-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA512
%
>>36
% Tricky? That was easy as fuck.
%% LyX 2.0.3 created this file. For more info, see
http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[[rem][/rem]11pt,oneside,english,reqno]{amsart}
\usepackage[[rem][/rem]T1]{fontenc}
\usepackage[[rem][/rem]latin9]{inputenc}
\usepackage{amsthm}
\usepackage{amstext}
\usepackage{amssymb}
\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
\numberwithin{figure}{section}
\theoremstyle{definition}
\newtheorem*{defn*}{\protect\definitionname}
\theoremstyle{plain}
\newtheorem*{prop*}{\protect\propositionname}
\newcounter{casectr}
\newenvironment{caseenv}
{\begin{list}{{\itshape\ \protect\casename} \arabic{casectr}.}{%
\setlength{\leftmargin}{\labelwidth}
\addtolength{\leftmargin}{\parskip}
\setlength{\itemindent}{\listparindent}
\setlength{\itemsep}{\medskipamount}
\setlength{\topsep}{\itemsep}}
\setcounter{casectr}{0}
\usecounter{casectr}}
{\end{list}}
\makeatother
\usepackage{babel}
\providecommand{\casename}{Case}
\providecommand{\definitionname}{Definition}
\providecommand{\propositionname}{Proposition}
\begin{document}
\begin{defn*}
Define the fibonacci sequence by $F_{0}=0,F_{1}=1,F_{n}=F_{n-2}+F_{n-1}$.\end{defn*}
\begin{prop*}
\begin{equation}
\forall n\in\mathbb{N},\: F_{n}^{2}+F_{n+1}^{2}=F_{2n+1}\label{eq:st}
\end{equation}
\begin{proof}
We shall prove this statement by induction on $n$.
\begin{caseenv}
\item Suppose $n=0$. Then (\ref{eq:st}) holds since
\[[rem][/rem]
F_{0}^{2}+F_{1}^{2}=0^{2}+1^{2}=1=F_{1}=F_{2\cdot0+1}
\]
\item Suppose $n=1$. Then (\ref{eq:st}) holds since
\[[rem][/rem]
F_{1}^{2}+F_{2}^{2}=1^{2}+1^{2}=2=F_{3}=F_{2\cdot1+1}
\]
\end{caseenv}
Suppose $n\geq2$ and assume that (\ref{eq:st}) holds for $n-2$
and $n-1$ (inductive hypothesis). Then we have:
\begin{eqnarray}
F_{\left(n-2\right)}^{2}+F_{\left(n-2\right)+1}^{2} & = & F_{2(n-2)+1}\nonumber \\
\iff F_{n-2}^{2}+F_{n-1}^{2} & = & F_{2n-3}\label{eq:st_2n-3}
\end{eqnarray}
and
\begin{eqnarray}
F_{\left(n-1\right)}^{2}+F_{\left(n-1\right)+1}^{2} & = & F_{2(n-1)+1}\nonumber \\
\iff F_{n-1}^{2}+F_{n}^{2} & = & F_{2n-1}\label{eq:st_2n-1}
\end{eqnarray}
Let's subtract (\ref{eq:st_2n-3}) from (\ref{eq:st_2n-1}):
\begin{eqnarray}
\left(F_{n-1}^{2}+F_{n}^{2}\right)-\left(F_{n-2}^{2}+F_{n-1}^{2}\right) & = & \left(F_{2n-1}\right)-\left(F_{2n-3}\right)\nonumber \\
F_{n}^{2}-F_{n-2}^{2} & = & F_{2n-2}\label{eq:st_2n-2}
\end{eqnarray}
since $F_{2n-1}=F_{2n-2}+F_{2n-3}$. Now we know that $F_{2n}=F_{2n-1}+F_{2n-2}$,
thus $F_{2n+1}=F_{2n}+F_{2n-1}$.
\begin{eqnarray*}
F_{2n+1} & = & 2F_{2n-1}+F_{2n-2}\\
& = & 2(F_{n-1}^{2}+F_{n}^{2})+F_{n}^{2}-F_{n-2}^{2}\qquad\text{from \eqref{eq:st_2n-1} and \eqref{eq:st_2n-2}}\\
& = & 3F_{n}^{2}+2F_{n-1}^{2}-F_{n-2}^{2}
\end{eqnarray*}
We know that $F_{n-2}=F_{n}-F_{n-1}$ and that $F_{n-1}=F_{n+1}-F_{n}$,
and by combining the two we obtain $F_{n-2}=F_{n}-\left(F_{n+1}-F_{n}\right)=2F_{n}-F_{n+1}$.
Using these facts, we obtain
\begin{eqnarray*}
F_{2n+1} & = & 3F_{n}^{2}+2F_{n-1}^{2}-F_{n-2}^{2}\\
& = & 3F_{n}^{2}+2(F_{n+1}-F_{n})^{2}-(2F_{n}-F_{n+1})^{2}\\
& = & 3F_{n}^{2}+2(F_{n+1}^{2}-2F_{n+1}F_{n}+F_{n}^{2})-(4F_{n}^{2}-4F_{n}F_{n+1}+F_{n+1}^{2})\\
& = & (3+2-4)F_{n}^{2}+(2-1)F_{n+1}^{2}+(-4+4)F_{n+1}F_{n}\\
& = & F_{n}^{2}+F_{n+1}^{2}
\end{eqnarray*}
\end{proof}
\end{prop*}
\end{document}
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.4.12 (GNU/Linux)
iF4EAREKAAYFAlDjc40ACgkQGRQwWY30ng3C/gD+JMRRy5nTTICNXkcsFO0V8FQp
NKruXHYeTqSLi90PYz0A/imjqbnAzoC9g4DLTtWmJgavowlD9AvKaaEXv00Z/8mN
=IDGw
-----END PGP SIGNATURE-----