3. Find all natural numbers $a, b, c$, satisfying the condition

$$
a^{b}+b^{c}=a b c
$$
Answer: $(1 ; 1 ; 2),(2 ; 2 ; 2),(2 ; 2 ; 3),(4 ; 2 ; 3),(4 ; 2 ; 4)$.

1) When $b=1$, we get the equation $a+1=a c \Rightarrow a(c-1)=1 \Rightarrow a=1, c=2$.
2) When $b=2$, the equation takes the form $(a-c)^{2}+2^{c}=c^{2}$. For $c=1$, it has no solutions, and by substituting $c=2,3,4$, we get the solutions $(2 ; 2 ; 2),(2 ; 2 ; 3)$ and $(4 ; 2 ; 3),(4 ; 2 ; 4)$, respectively. For $c>4$, there are no solutions, as in this case $2^{c}>c^{2}$ (proof by induction).
3) It remains to consider the case $b \geq 3$. When $a=1$, dividing both sides of the given equation by $b$, we get the equation $b^{c-1}=c-1 / b$, which clearly has no natural solutions. Let's now consider $a \geq 2$. We will prove that for these $a, b \geq 3$ and $c \geq 1$, the inequalities $a^{b} \geq \frac{2}{3} a^{2} b$ and $b^{c} \geq \frac{2}{3} b c^{2}$ hold. The first inequality:

$$
a^{b}=a^{b-2} a^{2} \geq 2 a^{2} \geq \frac{2}{3} a^{2} b
$$

The second inequality is verified directly for $c \leq 3$, and for $c \geq 4$ we have:

$$
b^{c} \geq b \cdot 3^{c-1}=\frac{b}{3}\left(\frac{3}{2}\right)^{c} \cdot 2^{c}>\frac{2}{3} b \cdot 2^{c} \geq \frac{2}{3} b c^{2}
$$

Finally, applying the inequality between the arithmetic mean and the geometric mean, we see that in the considered case, the left side of the equation is greater than the right side, i.e., there are no solutions:

$$
a^{b}+b^{c} \geq \frac{2}{3} b\left(a^{2}+c^{2}\right) \geq \frac{4}{3} b a c>a b c
$$