Solve the following system of equations:

$$
\begin{aligned}
x+y & =2 a \ldots \\
x y\left(x^{2}+y^{2}\right) & =2 b^{4} \ldots
\end{aligned}
$$

What is the condition for obtaining real solutions?

Num. values: $\quad a=10, \quad b^{4}=9375$.
I. Solution. According to (1), the sum of the two unknowns is given. Let's add the difference of the unknowns, so that $x-y=2 z$. If we determine $z$, then $x$ and $y$ can be easily calculated.

$$
x+y=2 a \ldots
$$

and

$$
x-y=2 z \ldots
$$

thus

$$
x=a+z, \quad y=a-z
$$

Substituting these into (2):

$$
\left(a^{2}-z^{2}\right)\left[(a+z)^{2}+(a-z)^{2}\right]=2 b^{4}, \quad \text { or } \quad\left(a^{2}-z^{2}\right) 2\left(a^{2}+z^{2}\right)=2 b^{4}
$$

Therefore:

$$
a^{4}-z^{4}=b^{4} \quad \text { and } \quad z= \pm \sqrt[4]{a^{4}-b^{4}} \ldots
$$

The value of $x$ and thus $x, y$ are real if $a^{4} \geq b^{4}$ or $a^{2} \geq b^{2}$.

Accordingly,

$$
x=a+\sqrt[4]{a^{4}-b^{4}}, \quad y=a-\sqrt[4]{a^{4}-b^{4}} \ldots
$$

The values of $x$ and $y$ can be swapped. (We are dealing with a symmetric system of equations!)

With the given numerical values:

$$
x=10+\sqrt[4]{10000-9375}=10+\sqrt[4]{625}=15 \text { and } y=5
$$

or

$$
x=5 \text { and } y=15
$$

Baán Sándor (Bencés g. VI. o. Kőszeg).

II. Solution. Equation (2) can also be written as:

$$
x y\left[(x+y)^{2}-2 x y\right]=x y\left(4 a^{2}-2 x y\right)=2 b^{4}
$$

$$
\text { If now } x y=u, \quad \text { then } \quad u^{2}-2 a^{2} u+b^{4}=0 \ldots
$$

The roots of this equation are real if $a^{4}-b^{4} \geq 0$, i.e., $a^{2} \geq b^{2}$.

From (6),

$$
u=x y=a^{2} \pm \sqrt{a^{4}-b^{4}}
$$

Accordingly, $x$ and $y$ are the roots of the following equations:

$$
\begin{aligned}
& X^{2}-2 a X+u_{1}=0 \ldots \\
& X^{2}-2 a X+u_{2}=0 \ldots
\end{aligned}
$$

where

$$
u_{1}=a^{2}+\sqrt{a^{4}-b^{4}} \quad \text { and } \quad u_{2}=a^{2}-\sqrt{a^{4}-b^{4}} \text {. }
$$

The roots of (7) are not real, because

$$
a^{2}-u_{1}=a^{2}-a^{2}-\sqrt{a^{4}-b^{4}}=-\sqrt{a^{4}-b^{4}}<0
$$

The roots of (8) are $x_{1,2}=a \pm \sqrt{\sqrt{a^{4}-b^{4}}}=a \pm \sqrt{a^{4}-b^{4}}$.

If $X_{1}=x$, then $X_{2}=y$, or $X_{1}=y, X_{2}=x$.

The numerical values are given in the first solution.

Pfeifer Béla (Izr. g. V.o. Bp.).

Note: $a^{4}-b^{4}=\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right) \geq 0$, if $a^{2}-b^{2} \geq 0$. Indeed, $a^{2}+b^{2} \geq 0$, unless $a$ and $b$ are real numbers.

$$
a^{2}-b^{2} \geq 0 \quad \text { if } \quad|a| \geq|b|
$$