Solve the following system of equations:

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

$$
x^{4}+y^{4}=97 \text {. }
$$
From (1)

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

From (2)

$$
\left(x^{2}+y^{2}\right)^{2}=97+2 x^{2} y^{2}
$$

Thus

$$
\frac{78^{2}}{x^{2} y^{2}}=97+2 x^{2} y^{2}
$$

or

$$
2(x y)^{4}+97(x y)^{2}-6084=0
$$

from which

$$
x^{2} y^{2}=36 \text { and } x^{2} y^{2}=-84.5
$$

so

$$
x y= \pm 6
$$

and

$$
x y= \pm \sqrt{-84.5}
$$

If we restrict ourselves to real roots, then from (1) and (3) the values of $x$ are: $\pm 3, \pm 2$; the corresponding values of $y$ are: $\pm 2, \pm 3$.

(Preisich Gusztáv, Beszterczebánya.)

The problem was also solved by: Ádámffy E., Bartók I., Demjén E., Deutsch I., Enyedi B., Haar A., Harsányi Z., Hirschfeld Gy., Kertész G., König D., Kürti I., Liebner A., Losonczy I., Messer P., Moskovits Zs., Pám M., Pivnyik I., Popoviciu M., Prékopa D., Rássy P., Riesz K., Schlesinger O., Schuster Gy., Schwarz Gy., Szántó H., Szücs A., Weisz P.