Note that $F_4{F}_6=24=F_5^2-1$ and $F_6{F}_8=168=F_7^2-1$ whereas $F_3{F}_5=10=F_4^2+1$ and $F_5{F}_7=65=F_6^2+1$. It seems that $F_{n+1}{F}_{n-1}=F_n^2+(-1{)}^n$ so we now prove this conjecture.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{F_{n+1}{F}_{n-1}}& =\frac{1}{5}({\alpha }^{n+1}-{\beta }^{n+1})({\alpha }^{n-1}-{\beta }^{n-1})\\ \multicolumn{1}{c}{}& =\frac{1}{5}[{\alpha }^{2n}+{\beta }^{2n}-({\alpha }^2+{\beta }^2)(\alpha \beta {)}^{n-1}]\\ \multicolumn{1}{c}{}& =\frac{1}{5}[({\alpha }^n-{\beta }^n{)}^2+2(\alpha \beta {)}^n-({\alpha }^2+{\beta }^2)(\alpha \beta {)}^{n-1}]\\ \multicolumn{1}{c}{}& =F_n^2-\frac{1}{5}(\alpha \beta {)}^{n-1}(\alpha -\beta {)}^2.\end{array}}$

At this point we use the fact that $\alpha \beta =-1$ and $\alpha -\beta =\sqrt{5}$ which gives the result

${\displaystyle F_{n+1}{F}_{n-1}=F_n^2+(-1{)}^n}$

thus proving the conjecture.