Let's determine all value systems of the unknowns $x_{1}, x_{2}, x_{3}, x_{4}$, and $x_{5}$ that satisfy the following system of equations, where $y$ is a parameter:

$$
\begin{aligned}
& \text { (1) } \quad x_{5}+x_{2}=y x_{1}, \quad \text { (2) } \quad x_{1}+x_{3}=y x_{2}
\end{aligned}
$$

(3) $x_{2}+x_{4}=y x_{3}$,
(4) $x_{3}+x_{5}=y x_{4}$,
(5) $x_{4}+x_{1}=y x_{5}$.
We can express the other unknowns simply in terms of $x_{1}$ and $x_{2}$: first from (1) $x_{5}$, from (2) $x_{3}$, and then using these, from (3) $x_{4}$:

$$
\begin{gathered}
x_{5}=y x_{1}-x_{2} \\
x_{3}=-x_{1}+y x_{2} \\
x_{4}=-y x_{1}+\left(y^{2}-1\right) x_{2}
\end{gathered}
$$

Substituting the obtained expressions into (4) and (5) and reducing the equations to 0 (taking advantage of the available factoring opportunities):

$$
\begin{gathered}
(y-1)\left(x_{1}+x_{2}\right)+y^{2} x_{1}-y\left(y^{2}-1\right) x_{2}= \\
=\left(y^{2}+y-1\right) x_{1}+(y-1)[1-y(y+1)] x_{2}=0 \\
(1-y) x_{1}+\left(y^{2}-1\right) x_{2}-y^{2} x_{1}+y x_{2}=\left(1-y-y^{2}\right) x_{1}+\left(y^{2}+y-1\right) x_{2}=0
\end{gathered}
$$

that is,

$$
\begin{gathered}
\left(y^{2}+y-1\right)\left[x_{1}-(y-1) x_{2}\right]=0 \\
\left(y^{2}+y-1\right)\left(-x_{1}-x_{2}\right)=0
\end{gathered}
$$

Let's first examine the case when

$$
y^{2}+y-1 \neq 0
$$

Then

$$
\begin{gathered}
x_{1}-(y-1) x_{2}=0 \\
x_{1}-x_{2}=0
\end{gathered}
$$

that is,

$$
\begin{gathered}
x_{1}=x_{2} \\
(y-2) x_{1}=0
\end{gathered}
$$

From this, if $y \neq 2$ also holds, it follows from (1a), (2a), and (6) that

$$
x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0
$$

This clearly satisfies equations (1)-(5).

In the case $y=2$ - when (9) also holds - (11) is satisfied by any $x_{1}$ number, and from (10), (1a), (2a), and (6) we get

$$
x_{2}=x_{3}=x_{4}=x_{5}=x_{1}
$$

which, together with $y=2$, satisfies the system of equations (1)-(5) for any value of $x_{1}$.

If

$$
y^{2}+y-1=0
$$

that is, $y$ is equal to one of the numbers

$$
\eta_{1}=\frac{\sqrt{5}-1}{2} \quad \text { and } \quad \eta_{2}=\frac{\sqrt{5}+1}{2}
$$

then the system of equations (7), (8) is satisfied by any pair of numbers $x_{1}, x_{2}$, and the appropriate values of the other three unknowns are provided by (1a), (2a), and (6). Since for the roots of (12)

$$
\eta_{i}^{2}-1=-\eta_{i}, \quad(i=1,2)
$$

the solution is:

$$
x_{1}, \quad x_{2}, \quad x_{3}=-x_{1}+\eta_{i} x_{2}, \quad x_{4}=-\eta_{i}\left(x_{1}+x_{2}\right), \quad x_{5}=\eta_{i} x_{1}-x_{2}
$$

Substitution shows that any value system (13) satisfies the given system of equations.

In summary: if the value of the parameter $y$ is different from 2, $\eta_{1}$, and $\eta_{2}$, then the only solution of the system is $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$.

If $y=2$, then every value system $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=u$ is a solution (where $u$ is any number);
if $y$ is equal to some $\eta$ number for which $\eta^{2}+\eta-1=0$, then every value system

$$
x_{1}=u, \quad x_{2}=v, \quad x_{3}=-u+\eta v, \quad x_{4}=-\eta(u+v), \quad x_{5}=\eta u-v
$$

is a solution, where $u, v$ are any numbers (this naturally differs from the case $y=2$ even when $v=u$).

Draskovits Pál (Budapest, Vörösmarty M. g. IV. o. t.)

Remark. Some, out of excessive caution, separately discussed the case $y=1$ after deriving equation (11). This was unnecessary, because (11) is a consequence of (7a) and (8a) even in this case (equivalent to (7a)), and the resulting solution (i.e., all $x_{i}=0$) could also have drawn attention to this.