FINAL ROUND

Answer: $(4;23)$, $(5;25)$, $(6;35)$. It is easy to see that the numbers $1!+505=506$, $2!+505=507$, $3!+505=511$ are not perfect squares. Further, $4!+505=529=23^2$, $5!+505=625=25^2$, $6!+505=1225=35^2$. So we have three pairs $(4;23)$, $(5;25)$, $(6;35)$ satisfying the problem conditions.

Show that there are no other solutions of the initial equation. Indeed, the numbers $7!+505=5545$ and $9!+505=363385$ are not perfect squares since they are divisible by $5$ but they are not divisible by $25$. The number $8!+505=40825=25\cdot 1633$ is not a perfect square since $1633$ is not a perfect square. At last, if $n\ge 10$, then the decimal representation of $n!$ ends by at least two zeros, so the decimal representation of the number $n!+505=m^2$ ends with the digits $05$. Therefore, this number is divisible by $5$ but is not divisible by $25$, so this number is not a perfect square.