## Task 15/61

All four-digit numbers are to be determined that have the following properties:

1. The sum of the first and second digit is equal to the square of the first digit.
2. The difference between the second and the third digit is equal to the first digit.
3. The sum of the third and the fourth digit is equal to the second digit.

How can these numbers be generally represented?
The task involves four unknown numbers, namely the four digits of the sought four-digit number. We denote the first digit with $x$, the second with $y$, the third with $z$, and the fourth with $u$. From the three conditions, we can set up three equations:
(1) $x+y=x^{2}$
1st condition
(2) $y-z=x$
2nd condition
(3) $z+u=y$
3rd condition

Since there are more unknowns than equations, the system of equations has infinitely many solutions. However, the number of solutions is strongly restricted by the problem statement: The solutions $x, y, z, u$ are digits of a four-digit number, so they must be one of the integers between 0 and 9 (both values inclusive). Therefore, $0 \leq x, y, z, u \leq 9$, and $x, y, z, u$ are integers.

To solve, we rearrange equation (1): $y=x(x-1)$ (1b). From this equation, we see that $x$ can only take the four values $x_{1}=0, x_{2}=1, x_{3}=2, x_{4}=3$; for $x=4$, $y$ would already be 12, which is not a single digit.

This gives us the $y$-values: $y_{1}=0, y_{2}=0, y_{3}=2, y_{4}=6$. Using equation (2), we get the $z$-values: $z_{1}=0, z_{2}=-1, z_{3}=0, z_{4}=3$. The value with index 2 is excluded because $z_{2}<0$. Using equation (3), we get $u_{1}=0, u_{3}=2, u_{4}=3$.

Thus, we have found the four-digit numbers 2202 and 3633 as solutions. The result 0000 is not considered a valid four-digit number.

To represent these numbers with a general expression, we write them in the form $1000 x + 100 y + 10 z + u$. Substituting $u=y-z$ and other relationships from above, we get

$$
1000 x + 100 y + 10 z + u = 1000 x + 101 y + 9 z = 991 x + 110 y = 881 x + 110 x^{2}
$$

This expression yields the number 2002 for $x=2$ and the number 3633 for $x=3$, which satisfy the given conditions. That there are no other numbers with the required properties follows from the solution approach, in which all possibilities have been exhausted.