Im not sure if this really works, but this is what I thought: rearranging the expression, we have $$$a_i \cdot a_j - (a_i+a_j) \equiv 0 \mod{p}$$$, which is equivalent to $$$(a_i-1)(a_j-1) \equiv 1 \mod{p}.$$$ This implies that $$$a_i \equiv a_j \equiv 0$$$ or $$$2 \mod{p}$$$. Then, let $$$x$$$ be the number of $$$a_k$$$ such that $$$a_k \equiv 0 \mod{p}$$$ and $$$y$$$ the number of $$$a_l$$$ such that $$$a_l \equiv 2 \mod{p}$$$. Answer is $$${x \choose 2} + {y \choose 2}$$$.

please give the problem link