Pythagorean Fibs
What have Fibonacci numbers got to do with Pythagorean triples?
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.
For a sequence of, mainly more elementary, problems on these topics see Golden Mathematics