## Task 3B - 321233B

For each natural number $n \geq 2$, determine all $n$-tuples $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ of positive integers $x_{1}, x_{2}, \ldots, x_{n}$ that satisfy the following system of equations:

$$
\begin{aligned}
& x_{1} \cdot x_{2}=3 \cdot\left(x_{1}+x_{2}\right) \\
& x_{2} \cdot x_{3}=3 \cdot\left(x_{2}+x_{3}\right) \\
& \cdots \\
& x_{n-1} \cdot x_{n}=3 \cdot\left(x_{n-1}+x_{n}\right) \\
& x_{n} \cdot x_{1}=3 \cdot\left(x_{n}+x_{1}\right)
\end{aligned}
$$
The above system of equations can also be simply rewritten as

$$
\begin{array}{rlr}
\left(x_{1}-3\right)\left(x_{2}-3\right)=9 & \left(x_{2}-3\right)\left(x_{3}-3\right)=9 \\
\left(x_{n-1}-3\right)\left(x_{n}-3\right)=9 & \cdots & \left(x_{n}-3\right)\left(x_{1}-3\right)=9
\end{array}
$$

If we look at the solution of the first equation

$$
\left(x_{1}-3\right)\left(x_{2}-3\right)=9
$$

it is immediately clear that the only solution pairs consisting of positive integers are

$$
\left(x_{1}, x_{2}\right) \in\{(4,12),(6,6),(12,4)\}
$$

The same applies to the other equations, but the solutions are coupled in the following way:

If we choose a pair $(a, b)$ like the ones above for $\left(x_{1}, x_{2}\right)$, then for the second equation the solution pair $\left(x_{2}, x_{3}\right)=(b, a)$, for the third equation $\left(x_{3}, x_{4}\right)=(a, b)$, and so on, must be taken. Specifically, for odd $n$, $a=b=6$ must hold, otherwise different solutions for $x_{1}$ would result from the first and the last equation.

In summary, for every $n \geq 2$, the solution

$$
\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right)=(6,6,6, \ldots, 6)
$$

exists, and for even $n \geq 2$, there are also the two additional solutions

$$
\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right) \in\{(4,12,4,12, \ldots, 4,12),(12,4,12,4, \ldots, 12,4)\}
$$