Race to end Fibonacci

$S = \left( {F_{2015} + F_{2013} } \right)\left( {F_{2013} + F_{2012} } \right) = \left( {F_{2015} + F_{2013} } \right)F_{2014}$ $S = \left( {\frac{{\varphi ^{2015} + \varphi ^{ - 2015} + \varphi ^{2013} + \varphi ^{ - 2013} }}{{\sqrt 5 }}} \right)\frac{{\varphi ^{2014} - \varphi ^{ - 2014} }}{{\sqrt 5 }}$ $S = \frac{1}{5}\left( {\varphi ^{4029} + \varphi ^{ - 4029} + \varphi ^{4027} + \varphi ^{ - 4027} } \right)$ $S \approx \frac{1}{5}\left( {\varphi ^{4029} + \varphi ^{4027} } \right) = \frac{{\varphi ^{4027} }}{{\sqrt 5 }}\frac{{\varphi ^2 + 1}}{{\sqrt 5 }}$ $S = \frac{{\varphi ^{4028} }}{{\sqrt 5 }} \approx \frac{{\varphi ^{4028} - \varphi ^{ - 4028} }}{{\sqrt 5 }} = F_{4028}$

Therefore, $a = 4028$ and:

$\frac{{a + 2}}{{10}} = \boxed{403}$

Considering the equation $Q^{2n}=(Q^n)^2$ for the Fibonacci Q-matrix , we find the equation $F_{2n}=(F_{n+1}+F_{n-1})F_n$ . Thus $S=(F_{2015}+F_{2013})F_{2014}=F_{4028}$ , with $\frac{a+2}{10}=\boxed{403}$