This solution is from Andrei from Tudor Vianu National College, Bucharest, Romania. Calculating $F_1$, $F_2,...,F_7$, I obtain:

${\displaystyle F_1=1;\hspace{0.5em}{F}_2=1;\hspace{0.5em}{F}_3=2;\hspace{0.5em}{F}_4=3;\hspace{0.5em}{F}_5=5;\hspace{0.5em}{F}_6=8;\hspace{0.5em}{F}_7=13.}$

Calculating $F_{n-1}{F}_{n+1}$ for some values of $n$:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{n=2:F_1{F}_3}& =2=1+1=(F_2{)}^2+1\\ \multicolumn{1}{c}{n=3:F_2{F}_4}& =3=4-1=(F_3{)}^2-1\\ \multicolumn{1}{c}{n=4:F_3{F}_5}& =10=9+1=(F_4{)}^2+1\\ \multicolumn{1}{c}{n=5:F_4{F}_6}& =24=25-1=(F_5{)}^2-1\\ \multicolumn{1}{c}{n=6:F_5{F}_7}& =65=64+1=(F_6{)}^2+1.\end{array}}$

Observing this, I try to prove the conjecture that:

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

Using the general formula for $F_{n-1}$ and $F_{n+1}$ in terms of $\alpha$ and $\beta$, the roots of the equation $x^2-x-1=0$, I find:

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

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.