We define a sequence of integers as follows:
\begin{equation}
f_0=0
\end{equation}
\begin{equation}
f_1=1
\end{equation}
\begin{equation}
f_n=f_{n-1}+f_{n-2}~~\mbox{for} ~~n\geq 2
\end{equation}
1.Prove that
\begin{equation}\label{eq:987654}
f_1+f_2+\cdots+f_n=f_{n+2}-1
\end{equation}for all positive integers $n$.
Proof:When $n=1$,
\begin{equation}
f_1=f_3-1
\end{equation}
Suppose when $n=k$,
\begin{equation}
f_1+f_2+\cdots+f_k=f_{k+2}-1
\end{equation}
Then
\begin{equation}
f_1+f_2+\cdots+f_k+f_{k+1}=f_{k+2}+f_{k+1}-1=f_{k+3}-1
\end{equation}.Done.
2.Prove that
\begin{equation}
f_{n+1}f_{n-1}-f_n^2=(-1)^n
\end{equation}for all positive integers $n$.
Proof:We just need to prove that
\begin{equation}
f_nf_{n-1}+f_{n-1}^2-f_n^2=(-1)^n
\end{equation}
When $n=1$,
\begin{equation}
f_1f_0+f_0^2-f_1^2=-1
\end{equation}
Suppose when $n=k$,
\begin{equation}
f_kf_{k-1}+f_{k-1}^2-f_k^2=(-1)^k
\end{equation}
Then
\begin{equation}
f_{k+1}f_k+f_k^2-f_{k+1}^2=(f_k+f_{k-1})f_k+f_k^2-(f_k+f_{k-1})^2=f_k^2-f_kf_{k-1}-f_{k-1}^2=(-1)^{k+1}
\end{equation}Done.
3.Prove that
\begin{equation}
f_n=f_{k+1}f_{n-k}+f_kf_{n-k-1}
\end{equation}for all $k=0,1,\cdots,n$.
Proof:When $k=0$,
\begin{equation}
f_n=f_1f_n+f_0f_{n-1}
\end{equation}
Suppose when $k=t$,the theorem holds,that is,
\begin{equation}
f_n=f_{t+1}f_{n-t}+f_tf_{n-t-1}
\end{equation}
That is ,
\begin{equation}
f_n=f_{t+1}(f_{n-t-1}+f_{n-t-2})+f_tf_{n-t-1}
\end{equation}
Then
\begin{equation}
f_{t+2}f_{n-t-1}+f_{t+1}f_{n-t-2}=(f_{t+1}+f_t)f_{n-t-1}+f_{t+1}f_{n-t-2}=f_n
\end{equation}
Done.
3.Prove that $f_n$ divides $f_{ln}$ for all positive integers $l$.
Proof:According to 3,
\begin{equation}
f_{ln}=f_{n}f_{(l-1)n+1}+f_{(l-1)n}f_{n-(l-1)n-1}
\end{equation}Done(Why?)
4.Prove that
\begin{equation}
\begin{pmatrix}
f_{n+1}&f_n\\
f_n&f_{n-1}
\end{pmatrix}=\begin{pmatrix}
1&1\\
1&0\\
\end{pmatrix}^n
\end{equation}
Proof:Prove it by induction.When $n=1$,
\begin{equation}
\begin{pmatrix}
f_2&f_1\\
f_1&f_0\\
\end{pmatrix}=\begin{pmatrix}
1&1\\
1&0\\
\end{pmatrix}
\end{equation}
Suppose when $n=k$,
\begin{equation}
\begin{pmatrix}
f_{k+1}&f_k\\
f_k&f_{k-1}\\
\end{pmatrix}=\begin{pmatrix}
1&1\\
1&0
\end{pmatrix}^k
\end{equation}
Then in case of $n=k+1$,
\begin{equation}
\begin{pmatrix}
f_{k+1}&f_k\\
f_k&f_{k-1}
\end{pmatrix}\begin{pmatrix}
1&1\\
1&0
\end{pmatrix}=\begin{pmatrix}
f_{k+1}+f_k&f_{k+1}\\
f_k+f_{k-1}&f_k\\
\end{pmatrix}=\begin{pmatrix}
f_{k+2}&f_{k+1}\\
f_{k+1}&f_k
\end{pmatrix}
\end{equation}
Done.
Remark1:4$\Rightarrow$ 2