In the following equation: $$5 F_n^2 \pm 4 = x,$$ where $F_n$ is a Fibonacci number, and the $\pm 4$ shall be treated as $+4$ for even $n$ and $-4$ for odd $n$.
Now, if the above requirements are satisfied, then why is $x$ always a perfect square? Is there a proof for the equation? If so, could I please know how to prove the equation? Also, are there any ideas which would investigate the above equation and why is $x$ always a perfect square?
[Edit: It was found that the $\sqrt{x}$ are Lucas numbers. But how could this be proved using only the LHS?]
Consider $$F_n =\frac{1}{\sqrt{5}} (\phi^n -\varphi^n),$$ where $$\phi =\frac{1+\sqrt{5}}{2}, \qquad \varphi =\frac{1-\sqrt{5}}{2}.$$ Note that $$\phi \varphi=-1.$$ Then \begin{align} 5 F_n^2 +4(-1)^n &=(\phi^n -\varphi^n)^2 +4(\phi\varphi)^n \\ &=\phi^{2n} -2\phi^n\varphi^n +\varphi^{2n} +4\phi^n\varphi^n \\ &=\phi^{2n} +2\phi^n\varphi^n +\varphi^{2n} \\ &=(\phi^n +\varphi^n)^2 \end{align} is actually a square. And the irrational parts of $\phi^n$ and $\varphi^n$ actually cancel each other because of the minus sign in $\varphi$, thus $\phi^n +\varphi^n$ is an integer.