More fun in 2015 (or not)

Using Fabonnaci Q-Matrix as suggested by Otto Bretscher :

$\begin{aligned} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^k & = \begin{pmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{pmatrix} \\ \Rightarrow \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^j \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^k & = \begin{pmatrix} F_{j+1} & F_j \\ F_j & F_{j-1} \end{pmatrix} \begin{pmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{pmatrix} \\ \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{j+k} & = \begin{pmatrix} F_{j+1}F_{k+1} + F_jF_k & F_{j+1}F_k + F_jF_{k-1} \\ F_jF_{k+1} +F_{j-1}F_k & F_jF_k +F_{j-1}F_{k-1} \end{pmatrix} \\ \begin{pmatrix} F_{j+k+1} & F_{j+k} \\ F_{j+k} & \color{#3D99F6}{F_{j+k-1}} \end{pmatrix} & = \begin{pmatrix} F_{j+1}F_{k+1} + F_jF_k & F_{j+1}F_k + F_jF_{k-1} \\ F_jF_{k+1} +F_{j-1}F_k & \color{#3D99F6}{F_jF_k+F_{j-1}F_{k-1}} \end{pmatrix} \\ \Rightarrow \color{#3D99F6}{F_{j+k-1}} & = \color{#3D99F6}{F_jF_k+F_{j-1}F_{k-1}} \\ \text{For } j=k \quad \Rightarrow F_{2k-1} & = F_k^2 + F_{k-1}^2 \\ \text{Therefore, } S = F_{2017} & = F_{1008}^2 + F_{1009}^2 \\ \Rightarrow n & = \boxed{2017} \end{aligned}$

Statement: $F_{a+b}=F_aF_{b+1}+F_{a-1}F_b$

Proven using Mathematical induction on $b$ .

Substitute $a-1=b=n$ : $F_{2n+1}=F_nF_n+F_{n+1}F_{n+1}$

$n=1008$ : $F_{2017}=F_{1008}^2+F_{1009}^2$

$S = F_{1008}^2 + F_{1009}^2$ $S = \left( {\frac{{\varphi ^{1008} - \varphi ^{ - 1008} }}{{\sqrt 5 }}} \right)^2 + \left( {\frac{{\varphi ^{1009} + \varphi ^{ - 1009} }}{{\sqrt 5 }}} \right)^2$ $S \approx \left( {\frac{{\varphi ^{1008} }}{{\sqrt 5 }}} \right)^2 + \left( {\frac{{\varphi ^{1009} }}{{\sqrt 5 }}} \right)^2$ $S = \frac{{\varphi ^{2016} + \varphi ^{2018} }}{5}$ $S = \frac{{\varphi ^{2016} }}{{\sqrt 5 }}\frac{{1 + \varphi ^2 }}{{\sqrt 5 }}$ $S = \frac{{\varphi ^{2017} }}{{\sqrt 5 }} \approx \frac{{\varphi ^{2017} - \varphi ^{ - 2017} }}{{\sqrt 5 }} = F_{2017}$ $n = \boxed{2017}$