(1) As
1
a
 = a-1

it follows that
a+ 1
a
 = 2a-1 = Ö5

. Similarly
b+ 1
b
 = -Ö5

.

(2)We shall use the formula for the nth Fibonacci number and the facts that ab = -1 and
a+ 1
a
 = -( b+ 1
b
 ) = Ö5.



Fn2+Fn+12
= 1
5
 [(an-bn)2+ (an+1-bn+1)2]
= 1
5
 [a2n+b2n-2(-1)n+a2n+2+b2n+2-2(-1)n+1]
= 1
5
 [a2n+1(a+ 1
a
 )+b2n+1(b+ 1
b
 )]
= 1
Ö5
 [a2n+1-b2n+1]
=F2n+1 .

(3) For n=1 the expression gives 32+42=52 = F52.

For n=2 the expression gives 52+122=132 = F72.

For n=3 the expression gives 162+302=342 = F92.

For n=4 the expression gives 392+802=892 = F112.

Conjecture : the result is F2n+32
(FnFn+3)2+(2Fn+1Fn+2)2
=[(b-a)(b+a)]2 + [2ab]2
= [b2-a2]2 +4a2b2
= [b2+a2]2
= [Fn+12 + Fn+22]2
= F2n+32
using the result from the second part of this question.