Pythagorean Fibs
What have Fibonacci numbers got to do with Pythagorean triples?
Given the formula for the $n$th Fibonacci number, namely $F_n={1\over\sqrt5}(\alpha^n-\beta^n)$ where $\alpha$ and $\beta$ are solutions of the quadratic equation $x^2-x-1=0$ and $\alpha > \beta$, prove that
(1) $(\alpha + {1\over \alpha})=-(\beta +{1\over \beta}) =\sqrt 5$,
(2) $F_n^2 + F_{n+1}^2 = F_{2n+1}$ where $F_n$ is the $n$th Fibonacci number and
(3) for any four consecutive Fibonacci numbers $F_n \ldots F_{n+3}$ the formula $$(F_nF_{n+3})^2+(2F_{n+1}F_{n+2})^2$$ is the square of another Fibonacci number giving a Pythagorean triple.
Parts (2) and (3) use the earlier parts of the question.
For part (3), as usual try small values of $n$ first, look for a pattern and make a conjecture about the result you expect might always be true.
To prove your conjecture take $$F_n=b-a,\ F_{n+1}=a,\ F_{n+2}= b,\ F_{n+3}= b+a $$ because this symmetry in the algebra will make the working simpler.
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 $n$th Fibonacci number, namely $F_n={1\over\sqrt5}(\alpha^n-\beta^n)$ where $\alpha$ and $\beta$ are solutions of the quadratic equation $x^2-x-1=0$ and $\alpha > \beta$.
Firstly, we can work out $\alpha$ and $\beta$, the two solutions of, $x^2-x-1=0$ using the quadratic equation formula. As $\alpha > \beta$ we have $\alpha = {1+\sqrt 5\over 2}$ and $\beta = {1-\sqrt 5 \over 2}$.
An interesting thing to point out here is that, using the formula for the sum and product of the roots of the quadratic equation, $\alpha + \beta = 1$ and $\alpha\beta = -1$.
Editor's note: All the results can be proved by number crunching using the values of $\alpha$ and $\beta$ and simplifying the expressions involving $\sqrt 5$ but they can be proved more simply using the expression $x^2-x-1=0$.]
(1) Dividing $\alpha^2 - \alpha - 1=0$ by $\alpha$ gives $\alpha -1- {1\over \alpha} = 0 $ and it follows that $\alpha + {1\over \alpha} = 2\alpha -1 =\sqrt 5$. Similarly $\beta + {1\over \beta} = 2\beta - 1 = -\sqrt 5$.
(2) From Part (1) we have
$$\begin{eqnarray} \alpha^2 -\alpha\sqrt 5 +1 &= 0 \cr \beta^2 + \beta \sqrt 5 + 1 &= 0. \end{eqnarray}$$
Substituting the definition of the n-th Fibonacci number and using these results from Part (1) and the identity $\alpha\beta=-1$ we get:
$$\begin{eqnarray} F_n^2 + F_{n+1}^2 - F_{2n+1}&= {1\over 5}[(\alpha^n - \beta^n)^2 + (\alpha^{n+1} - \beta^{n+1})^2 - \sqrt 5(\alpha^{2n+1} - \beta^{2n+1})] \cr &= [(\alpha^{2n} + \alpha^{2n+2} -\sqrt 5\alpha^{2n+1})+(\beta^{2n} +\beta^{2n+2} + \sqrt 5 \beta^{2n+1})-2(\alpha\beta)^n - 2(\alpha\beta)^{n+1} ]\cr &={1\over 5}[\alpha^{2n}(\alpha^2 -\alpha\sqrt 5 +1) + \beta^{2n}(\beta^2 + \beta \sqrt 5 + 1)-2(\alpha\beta)^n(1 + \alpha\beta)]=0 \end{eqnarray}$$
Hence $F_n^2 + F_{n+1}^2 = F_{2n+1}$.
(3)To show that for any four consecutive Fibonacci numbers $F_n,\ldots ,F_{n+3}$ the formula $$(F_nF_{n+3})^2+(2F_{n+1}F_{n+2})^2$$ 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: $F_n = F_{n-1} + F_{n-2}$ . Any four consecutive Fibonacci numbers can hence be expressed as $x-y,\ y,\ x,\ x+y$. (Check this for yourself!)
If we therefore substitute $F_n = x-y$, $F_{n+1}=y$, $F_{n+2} = x$ and $F_{n+3}= x + y$ , the formula $$(F_nF_{n+3})^2+(2F_{n+1}F_{n+2})^2$$ becomes $[(x-y)(x+y)]^2+ [2yx]^2$. This gives $$[(x-y)(x+y)]^2+ [2yx]^2 = x^4 + y^4 -2x^2y^2 + 4 x^2y^2 = (y^2 + x^2)^2.$$ However, we have already shown that for any two consecutive Fibonacci numbers, $F_n^2 + F_{n+1}^2 = F_{2n+1}$ . We defined $y$ and $x$ earlier to be two consecutive Fibonacci numbers, so $y^2 + x^2$ must be the Fibonacci number $F_{2n+3}$. Hence $(y^2 + x^2)^2$ , which we have already shown to be the sum of the squares of two different Fibonacci numbers, is the square of another Fibonacci number $F_{2n+3}$, and we have therefore made a Pythagorean triple. Therefore the proposition made in Part (3) of the question is true, QED!
The Fibonacci sequence is defined by the recurrence relation (sometimes called 'difference equation') $$F_n + F_{n+1}=F_{n+2}.$$ This is the simplest possible second order recurrence relation with constant coefficients as all the coefficients are one. The method of solving recurrence relations like this is to let $F_n=x^n$. Then $x^n+x^{n+1}=x^{n+2}$ and hence (dividing by $x^n$), $1 + x = x^2$ giving the quadratic equation $x^2-x-1=0$. So the quadratic equation has solutions $x={1 \pm \sqrt5\over 2}$. Hence the solutions of the recurrence relation are $$F_n=A\left({1+\sqrt5\over 2}\right)^n +B \left({1-\sqrt 5\over 2}\right)^n$$ where we have to find the values of the constants $A$ and $B$.
Putting $n=1$ and $F_1 = 1$ and multiplying by 2 $$2 = A(1 + \sqrt 5)+B(1-\sqrt 5)$$ and putting $n=2$ and $F_2=1$ and multiplying by 4 $$ 4 = A(1 + \sqrt 5)^2 + B(1-\sqrt 5)^2.$$ Solving these simultaneous equations for $A$ and $B$ we get $$A={1\over \sqrt 5}, \quad B =-{1\over \sqrt 5}.$$ Hence the solution of the recurrence relation is $$F_n = {1\over \sqrt 5}\left({1+\sqrt 5\over 2}\right)^n - {1\over \sqrt 5}\left(1-\sqrt 5\over 2\right)^n.$$ \par Note that the formula for $F_n$ is given in terms of the roots of the quadratic equation $x^2-x-1=0$ and one of the roots is the Golden Ratio which accounts for the many connections between Fibonacci numbers and the Golden Ratio.
This problem complements the material in the article
For a sequence of, mainly more elementary, problems on these topics see Golden Mathematics