $\pm4$

Where $\phi = (1+\sqrt5)/2 $

Where $\psi = (1-\sqrt5)/2 $

When n=even number, use +4

Whenn n=odd number, use -4

If anyone could show me how to simplify this further it would be greatly appreciated.

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You are dealing with Lucas numbers, defined by $L_0=2$, $L_1=1$, $L_n=L_{n-1}+L_{n-2}$. We have $L_n=\varphi^n+\psi^n$.

Observe that $L_n^2=\varphi^{2n}+2\varphi^n\psi^n+\psi^{2n}=L_{2n}+2(-1)^n$.