Congratulations on your solutions to David from Sevenoaks School and Andrei from Tudor Vianu National College, Bucharest, Romania. The following solution takes parts of David's and parts of Andrei's solution.

We are 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.

Firstly, we can work out a and b, the two solutions of, x2-x-1=0 using the quadratic equation formula. As a > b we have
a = 1+Ö5
2

and
b = 1-Ö5
2

.

An interesting thing to point out here is that, using the formula for the sum and product of the roots of the quadratic equation, a+ b = 1 and ab = -1.

[ Editor's note: All the results can be proved by number crunching using the values of a and b and simplifying the expressions involving Ö5 but they can be proved more simply using the expression x2-x-1=0.]

(1) Dividing a2 - a- 1=0 by a gives
a-1- 1
a
 = 0

and it follows that
a+ 1
a
 = 2a-1 = Ö5

. Similarly
b+ 1
b
 = 2b- 1 = -Ö5

.

(2) From Part (1) we have
a2 -aÖ5 +1
= 0
b2 + bÖ5 + 1
= 0.
Substituting the definition of the n-th Fibonacci number and using these results from Part (1) and the identity ab = -1 we get:
Fn2 + Fn+12 - F2n+1
= 1
5
 [(an - bn)2 + (an+1 - bn+1)2 - Ö5(a2n+1 - b2n+1)]
= [(a2n + a2n+2 -Ö5a2n+1)+(b2n +b2n+2 + Ö5 b2n+1)-2(ab)n - 2(ab)n+1 ]
= 1
5
 [a2n(a2 -aÖ5 +1) + b2n(b2 + bÖ5 + 1)-2(ab)n(1 + ab)]=0
Hence Fn2 + Fn+12 = F2n+1.

(3)To show that for any four consecutive Fibonacci numbers Fn,¼,Fn+3 the formula
(FnFn+3)2+(2Fn+1Fn+2)2
is the square of another Fibonacci number it is much easier to consider the numbers with regard to the nature of the Fibonacci series that most people know and love, i.e. using the recursive formula for the series: Fn = Fn-1 + Fn-2 . Any four consecutive Fibonacci numbers can hence be expressed as x-y, y, x, x+y. (Check this for yourself!)

If we therefore substitute Fn = x-y, Fn+1=y, Fn+2 = x and Fn+3 = x + y , the formula
(FnFn+3)2+(2Fn+1Fn+2)2
becomes [(x-y)(x+y)]2+ [2yx]2. This gives
[(x-y)(x+y)]2+ [2yx]2 = x4 + y4 -2x2y2 + 4 x2y2 = (y2 + x2)2.
However, we have already shown that for any two consecutive Fibonacci numbers, Fn2 + Fn+12 = F2n+1 . We defined y and x earlier to be two consecutive Fibonacci numbers, so y2 + x2 must be the Fibonacci number F2n+3. Hence (y2 + x2)2 , which we have already shown to be the sum of the squares of two different Fibonacci numbers, is the square of another Fibonacci number F2n+3, and we have therefore made a Pythagorean triple. Therefore the proposition made in Part (3) of the question is true, QED!