Plus or minus

This solution is from Andrei from Tudor Vianu National College, Bucharest, Romania. Calculating $F_1$, $F_2,. . . , F_7$, I obtain: $$F_1=1;F_2=1;F_3=2;F_4=3;F_5=5;F_6=8;F_7=13.$$ Calculating $F_{n-1}F_{n+1}$ for some values of $n$: $$\begin{array}{rc}n=2:F_1{F}_3& =2=1+1=(F_2{)}^2+1\\ n=3:F_2{F}_4& =3=4-1=(F_3{)}^2-1\\ n=4:F_3{F}_5& =10=9+1=(F_4{)}^2+1\\ n=5:F_4{F}_6& =24=25-1=(F_5{)}^2-1\\ n=6:F_5{F}_7& =65=64+1=(F_6{)}^2+1.\end{array}$$ Observing this, I try to prove the conjecture that: $$F_{n-1}{F}_{n+1}=(F_n{)}^2+(-1{)}^n.$$ 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: $$\begin{array}{rc}{F}_{n-1}{F}_{n+1}& =\frac{1}{5}({\alpha }^{n-1}-{\beta }^{n-1})({\alpha }^{n+1}-{\beta }^{n+1})\\ & =\frac{1}{5}[{\alpha }^{2n}+{\beta }^{2n}-({\alpha }^2+{\beta }^2)(\alpha \beta {)}^{n-1}]\\ & =\frac{1}{5}[({\alpha }^n-{\beta }^n{)}^2+2(\alpha \beta {)}^n-({\alpha }^2+{\beta }^2)(\alpha \beta {)}^{n-1}]\\ & =F_n^2-\frac{1}{5}(\alpha \beta {)}^{n-1}(\alpha -\beta {)}^2.\end{array}$$ At this point we use the fact that $\alpha\beta=-1$ and $\alpha-\beta=\sqrt 5$ which gives the result $$F_{n+1}{F}_{n-1}=F_n^2+(-1{)}^n$$ thus proving the conjecture.