(2)We shall use the formula for the $n$th Fibonacci number and the facts that $\alpha \beta =-1$ and $\alpha +\frac{1}{\alpha }=-(\beta +\frac{1}{\beta })=\sqrt{5}.$

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

(3) For $n=1$ the expression gives $3^2+4^2=5^2=F_5^2.$

For $n=2$ the expression gives $5^2+{12}^2={13}^2=F_7^2.$

For $n=3$ the expression gives ${16}^2+{30}^2={34}^2=F_9^2.$

For $n=4$ the expression gives ${39}^2+{80}^2={89}^2=F_{11}^2.$

Conjecture : the result is $F_{2n+3}^2$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{(F_n{F}_{n+3}{)}^2+(2F_{n+1}{F}_{n+2}{)}^2}& =[(b-a)(b+a{)]}^2+[2\mathrm{ab}{]}^2\\ \multicolumn{1}{c}{}& =[b^2-a^2{]}^2+4a^2{b}^2\\ \multicolumn{1}{c}{}& =[b^2+a^2{]}^2\\ \multicolumn{1}{c}{}& =[F_{n+1}^2+F_{n+2}^2{]}^2\\ \multicolumn{1}{c}{}& =F_{2n+3}^2\end{array}}$

using the result from the second part of this question.