
\begin{enumerate}
\item La liste des nombres premiers inférieurs à 50 est:
\begin{center}
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
\end{center}

\item $\sqrt{\numprint{1517}}\approx38,95$.

On teste alors si les nombres premiers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 divisent \numprint{1517}.

\begin{console}
>>> L = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
>>> for i in L:
    	if 1517%i == 0: print(i)
 37
\end{console}

On trouve que 37 divise \numprint{1517} donc \numprint{1517} n'est pas premier.

\item $\begin{aligned}[t]
a^2=b^2+\numprint{1517} & \iff a^2-b^2=\numprint{1517}\\
                        & \iff (a-b)(a+b)=37\times41\\
                        & \iff \begin{cases}
                        a-b=37\\
                        a+b=41
                        \end{cases}\\
                        & \iff a = 39 \text{ et } b=2.
\end{aligned}$

\medskip

\uline{Ainsi, le seul couple $\coord{a}{b}\in\N^2$ tel que $a^2=b^2+\numprint{1517}$ est \coord{39}{2}.}
\end{enumerate}
