Baltic Way 2023 Shortlist

Answer: $(a,b)\in \{(1,b)\mid b\in {\mathbb{Z}}^{+}\}\cup \{(3,2),(9,1)\}$.

Let us consider the case where $k\ge 2$. Then $k\le 2(k-1)$ and note that $b+1\le 2b$ as $b\ge 1$. Put $(k-1)b=:x$, then $k(b+1)\le 4(k-1)b=4x$. So $10^x\le 36x$. It is obvious that the only solutions in nonnegative integers to this inequality are $x=0$ and $x=1$. Indeed, for $x\ge 2$, 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 $S(a^2)=a$, for $10\le a<100$. Then $a^2<10^4$, so $a=S(a^2)\le 9\cdot 4=36$. Moreover, taking into account the fact that the sum of digits does not change the number modulo 9, $a^2\equiv a(\mathrm{mod}9)$, i.e., $a(a-1)\equiv 0(\mathrm{mod}9)$, therefore $a\equiv 0(\mathrm{mod}9)$ or $a\equiv 1(\mathrm{mod}9)$. 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, by checking modulo 9, we get that $a^{b+1}\equiv a^b(\mathrm{mod}9)$ implies $a^b(a-1)\equiv 0(\mathrm{mod}9)$. 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 $S(a^{b+1})\le 9(b+1)$. Therefore, $3^b\le a^b=S(a^{b+1})\le 9(b+1)$. But from $3^b\le 9(b+1)$, we can conclude $b\le 3$. Indeed, for $b\ge 4$, the left hand side increases faster. So we are left with $a\in \{3,6,9\}$ and $b\le 3$. We check all these cases to determine that only $(a,b)=(3,2)$ or $(a,b)=(9,1)$ are solutions.