3. In the set of natural numbers, solve the equation

$$
k!l!=k!+l!+m!
$$
Solution. Without loss of generality, we can assume $k \geq l$. The given equation is equivalent to the equation $k!=\frac{k!}{l!}+1+\frac{m!}{l!}$. In the last equation, all three terms are natural numbers, so $\frac{m!}{l!}$ must also be a natural number, which means $m \geq l$. Furthermore, the sum on the right side is greater than or equal to 3, so $k \geq 3$, which means $k!$ is an even number. The latter means that exactly one of the numbers $\frac{k!}{l!}, \frac{m!}{l!}$ is an odd number.

1) Let's assume that $\frac{k!}{l!}$ is odd and $\frac{m!}{l!}$ is even. Then either $k=l$ or $k=l+1$ and $l$ is even, with $m \geq l+1$. If $k=l$, then $k!=2+\frac{m!}{l!}$.

- If $k=3$, then the solution is $k=l=3$ and $m=4$. If $k>3$, then $k!$ is divisible by 3, from which it follows that $k!-2$ is not divisible by 3, which means $m=k+1$ or $m=k+2$. From this, it follows that

$$
\frac{m!}{k!}=k+1 \text { or } \frac{m!}{k!}=(k+1)(k+2)
$$

so $k!=k+3$ or $k!=2+(k+1)(k+2)$. By direct verification, it is found that $k=3$ and $k=4$ are not solutions, and for $k>4$ the left side of the equations is greater than the right side.

- If $k=l+1$ and $l$ is even, then the equation $(l+1)!=l+2+\frac{m!}{l!}$ needs to be solved. The left side of the equation and $\frac{m!}{l!}$ are divisible by $l+1$, so $l+2$ must also be divisible by $l+1$, which is not possible.

2) Let $\frac{k!}{l!}$ be even and $\frac{m!}{l!}$ be odd. Then either $m=l$ or $m=l+1$ and $l$ is even.

- If $m=l$, then the equation reduces to $k!l!=k!+2l!$, or $\frac{k!}{l!}(l!-1)=2$. Since $\frac{k!}{l!}$ is even, we get $l!-1=1$, i.e., $l=2$ and $k!=4$, which is not possible.
- If $m=l+1$ and $l$ is even, then the equation is $k!(l!-1)=(l+2)l!$. Since $l!$ and $l!-1$ are coprime, we get that $l+2$ must be divisible by $l!-1$. The latter is only possible for $l=2$ and then we get $k!=8$, which is not possible.

From the previous considerations, it follows that the only solution to the given equation is $k=l=3$ and $m=4$.