9.5. Find all pairs of natural numbers $m, n$ for which $n!+4!=m^{2}$ (where $n!=1 \cdot 2 \cdot \ldots \cdot n$ ).
Answer: Two pairs $n=1, m=5$ and $n=5, m=12$. Solution. We will show that $n<6$. Indeed, for $n \geq 6$ the number $n!$ is divisible by 16. Therefore, the sum $n!+4$ can be represented as $8 \cdot(2 m+3)$. This number cannot be a square of an integer, since in its prime factorization, the factor 2 appears in an odd power. Thus, $n$ can take values from 1 to 5, of which only 1 and 5 are suitable. (A contradiction can also be obtained by considering divisibility by 3: for $n \geq 6$ the number $n!$ is divisible by 9, and $4!=24$ is divisible by 3 but not by 9, so the number on the left side of the equation cannot be a square of an integer).