Originally Posted by

**ThePerfectHacker** Okay,

$\displaystyle F(n)^2-F(n+1)F(n-1)$=$\displaystyle F(n)(F(n-1)+F(n-2))-F(n+1)F(n-1)$*

$\displaystyle =F(n)F(n-1)+F(n)F(n-2)-F(n+1)F(n-1)$

$\displaystyle =(F(n)-F(n+1))F(n-1)+F(n)F(n+2)$

$\displaystyle =(-1)(F(n-1)^2-F(n)F(n-2))$*

*)Used the fibonacci recrusion formula,

Thus,

$\displaystyle F(n)^2-F(n+1)F(n-1)$=$\displaystyle (-1)(F(n-1)^2-F(n)F(n-2))$

Note, the subscripts decrease by one with a negative in front.

Thus,

$\displaystyle (-1)(F(n-1)^2-F(n)F(n-2))$=$\displaystyle (-1)^2(F(n-2)^2-F(n-1)F(n-3))$

Thus,

$\displaystyle F(n)^2-F(n+1)F(n-1)$=$\displaystyle (-1)^2(F(n-2)^2-F(n-1)F(n-3))$

Countinuing this $\displaystyle n-2$ times we have,

$\displaystyle F(n)^2-F(n+1)F(n-1)$=$\displaystyle (-1)^{n-2}(F(2)^2-F(3)F(1))$=$\displaystyle (-1)^{n-2}(-1)=(-1)^{n-1}$

Q.E.D.