>>512
In \begin{math}
 n!+1=m^2 \end{math}, m can be rewrite into \begin{math} \dot{m}+n \end{math} and \begin{math} n!+1=(n+\dot{m})^2 \end{math} have the same integer solution of \begin{math} n!+1=(n+(n+1)^2)^2 \lor n!+1=(n-(n-1)^2)^2 \end{math} for \begin{math} n>1 \end{math}