Given the formula for the nth Fibonacci number, namely
Fn= 1
Ö5
 (an-bn)

where a and b are solutions of the quadratic equation x2-x-1=0 and a > b, prove that

(1)
(a+ 1
a
 )=-(b+ 1
b
 ) = Ö5

,

(2) Fn2 + Fn+12 = F2n+1 where Fn is the nth Fibonacci number and

(3) for any four consecutive Fibonacci numbers Fn ¼Fn+3 the formula
(FnFn+3)2+(2Fn+1Fn+2)2
is the square of another Fibonacci number giving a Pythagorean triple.