Which are those positive integers that are 1 greater than the sum of the squares of their digits in their decimal representation?
If the sought number in decimal notation is represented by the digits $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$, then on the one hand $0 \leqq a_{i} \leqq 9, a_{n} \neq 0$, and on the other hand, according to the conditions,

$$
a_{n} \cdot 10^{n}+a_{n-1} \cdot 10^{n-1}+\ldots+a_{1} \cdot 10+a_{0}=a_{n}^{2}+a_{n-1}^{2}+\ldots+a_{0}^{2}+1
$$

After rearranging,

$$
a_{n}\left(10^{n}-a_{n}\right)+a_{n-1}\left(10^{n-1}-a_{n-1}\right)+\ldots+a_{1}\left(10-a_{1}\right)=a_{0}\left(a_{0}-1\right)+1
$$

follows. Here, the right-hand side is largest when $a_{0}=9$, in which case its value is 73. The left-hand side is at least $a_{n}\left(10^{n}-a_{n}\right)$. Since $1 \leqq a_{n} \leqq 9$, we have

$$
\left(a_{n}-1\right)\left(10^{n}-a_{n}-1\right) \geqq 0, \quad \text { if } \quad n \geqq 1
$$

After multiplying out and rearranging, we get

$$
a_{n}\left(10^{n}-a_{n}\right) \geqq 10^{n}-1
$$

so, according to (1),

$$
73 \geqq 10^{n}-1
$$

thus $n$ can be at most 1, so the sought number can have at most two digits. From (1) we then get

$$
a_{1}\left(10-a_{1}\right)=a_{0}\left(a_{0}-1\right)+1
$$

The right-hand side of (2) is an odd number, so $a_{1}$ must be odd. The possible values of $a_{1}\left(10-a_{1}\right)-1$ are then 8, 20, or 24. Of these, only 20 can be written as the product of two consecutive integers, $20=5 \cdot 4$, so $a_{0}=5$, from which $a_{1}=3$ or $a_{1}=7$.

Thus, the problem has two solutions, 35 and 75.