## Task 5 - 271235

Determine all triples $(x, y, z)$ of integers that satisfy the following equation (1):

$$
1243 \cdot(1+y z)=65 \cdot(x y z+x+z)
$$
I. If $x, y, z$ are integers that satisfy (1), then:

Since 1243 is coprime to 65, $1 + yz$ must be divisible by 65, i.e., there must exist an integer $k$ such that

$$
1 + yz = 65 \cdot k
$$

From (1), it follows that

$$
65 \cdot (xyz + x + z) = 1243 \cdot 65 \cdot k \quad \text{thus} \quad (1 + yz) \cdot x + z = 1243 \cdot k
$$

With (2), this gives $65kx + z = 1243k$ and thus

$$
z = (1243 - 65 \cdot x) \cdot k
$$

Substituting this into the equation $65k - yz = 1$ derived from (2), we get

$$
(65 - y(1243 - 65 \cdot x)) \cdot k = 1
$$

This can only be satisfied due to the integer nature of the factors with

$$
65 - y(1243 - 65 \cdot x) = k = \pm 1
$$

Thus, it follows that

$$
y(1243 + 65 \cdot x) = 64 \quad \text{or} \quad y(1243 + 65 \cdot x) = 66
$$

i.e., $1243 - 65x$ must be a divisor of 64 or 66.

(6)

This particularly implies

$$
\begin{array}{rrrr}
-66 \leq 1243 - 65 \cdot x \leq 66 \quad \Rightarrow \quad 1177 \leq 65 \cdot x \Rightarrow 66 \quad \Rightarrow x = 19 \quad \text{or} \quad x = 20 \\
1243 - 65 \cdot x = 8 \quad \text{or} \quad 1243 - 65 \cdot x = -57 &
\end{array}
$$

The conditions (6) and (7) are only satisfied by $1243 - 65 \cdot x = 8$, i.e., $x = 19$. From (5), together with $y = 8$, (4) leads to $k = 1$ and thus (3) to $z = 8$.

Therefore, among all triples of integers, only

$$
(x, y, z) = (19, 8, 8)
$$

can satisfy equation (1).

II. As evident from $1243(1 + 8 \cdot 8) = 1243 \cdot 65 = 65 \cdot (19 \cdot 8 \cdot 8 + 19 + 8)$, this triple satisfies the equation. With I. and II., it is shown that exactly the triple mentioned in (8) meets the requirements of the problem.

## Adopted from $[5]$