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13. (2005 Croatian Mathematical Competition) In the set of positive integers, find all solutions to the equation $k!l!=k!+l!+m!$.

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13. (2005 Croatian Mathematical Competition) In the set of positive integers, find all solutions to the equation $k!l!=k!+l!+m!$.

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Note: The instruction at the end is not part of the original text and is provided for context.
13. Without loss of generality, let $k \geqslant l$, then
$k!=\frac{k!}{l!}+1+\frac{m!}{l!}$.
Since the three terms in the equation are integers, the last term must be an integer, i.e., $m \geqslant l$.
The sum of the three terms on the left side of the equation is at least 3, so $k \geqslant 3$. Therefore, $k!$ is even.
Thus, $\frac{k!}{l!}, \frac{m!}{l!}$ must have exactly one that is odd.
We consider two cases:
(1) $\frac{k!}{l!}$ is odd, $\frac{m!}{l!}$ is even, so $k=l+1$ and $l$ is even, or $k=l$. At the same time, $m \geqslant l+1$.
(i) $k=l$, then we have $k!=2+\frac{m!}{k!}$.

If $k=3$, the solution to the equation is $k=l=3, m=4$.
If $k>3$, then $k!$ is divisible by 3, but $k!-2$ is not divisible by 3.
Therefore, $m=k+1$ or $m=k+2$.
Thus, $\frac{m!}{k!}=k+1$ or $\frac{m!}{k!}=(k+1)(k+2)$, i.e.,
$k!=k+3$ or $k!=2+(k+1)(k+2)$.
Substituting $k=4$ and $k=5$, we find that neither is a solution to the equation.
For larger values of $k$, the left side of the equation is greater than the right side.
(ii) $k=l+1, l$ is even, then
$(l+1)!=l+2+\frac{m!}{l!}$.
Since $(l+1)!$ and $\frac{m!}{l!}$ are both divisible by $l+1$, $l+2$ must also be divisible by $l+1$, but this is impossible.
(2) $\frac{k!}{l!}$ is even, $\frac{m!}{l!}$ is odd, so $m=l+1$ and $l$ is even, or $m=l$. At the same time, $k \geqslant l+1$.
(i) If $m=l$, the equation simplifies to $k!l!=k!+2 l!$, i.e.,
$\frac{k!}{l!}(l!-1)=2$.
Since $\frac{k!}{l!}$ is even, we get $l!-1=1$.
So, $l=2, k!=4$, but this is impossible.
(ii) If $m=l+1, l$ is even, the equation simplifies to
$k!l!=k!+(l+2) l!$, i.e., $k!(l!-1)=(l+2) l!$.
Since $l!$ and $l!-1$ are coprime, $l+2$ must divide $l!-1$. Only when $l=2, k!=8$ is it possible, but this is impossible.
In conclusion, the equation has a unique solution $k=l=3, m=4$.