Problem 1610

For any positive integer $a$, let $s(a)$ denote the sum of digits of $a$. Clearly $s(n + 1) = s(n) + 1$ unless the last digit of $n + 1$ is zero. The only two consecutive integers that are both perfect squares are $0$ and $1$, but $s(n) = 0$ is impossible for a positive $n$. The contradiction shows that the last digit of $n + 1$ must be $0$ and the last digit of $n$ must be $9$.

Let $n$ be an interesting positive integer that ends with exactly $m$ digits $9$. Clearly $s(n) - s(n + 1) = 9m - 1$. Note that a perfect square is congruent to $0$, $1$, $4$ or $7$ modulo $9$. As $s(n) - s(n + 1) = 9m - 1 \equiv 8 \pmod 9$, we must have $s(n) \equiv 0 \pmod 9$ and $s(n + 1) \equiv 1 \pmod 9$. But $s(n + 1) = 1$ is impossible, because this would require $n + 1$ to be a power of $10$, in the case of which $n$ and $n + 1$ would not consist of the same number of digits. Thus $s(n + 1) \ge 8^2 = 64$, as $8^2$ is the next smallest perfect square congruent to $1$ modulo $9$. But as $s(n) > s(n + 1)$, we have $s(n) \ge 9^2 = 81$. This implies that $n$ must contain at least $9$ digits. Containing exactly $9$ digits would be possible only if $n$ consisted entirely of nines, but in such case $n + 1$ would contain one more digit. Hence $n$ must contain at least $10$ digits.

Consider now the $10$-digit number $n = 788888899$. In this case $m = 2$, $s(n) = 81 = 9^2$ and $s(n + 1) = 81 - 9 \cdot 2 + 1 = 64 = 8^2$, thus $n$ is interesting. Inserting zeros between $7$ and $8$ does not influence $s(n)$ and $s(n + 1)$. So we can obtain an $m$-digit interesting number for every $m \ge 10$.