5. Find all pairs of natural numbers $(m, n)$ for which the equality

$$
n \cdot(n-1) \cdot(n-2) \cdot(n-3)=m \cdot(m-1)
$$

holds.
Answer: $(1,1),(2,1);(3,1)$.

Solution: Rewrite the equation as

$$
\left(n^{2}-3 n\right) \cdot\left(n^{2}-3 n+2\right)=m^{2}-m
$$

and let $N=n^{2}-3 n+1$, then, by completing the squares, we get $N^{2}-1=\left(m-\frac{1}{2}\right)^{2}-\frac{1}{4}$. Multiplying by 4, we obtain $(2 N)^{2}-(2 m-1)^{2}=3$. The number $2 N$ is an even integer, and $2 m-1$ is an odd integer. Therefore, $2 N= \pm 2, 2 m-1= \pm 1$. From the first equation, we get $n=0,1,2,3$, and from the second $m=0,1$. Since the problem asks for only natural solutions, we have $n=1,2,3$ and $m=1$.

The given equation is possible only when both sides are equal to zero.