Let's calculate $x$, if

$$
x^{2}+(2 m p+2 n q)^{2}+(2 m q-2 n p)^{2}=\left(m^{2}+n^{2}+p^{2}+q^{2}\right)^{2}
$$
Expanding the squares on the right side, the sum of the two "double products" is $\operatorname{mnpq}(8-8)=0$, and the sum of the remaining square terms can be factored as:

$$
\begin{aligned}
& (2 m p+2 n q)^{2}+(2 m q-2 n p)^{2}=4 m^{2} p^{2}+4 n^{2} q^{2}+4 m^{2} q^{2}+4 n^{2} p^{2}= \\
& \quad=4\left(m^{2}+n^{2}\right)\left(p^{2}+q^{2}\right)
\end{aligned}
$$

On the right side, it is appropriate to combine $2-2$ terms accordingly:

$$
\left(\left(m^{2}+n^{2}\right)+\left(p^{2}+q^{2}\right)\right)^{2}=\left(m^{2}+n^{2}\right)^{2}+2\left(m^{2}+n^{2}\right)\left(p^{2}+q^{2}\right)+\left(p^{2}+q^{2}\right)^{2}
$$

Based on this, from (1)

$$
x^{2}=\left(m^{2}+n^{2}\right)^{2}-2\left(m^{2}+n^{2}\right)\left(p^{2}+q^{2}\right)+\left(p^{2}+q^{2}\right)^{2}=\left(m^{2}+n^{2}-p^{2}-q^{2}\right)^{2}
$$

From here, the possible values of $x$ are

$$
x_{1}=m^{2}+n^{2}-p^{2}-q^{2}, \quad x_{2}=-x_{1}=-m^{2}-n^{2}+p^{2}+q^{2}
$$

These are solutions to (1), since our transformations can also be performed in reverse.

Iván Marschik (Budapest, Radnóti M. Gyak. G. III. o. t.)

Remark. Based on the identity obtained by writing $x$, we can state that if the values of the parameters $m, n, p, q$ are positive integers, the

$$
2 m p+2 n q, \quad|2 m q-2 n p|, \quad\left|m^{2}+n^{2}-p^{2}-q^{2}\right|, \quad m^{2}+n^{2}+p^{2}+q^{2}
$$

formula quartet forms a (spatial) Pythagorean number quartet, that is, if the first three numbers are taken as the measurements of the edges meeting at a vertex of a rectangular parallelepiped, then the fourth number is the measurement of the space diagonal.