Czech-Slovak-Polish Match

Let us label the $p^2$ rows and the $p^2$ columns of the chessboard by all the pairs $(a,b)$ and $(c,d)$, respectively, where $a,b,c,d\in \{0,1,\dots ,p-1\}$. A field lying in the row $(a,b)$ and in the column $(c,d)$ will be called good if and only if $$ac\equiv b+d(\mathrm{mod}p).(1)$$ Given a pair $(a,b)$, (1) holds for exactly $p$ pairs $(c,d)$. Thus the number of good fields is $p$ in any row and $p^3$ in total.

It remains to show that there are no four good fields with the prohibited property. Suppose on the contrary that $$a_i{c}_j\equiv b_i+d_j(\mathrm{mod}p)\text{whenever }i,j\in \{1,2\},(2)$$ for some pairs $(a_1,b_1)\ne (a_2,b_2)$ and $(c_1,d_1)\ne (c_2,d_2)$. Subtracting (2) with a fixed $i$ yields $$a_i(c_2-c_1)\equiv d_2-d_1(\mathrm{mod}p)\text{for }i\in \{1,2\},(3)$$ which (in the same way) leads to $$(a_2-a_1)(c_2-c_1)\equiv 0(\mathrm{mod}p).$$ Thus $a_1=a_2$ or $c_1=c_2$. In view of symmetry, we can assume that $c_1=c_2$. Then (3) implies that $d_1=d_2$, hence $(c_1,d_1)=(c_2,d_2)$, a contradiction.