Note that F₄F₆=24=F₅²-1 and F₆F₈=168=F₇²-1 whereas F₃F₅=10=F₄²+1 and F₅F₇=65=F₆²+1. It seems that F_n+1F_n-1=F_n² +(-1)ⁿ so we now prove this conjecture.

Fn+1Fn-1 = 1 5 (an+1-bn+1)(an-1-bn-1) = 1 5 [ a2n + b2n -(a2+b2)(ab)n-1] = 1 5 [(an-bn)2+ 2(ab)n -(a2+b2)(ab)n-1] = Fn2 - 1 5 (ab)n-1(a-b)2 .

At this point we use the fact that ab = -1 and a-b = Ö5 which gives the result

Fn+1Fn-1=Fn2 +(-1)n

thus proving the conjecture.