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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.
Pythagorean Fibs
Age 16 to 18
Challenge Level 
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.
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