Solve the following system of equations:

$$
\begin{aligned}
& x+y+z+u=8 \\
& x^{2}+y^{2}+z^{2}+u^{2}=20 \\
& x y+x u+z y+z u=16 \\
& x y z u=9
\end{aligned}
$$
Our equations, with the exception of (3), are symmetric in $x, y, z$, and $u$, meaning each unknown plays the same role: in (1), each appears as a first-degree term; in (2), each appears as a squared term on the same side; in (4), each appears as a first-degree factor in the same term. In (3), however, each term is the product of two different unknowns, but the possible products $x z$ and $y u$ are missing, making this equation symmetric in the pairs $x, z$ and $y, u$. By leveraging these observations, we can break down the solution of our system into several steps.

From (3) and (1), by appropriate grouping:

$$
\begin{gathered}
(x+z)(y+u)=16 \\
(x+z)+(y+u)=8
\end{gathered}
$$

This is the simplest type of quadratic system for the unknowns $x+z=\alpha$ and $y+u=\beta$ (the product and the sum are given), so their values can be found from the equation $\xi^{2}-8 \xi+16=(\xi-4)^{2}=0$ (where $\xi$ can represent either $\alpha$ or $\beta$), yielding the single solution $\alpha=\beta=4$.

On the other hand, by squaring (1), we almost get all the terms of (2) and (3) doubled on the left side, i.e.,

$$
\begin{gathered}
(x+y+z+u)^{2}=\left(x^{2}+y^{2}+z^{2}+u^{2}\right)+2[(x y+x u+z y+z u)+ \\
+(x z+y u)]
\end{gathered}
$$

From this, we get an equation for the sum of the unknowns $x z=\gamma$ and $y u=\delta$, whose product is also known from (4). Considering (1)-(3), from (5) we have

$$
64=20+32+2(x z+y u)
$$

Thus,

$$
\begin{gathered}
\gamma+\delta=6 \quad \text { and } \\
\gamma \delta=9
\end{gathered}
$$

From the equation $\eta^{2}-6 \eta+9=(\eta-3)^{2}=0$, the single solution is $\gamma=\delta=3$.

Thus, we have obtained both the sums and the products of the unknown pairs $x, z$ and $y, u$, forming a third "sum and product" type system. Moreover, since $\alpha=\beta$ and $\gamma=\delta$, the two systems differ only in the notation of the unknowns, and $x$ and $z$ are roots of the same quadratic equation as $y$ and $u$, namely the equation $\zeta^{2}-\alpha \zeta+\gamma=0$ or $\zeta^{2}-4 \zeta+3=0$.

The solutions are:

$\left(A_{1}\right): x_{1}=1 \quad$ and $z_{1}=3, \quad\left(A_{2}\right): x_{2}=3$ and $z_{2}=1$;

$\left(B_{1}\right): y_{1}=1 \quad$ and $u_{1}=3, \quad\left(B_{2}\right): y_{2}=3 \quad$ and $u_{2}=1$.

Now we can obtain all the solutions of the complete system by writing down all $2 \cdot 2=4$ possible pairs from $A_{1}$ and $A_{2}$, as well as $B_{1}$ and $B_{2}$ (Table 1).

|  | $x$ | $z$ | $y$ | $u$ |
| ---: | :---: | :---: | :---: | :---: |
| I. | 1 | 3 | 1 | 3 |
| II. | 1 | 3 | 3 | 1 |
| III. | 3 | 1 | 1 | 3 |
| IV. | 3 | 1 | 3 | 1 |

George Muszély (Bp. VIII., Vörösmarty M. g. II. o. t.)

Remarks. 1. If we assume that the roots are integers, then the solutions of our system can also be found with the following considerations. Due to (2), the absolute value of none of the unknowns can be greater than 4. Due to (4), none can be even (0 is also not possible), and the absolute value of two must be 3, and of two must be 1. None can be negative, because due to (4), the number of negatives can only be even, but due to (1), the negativity of all four is not possible, and two cannot be negative, because then, according to our previous finding (by making the two with absolute value 3 positive and the two with absolute value 1 negative), the left side of (1) would be maximally $3+3-1-1=4$. Therefore, the values of the unknowns can only be 1, 1, 3, 3 in some order, which also satisfies (2). The arrangement of these numbers into the roles of $x, y, z, u$ can be done based on (3). Out of the six possible orders, 1,3,1,3 and 3,1,3,1 (be careful with the alphabetical order!) do not meet the criteria.

Ildikó Marót (Bp. V., Veres Pálné lg. II. o. t.)

2. The above "search" cannot be considered a complete solution, partly because it is based on the often heard but incorrect notion that numbers only mean the best-known numbers, the natural numbers (it does not matter that the right-hand sides of (1)-(4) are integers), and partly because even systematic trial and error does not always guarantee the production of all root systems, and is always tailored to a specific case, not universally valid.

The solution presented also has a certain trial-and-error element, but this pertains to the relationships among the unknowns, not to the positive integer nature of the known numbers. The method presented could solve the system even with other, non-integer numbers, or even with $a, b, c, d$, in which case we would need to proceed with the two roots of the equations for $\xi$ and $\eta$.