Gazeta Matematica 2/2015, Problem S:L15.52

Problem:
Solve in natural numbers:
$$x!+112482225=y^2.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $112482225 \equiv 11 \pmod{13}$ and the perfect squares modulo $13$ are $0,1,3,4,9,10,12$. So, we have no solutions for $x \geq 13$. Hence, $x<13$. If $x=1$, we have $112482226$, which is divisible by $2$, but not by $4$, so it is not a perfect square.
If $x=2$, we have $112482227$, whose last digit is $7$, so it not a perfect square. If $x=3$, we have $112482231$, which has the last two digit odd, so it is not a perfect square. If $x=4$, we have $112482249$, which is divisible by $3$, but not by $9$, so it is not a perfect square. If $x \in \{5,6,7,8,9\}$, we obtain numbers which are divisible by $5$, but not by $25$, so they are not perfect squares. If $x=10$, we obtain $116111025=3^2\cdot5^2\cdot516049$, which is not a perfect square. If $x=11$, we have $152399025=12345^2$, so $y=12345$. If $x=12$, we have $591483825=3^2\cdot5^2\cdot2628817$, which is not a perfect square. In conclusion, the equation has a unique solution: $x=11,y=12345$.