## Task 5 - 241235

Determine all triples $(a, b, c)$ of positive natural numbers for which $a^{b}+b^{c}=a b c$ holds.
We start with two simple auxiliary statements.

Lemma 1:

$$
\forall a \geq 2 \forall b \geq 4: \quad a^{b} \geq a^{2} b
$$

Proof: It suffices to show

$$
\forall b \geq 4: \quad 2^{b-2} \geq b
$$

This is trivially true for $b=4$, and if it holds for some $b \geq 4$, then because

$$
2^{(b+1)-2}=2 \cdot 2^{b-2} \geq 2 b > b+1
$$

it also holds for $b+1$, q.e.d.

Lemma 2:

$$
\forall b \geq 4 \forall c \geq 3: \quad b^{c} > b c^{2}
$$

Proof: This can be immediately reduced to the simpler claim

$$
\forall c \geq 3: \quad 4^{c-1} > c^{2}
$$

which we can prove by induction. The claim is directly verifiable for $c=3$, and from the validity for some $c \geq 3$, the validity for $c+1$ follows immediately:

$$
4^{c} = 4 \cdot 4^{c-1} > (2c)^{2} > (c+1)^{2}
$$

q.e.d.

Thus, from the well-known inequality for the arithmetic and geometric means, it immediately follows that under the assumptions of the two auxiliary lemmas, there can be no solution to the equation in the problem:

$$
\forall a \geq 2 \forall b \geq 4 \forall c \geq 3: \quad a^{b} + b^{c} > 2 \sqrt{\left(a^{2} b\right)\left(b c^{2}\right)} = 2 a b c > a b c \quad(*)
$$

We only need to check the triples $(a, b, c)$ not covered by $(*)$, which ultimately leads to the following five solutions to the problem:

$$
(a, b, c) \in \{(1,1,2),(2,2,2),(2,2,3),(4,2,3),(4,2,4)\}
$$