A certain three-digit number, when the first and third digits are swapped, increases by 198; when the second and third digits are swapped, it increases by 9; and the sum of the squares of the digits is 2 more than four times their sum. Which number is this $?$
I. Solution: Let the sought number be $100 x+10 y+z$, then, according to the problem,

\[
\begin{aligned}
& 100 x+10 y+z=100 z+10 y+x-198 \\
& 100 x+10 y+z=100 x+10 z+y-9 \\
& x^{2}+y^{2}+z^{2}-2=4(x+y+z) \\
& x=z-2 \\
& y=z-1
\end{aligned}
\]

Substituting these values into (3),

\[
(z-2)^{2}+(z-1)^{2}+z^{2}-2=4[(z-2)+(z-1)+z]=12(z-1)
\]

Rearranging and simplifying,

\[
z^{2}-6 z+5=0
\]

from which

\[
z_{1}=5, \quad\left[z_{2}=1 .\right]
\]

The latter root cannot be used in the calculation, because in this case $x=-1$.

Thus,

\[
z=5, \quad x=z-2=3, \quad y=z-1=4
\]

so the sought number is 345.

Gábor Halász (Bp. II., Rákóczi g. II. o. t.)

II. Solution: We can reach the goal with fewer calculations if we start from equation (3) and use the fact that $x, y, z$ are positive (single-digit) integers.

(3) can be written as:

\[
(x-2)^{2}+(y-2)^{2}+(z-2)^{2}=14
\]

so 14 must be decomposed into the sum of three squares of integers.

The largest square number must be 9, otherwise the sum would be at most 12, so only the decomposition $1^{2}+2^{2}+3^{2}=14$ is possible, which means the digits of the sought number are: $3,4,5$.

From the first two conditions, it follows that $z>x$, and $z>y$, so $z=5$, and thus the sought number is either 345 or 435. However, 435 does not satisfy the requirements of (1) and (2), so the solution can only be 345, and 345 indeed meets the conditions.