Pythagorean Fibs
What have Fibonacci numbers got to do with Pythagorean triples?
Age
16 to 18
Challenge level
Problem
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.
Getting Started
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.
Teachers' Resources
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