Solve the following system of equations:

$$
\begin{aligned}
\left(x^{3}+y^{3}\right)\left(x^{2}+y^{2}\right) & =64 \\
x+y & =2
\end{aligned}
$$
I. solution: The factors on the left side of (1) are symmetric functions of $x$ and $y$, so it is expected that they can be expressed with $x+y$ and $xy$, the elementary symmetric functions. Indeed,

$x^{3}+y^{3}=(x+y)^{3}-3 x^{2} y-3 x y^{2}=(x+y)^{3}-3 x y(x+y)$,

$x^{2}+y^{2}=(x+y)^{2}-2 x y$

Thus, considering (2), from (1) we get a quadratic equation for $xy=w$:

$$
(8-6 x y)(4-2 x y)=64, \quad 3 w^{2}-10 w-8=0
$$

From this, $w_{1}=-2 / 3, w_{2}=4$.

Substituting these into (2), $x, y$ are given by the roots of the equation

$$
u^{2}-2 u+w=0
$$

where $u$ can represent either $x$ or $y$. From this, $u_{1}=1 \pm \sqrt{1-w}$. Only with $w_{1}$ do we get real solutions, the two pairs of roots are

$$
\left\{\begin{array}{l}
x_{1}=1+\sqrt{5 / 3}, \\
y_{1}=1-\sqrt{5 / 3};
\end{array} \quad \text { and } \quad \left\{\begin{array}{l}
x_{2}=y_{1} \\
y_{2}=x_{1}
\end{array}\right.\right.
$$

István Bencsik (Bp. V., Eötvös J. g. III. o. t.)

II. solution: The requirement of (2) can also be stated as: the arithmetic mean of $x$ and $y$ is 1. This also means that one is as much larger than 1 as the other is smaller than 1.

Denoting the difference by $v$:

$$
x=1+v, \quad y=1-v
$$

and substituting these into (1), we get a reducible fourth-degree equation in $v$:

$$
\begin{gathered}
\left(2+6 v^{2}\right)\left(2+2 v^{2}\right)=64 \\
3 v^{4}+4 v^{2}-15=0
\end{gathered}
$$

From this, $\left(v^{2}\right)_{1}=-3,\left(v^{2}\right)_{2}=5 / 3$. We only get real values of $v$ from the latter, with $v= \pm \sqrt{5 / 3}$, the above pairs of values for $x, y$ are obtained.

Péter Grünfeld (Bp. IX., József A. gépip. t. III. o. t.)

III. solution: Substituting $y=2-x$ from (2) into (1), we get the fourth-degree equation for $x$:

$$
x^{4}-4 x^{3}+\frac{22}{3} x^{2}-\frac{20}{3} x-\frac{8}{3}=0
$$

Here, the coefficients satisfy the condition that the left side can be transformed into the form $z^{4}+A z^{2}+B$ by a substitution of the form $x=z+\lambda$ $^{1}$, indeed

$$
a^{3}-4 a b+8 c=-64+\frac{352}{3}-\frac{160}{3}=0
$$

and the appropriate value of $\lambda$ is: $\lambda=-a / 4=+1$. With the substitution $x=z+1$, we essentially arrive at (5) again.

László Simonfai (Bp. II., Rákóczi F. g. IV. o. t.)[^0]


[^0]:    ${ }^{1} \mathrm{The}$ function $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$ can be written in the form $\varphi(z)=z^{4}+A z^{2}+B$ by a substitution of the form $x=z+\lambda$ if and only if $a^{3}-4 a b+8 c=0$, and the appropriate value of $\lambda$ is: $\lambda=-a / 4$ (KML. XI. volume 57. p., 656. problem, October 1955).