In the following product, convert each of the two factors into fractions and then transform the numerator into the product of four numbers. How can we write the numerator if we replace $a+b+c+d$ with $2s$?

$$
\left(1+\frac{a^{2}+b^{2}-c^{2}-d^{2}}{2(a b+c d)}\right) \cdot\left(1-\frac{a^{2}+b^{2}-c^{2}-d^{2}}{2(a b+c d)}\right)
$$
The given expression can be written as:

$$
\frac{a^{2}+2 a b+b^{2}-\left(c^{2}-2 c d+d^{2}\right)}{2(a b+c d)} \cdot \frac{c^{2}+2 c d+d^{2}-\left(a^{2}-2 a b+b^{2}\right)}{2(a b+c d)}
$$

The numerator takes the following form:

$$
\begin{gathered}
{\left[(a+b)^{2}-(c-d)^{2}\right] \cdot\left[(c+d)^{2}-(a-b)^{2}\right]=} \\
=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)= \\
=(2 s-2 d)(2 s-2 c)(2 s-2 b)(2 s-2 a)= \\
=16(s-a)(s-b)(s-c)(s-d)
\end{gathered}
$$

(Note by Pál Ligeti, Budapest)

Note. The area of a cyclic quadrilateral:

$$
F=\sqrt{(s-a)(s-b)(s-c)(s-d)}
$$

Number of solutions: 8.