\section*{Problem 4 - 261044}

Determine for each natural number \(k \geq 2\) the number of all solutions \((x, y, z, t)\) of the equation \(\overline{x y} + \overline{z t} = \overline{y z}\), where \(x, y, z, t\) are natural numbers with

\[
1 \leq x \leq k-1, \quad 1 \leq y \leq k-1 \quad 1 \leq z \leq k-1 \quad 1 \leq t \leq k-1
\]

\section*{are allowed!}

Here, \(\overline{p q}\) denotes the number that is written in the positional system of base \(k\) with the digits \(p, q\) (in this order).
}

I. If \(k, x, y, z, t\) are natural numbers such that \(k \geq 2\) and \(1 \leq x, y, z \leq k-1\) and \(0 \leq t \leq k-1\) and the required equation, i.e.,

\[
k \cdot x + k \cdot z + t = k \cdot y + z
\]

holds, then it follows that:

\[
k \cdot x + t = (k-1) \cdot (y - z)
\]

Due to \(x \geq 1\) and \(t \geq 0\), it follows that

\[
y - z \geq \frac{k \cdot 1 + 0}{k-1} > 1 \quad \text{and hence} \quad y - 1 > z
\]

Since \(z \geq 1\), it follows that \(y > 2\) (4) and since \(k-1 \geq y\), therefore \(k > 3\). According to (4), \(y\) is one of the numbers \(3, 4, \ldots, k-1 \quad (5)\)

According to (3), \(z\) is one of the numbers \(1, 2, \ldots, y-2\) (6).

II. Conversely, if \(k > 3\) and (5), (6), and (2) hold: \(y\) and \(z\) satisfy the conditions \(1 \leq y, z \leq k-1\); hence, on the one hand,

\[
(k-1) \cdot (y - z) \geq k - 1
\]

so \((k-1) \cdot (y - z)\) is a two-digit number in the positional system with base \(k\). By (2), natural numbers \(x, t\) with \(1 \leq x \leq k-1\) and \(0 \leq t \leq k-1\) are uniquely determined for each \(y, z\), and for these \(y, z, x, t\), the required equation (1) is also satisfied.

III. From I. and II. it follows: For \(k = 2\) and for \(k = 3\), the number of solutions sought is 0; in the case \(k \geq 4\), the number of solutions sought is equal to the number of all pairs \((y, z)\) that can be formed according to (5) and (6). Here, \(z\) takes the values

for \(y = 3\), the value \(z = 1\),

for \(y = 4\), the values \(z = 1, 2\)

...

for \(y = k-1\), the values \(z = 1, 2, \ldots, k-3\)

The number sought is thus \(1 + 2 + \ldots + (k-3) = \frac{1}{2}(k-3)(k-2)\).

Adapted from \([5]\)