Let's determine the real roots of the following system of equations

$$
x y\left(x^{2}+y^{2}\right)=78, \quad x^{4}+y^{4}=97
$$
Let's square the first equation:

$$
x^{2} y^{2}\left(x^{4}+2 x^{2} y^{2}+y^{4}\right)=78^{2}
$$

Substitute the value of $x^{4}+y^{4}$ from the second equation, then

$$
2 x^{4} y^{4}+97 x^{2} y^{2}-78^{2}=0
$$

from which

$$
x^{2} y^{2}=\frac{-97 \pm \sqrt{9409+48672}}{4}=\frac{-97 \pm 241}{4}
$$

Only the positive value provides a real root:

$$
x^{2} y^{2}=36
$$

From this, since according to the first equation $x y$ must be positive,

$$
x y=6
$$

and from the first equation,

$$
x^{2}+y^{2}=\frac{78}{6}=13
$$

From the last two equations, it follows that

$$
(x+y)^{2}=x^{2}+y^{2}+2 x y=25, (x-y)^{2}=x^{2}+y^{2}-2 x y=1
$$

It is clear that only real numbers can satisfy our initial equations where $x$ and $y$ have the same sign, and that thus, along with the $x, y$ value pair, the pairs $-x, -y ; y, x ; -y, -x$ also provide solutions. It is sufficient to find a solution where $x>y>0$, and for these, the above 3 additional value pairs. Under this condition, $x+y$ and $x-y$ are positive, so

$$
x+y=5, x-y=1, \text { from which } x_{1}=3, y_{1}=2
$$

and the other 3 root pairs

$$
x_{2}=2, y_{2}=3 ; x_{3}=-3, y_{3}=-2 ; x_{4}=-2, y_{4}=-3
$$

Parlagh Gyula (Kecskemét, Katona J. g. IV. o. t.)