: Let $S(m)$ be the sum of the digits of the positive integer $m$. Find all pairs $(a, b)$ of positive integers such that $S\left(a^{b+1}\right)=a^{b}$.
Answer: $(a, b) \in\left\{(1, b) \mid b \in \mathbb{Z}^{+}\right\} \cup\{(3,2),(9,1)\}$.

Let $k$ denote the number of digits of $a$. Then $10^{k-1} \leq a<10^{k}$ and, therefore, $10^{(k-1) b} \leq a^{b}$ and $a^{b+1}<10^{k(b+1)}$. Of course, the digits are at most 9, so $S\left(a^{b+1}\right) \leq 9 \cdot k(b+1)$. We get

$$
10^{(k-1) b} \leq a^{b}=S\left(a^{b+1}\right) \leq 9 \cdot k(b+1) \text {, i.e. } 10^{(k-1) b} \leq 9 \cdot k(b+1)
$$

Let us consider the case where $k \geq 2$. Then $k \leq 2(k-1)$ and note that $b+1 \leq 2 b$ as $b \geq 1$. Put $(k-1) b=: x$, then $k(b+1) \leq 4(k-1) b=4 x$. So $10^{x} \leq 36 x$. It is obvious that the only solutions in nonnegative integers to this inequality are $x=0$ and $x=1$. Indeed, for $x \geq 2$, the left hand side grows faster. Therefore, either $k=1$ or $k=2$ and $b=1$.

Now we have only two cases left.

Case 1: $b=1$ and $k=2$. We are left with the equation $S\left(a^{2}\right)=a$, for $10 \leq a<100$. Then $a^{2}<10^{4}$, so $a=S\left(a^{2}\right) \leq 9 \cdot 4=36$.

Moreover, taking into account the fact that the sum of digits does not change the number modulo $9, a^{2} \equiv a(\bmod 9)$, i.e., $a(a-1) \equiv 0(\bmod 9)$, therefore $a \equiv 0(\bmod 9)$ or $a \equiv 1(\bmod 9)$. So now we are left only with numbers $a \in$ $\{10,18,19,27,28,36\}$, which we can easily check by substitution and see that there are no solutions.

Case 2: $k=1$. In the same way, looking modulo 9, we get that $a^{b+1} \equiv a^{b}(\bmod 9)$ implies $a^{b}(a-1) \equiv 0(\bmod 9)$. Therefore either $a=1$ or $a$ is divisible by 3 . $a=1$ is an obvious solution with all $b \in \mathbb{Z}^{+}$.

Otherwise, $a \in\{3,6,9\}$. But then $a^{b+1}<10^{b+1}$ and $S\left(a^{b+1}\right) \leq 9(b+1)$. Therefore, $3^{b} \leq a^{b}=S\left(a^{b+1}\right) \leq 9(b+1)$. But from $3^{b} \leq 9(b+1)$, we can conclude $b \leq 3$. Indeed, for $b \geq 4$, the left hand side increases faster. So we are left with $a \in\{3,6,9\}$ and $b \leq 3$. We check all these cases to determine that only $(a, b)=(3,2)$ or $(a, b)=(9,1)$ are solutions.