We are given the formula for the $n$th Fibonacci number, namely $F_n=\frac{1}{\sqrt{5}}({\alpha }^n-{\beta }^n)$ 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 $\alpha =\frac{1+\sqrt{5}}{2}$ and $\beta =\frac{1-\sqrt{5}}{2}$.

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 $\alpha -1-\frac{1}{\alpha }=0$ and it follows that $\alpha +\frac{1}{\alpha }=2\alpha -1=\sqrt{5}$. Similarly $\beta +\frac{1}{\beta }=2\beta -1=-\sqrt{5}$.

(2) From Part (1) we have

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\alpha }^2-\alpha \sqrt{5}+1}& =0\\ \multicolumn{1}{c}{{\beta }^2+\beta \sqrt{5}+1}& =0.\end{array}}$

Substituting the definition of the n-th Fibonacci number and using these results from Part (1) and the identity $\alpha \beta =-1$ we get:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{F_n^2+F_{n+1}^2-F_{2n+1}}& =\frac{1}{5}[({\alpha }^n-{\beta }^n{)}^2+({\alpha }^{n+1}-{\beta }^{n+1}{)}^2-\sqrt{5}({\alpha }^{2n+1}-{\beta }^{2n+1})]\\ \multicolumn{1}{c}{}& =[({\alpha }^{2n}+{\alpha }^{2n+2}-\sqrt{5}{\alpha }^{2n+1})+({\beta }^{2n}+{\beta }^{2n+2}+\sqrt{5}{\beta }^{2n+1})-2(\alpha \beta {)}^n-2(\alpha \beta {)}^{n+1}]\\ \multicolumn{1}{c}{}& =\frac{1}{5}[{\alpha }^{2n}({\alpha }^2-\alpha \sqrt{5}+1)+{\beta }^{2n}({\beta }^2+\beta \sqrt{5}+1)-2(\alpha \beta {)}^n(1+\alpha \beta )]=0\end{array}}$

Hence $F_n^2+F_{n+1}^2=F_{2n+1}$.

(3)To show that for any four consecutive Fibonacci numbers $F_n,\dots ,F_{n+3}$ the formula

${\displaystyle (F_n{F}_{n+3}{)}^2+(2F_{n+1}{F}_{n+2}{)}^2}$

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 $x-y,\hspace{0.5em}y,\hspace{0.5em}x,\hspace{0.5em}x+y$. (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

${\displaystyle (F_n{F}_{n+3}{)}^2+(2F_{n+1}{F}_{n+2}{)}^2}$

becomes $[(x-y)(x+y{)]}^2+[2\mathrm{yx}{]}^2$. This gives

${\displaystyle [(x-y)(x+y{)]}^2+[2\mathrm{yx}{]}^2=x^4+y^4-2x^2{y}^2+4x^2{y}^2=(y^2+x^2{)}^2.}$

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!