Czech-Slovak-Polish Match

First part. Assume that the equations (2) have real roots satisfying $x_1{y}_1-x_2{y}_2=1$. By the familiar formula, the roots of the quadratic equations are given by $$x_{1,2}=\frac{-p\pm K}{2}\text{and}{y}_{1,2}=\frac{p\pm L}{2},$$ where the real numbers $K,L$ satisfy $K^2=p^2-4q$ and $L^2=p^2-4r$ (we choose the signs of $K,L$ in accordance with the labelling of the roots). Thus $$1=x_1{y}_1-x_2{y}_2=\frac{(-p+K)(p+L)-(-p-K)(p-L)}{4}=\frac{p(K-L)}{2},$$ whence $p\ne 0$ and $K-L=2/p$. Substituting this into the equality $$(K+L)(K-L)=K^2-L^2=(p^2-4q)-(p^2-4r)=4(r-q)$$ yields $K+L=2p(r-q)$. From these values of $K+L$ and $K-L$ we obtain $K=1/p-p(q-r)$, so upon squaring, $K^2=1/p^2-2(q-r)p^2+(q-r{)}^2$. Comparing this with the equality $K^2=p^2-4q$, an easy manipulation leads to the desired equation (1).

Second part. Assume that (1) holds. Then clearly $p\ne 0$. The equation (1) can be rewritten in either of the following two forms, $$p^4(q-r{)}^2+2p^2(q-r)+1=p^4-4p^{2}r\text{and}{p}^4(r-q{)}^2+2p^2(r-q)+1=p^4-4p^{2}q.$$ Upon dividing by $p^2$ we find that the discriminants of the equations (2) are equal to $$p^2-4q={\left(\frac{p^2(r-q)+1}{p}\right)}^2\text{and}{p}^2-4r={\left(\frac{p^2(q-r)+1}{p}\right)}^2;$$ hence, they are nonnegative and the (real) roots of (2) have the form (3), where $$K=\frac{p^2(r-q)+1}{p}\text{and}L=-\frac{p^2(q-r)+1}{p}.$$ The signs of the numbers $K$ and $L$ have been chosen so that (see First Part) $$x_1{y}_1-x_2{y}_2=\frac{p(K-L)}{2}=\frac{p}{2}\cdot \left(\frac{p^2(r-q)+1}{p}+\frac{p^2(q-r)+1}{p}\right)=1.$$