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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{eqnarray} n = 2: F_1F_3 &= 2 = 1 + 1 = (F_2)^2 + 1 \\ n = 3: F_2F_4 &= 3 = 4 - 1 = (F_3)^2 - 1 \\ n = 4: F_3F_5 &= 10 = 9 + 1 = (F_4)^2 + 1 \\ n = 5: F_4F_6 &= 24 = 25 - 1 = (F_5)^2 - 1 \\ n = 6: F_5F_7 &=65 = 64 + 1 = (F_6)^2 + 1 . \end{eqnarray} 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{eqnarray} F_{n-1}F_{n+1} &= {1\over 5}(\alpha^{n-1}-\beta^{n-1})(\alpha^{n+1}-\beta^{n+1}) \\ &= {1\over 5}[ \alpha ^{2n} + \beta^{2n}-(\alpha^2+\beta^2)(\alpha\beta)^{n-1}] \\ &= {1\over 5}[(\alpha^n-\beta^n)^2+ 2(\alpha\beta)^n -(\alpha^2+\beta^2)(\alpha\beta)^{n-1}] \\ &= F_n^2 -{1\over 5}(\alpha\beta)^{n-1}(\alpha-\beta)^2 . \end{eqnarray} 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.