Problem 10.6. Determine all natural numbers $m$ and $n$ that satisfy the equation $(m!-1)^{2}=n!+1$.

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Solution. $\quad(m!-1)^{2}=n!+1 \Leftrightarrow n!=m!\cdot(m!-2) \Leftrightarrow \frac{n!}{m!}=m!-2$.

$n!+1 \geq 2 \Rightarrow m \geq 3 \Rightarrow \frac{n!}{m!} \geq 4 \Rightarrow n>m$, so the equation $\left({ }^{*}\right)$ can be written as $(m+1) \cdot(m+2) \cdot \ldots \cdot n=m!-2$.

If the left side of this equality has at least 3 factors, then the left side of the equality is divisible by 3. We observe that the right side is not divisible by 3, so the equality is impossible.

There are two cases left to examine:

1. $n=m+1$. We get the equation $m+1=m!-2 \Leftrightarrow m!=m+3 \Leftrightarrow(m-1)!=1+\frac{3}{m}$.

The only solution to this equation in $N$ is $m=3$. Then $n=4$.

2. Let $n=m+2$. We get the equation $(m+1)(m+2)=m!-2 \Leftrightarrow m^{2}+3 m+4=m$.

$m^{2}+3 m+4 \geq 22$, then $m \geq 4$. For $m=4$, the equality is false. For $m \geq 5$, the last digit of the number $m!$ is equal to zero, while the last digit of the number $m^{2}+3 m+4$ can be 2, 4, or 8. Answer: $m=3, n=4$