\section*{Exercise 4 — Candidates who have followed the specialization course}
We consider the equation (E)
$$x ^ { 2 } - 5 y ^ { 2 } = 1$$
where $x$ and $y$ are natural integers.

\section*{Part A}
We suppose that ( $x ; y$ ) is a solution pair of equation (E).
\begin{enumerate}
  \item Can $x$ and $y$ have the same parity? Justify.
  \item Prove that $x$ and $y$ are coprime.
  \item Let $k$ be a natural integer.\\
Copy and complete the following table:
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | }
\hline
\begin{tabular}{ l }
Remainder of the euclidean \\
division of $k$ by 5 \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 \\
\hline
\begin{tabular}{ l }
Remainder of the euclidean \\
division of $k ^ { 2 }$ by 5 \\
\end{tabular} &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}
  \item Deduce that $x \equiv 1$ [5] or $x \equiv 4$ [5].
\end{enumerate}

\section*{Part B}
Let $A$ be the matrix $\left( \begin{array} { c c } 9 & 20 \\ 4 & 9 \end{array} \right)$.\\
We consider the sequences $\left( x _ { n } \right)$ and $\left( y _ { n } \right)$ defined by
$$x _ { 0 } = 1 \text { and } y _ { 0 } = 0 \text {, and for all natural integer } n , \binom { x _ { n + 1 } } { y _ { n + 1 } } = A \binom { x _ { n } } { y _ { n } }$$
\begin{enumerate}
  \item For all natural integer $n$, express $x _ { n + 1 }$ and $y _ { n + 1 }$ in terms of $x _ { n }$ and $y _ { n }$.
  \item Prove by induction that, for all natural integer $n$, $\left( x _ { n } , y _ { n } \right)$ is a solution of equation (E).
  \item a. Determine $A ^ { 2 }$, then deduce $x _ { 2 }$ and $y _ { 2 }$.\\
b. Let $p$ be a natural integer.\\
Prove that if $y _ { p }$ is a multiple of 9 then $y _ { p + 2 }$ is also a multiple of 9.\\
c. Deduce that $y _ { 2020 }$ is a multiple of 9.
\end{enumerate}