Romanian Mathematical Olympiad

Consider the relation $(a+c)(b+c)=ab+ac+bc+c^2=(a+b+c+d)c+ab-cd=pc+ab-cd$. If $p$ divides $ab-cd$, then $p$ divides $a+c$ or $b+c$. On the other hand, $0<a+c<p$, and $0<b+c<p$, a contradiction.