This solution is from Andrei from Tudor Vianu National College, Bucharest, Romania. Calculating F₁, F₂,. . . , F₇, I obtain:

F1 = 1; F2 = 1; F3 = 2; F4 = 3; F5 = 5; F6 = 8; F7 = 13.

Calculating F_n-1F_n+1 for some values of n:

n = 2: F1F3 = 2 = 1 + 1 = (F2)2 + 1 n = 3: F2F4 = 3 = 4 - 1 = (F3)2 - 1 n = 4: F3F5 = 10 = 9 + 1 = (F4)2 + 1 n = 5: F4F6 = 24 = 25 - 1 = (F5)2 - 1 n = 6: F5F7 =65 = 64 + 1 = (F6)2 + 1 .

Observing this, I try to prove the conjecture that:

Fn-1Fn+1 = (Fn)2 + (-1)n.

Using the general formula for F_n-1 and F_n+1 in terms of a and b, the roots of the equation x²-x-1=0, I find:

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.