Suppose that there are $n$ subjects in total and consider bipartite graph $G$ between set $M$ of $m$ points corresponding to $m$ students and set $N$ of $n$ points corresponding to $n$ subject We count number of $(A,B,C)$ such that $A,B\in M,C\in N,AC,BC\in G$ easy to see that it's equal to $\binom{m}{2}$ and $\sum_{C\in N}{\binom{deg(C)}{2}} \leq n\cdot \binom{p}{2}$ And let $e$ is the number of edges in $G$, easy to see that $e\geq p\cdot n$ and $e\leq (p+1)m$ Combining two inequalities give us $m\leq p^2$ and equality hold when $n=p(p+1)$ For construction part, we denote each students with number $1,2,...,p^2$ and each subjects with number $1,2,...,p(p+1)$ For all positive integer $x$, let $T_x$ be integer such that $T_x \equiv_p x$ and $T_x\in \{ 1,2,...,p\}$ Then for student with number $p\cdot q+r$ where $q\geq 0,0<r\leq p$ let he/she took subjects number $p\cdot i+T_{r+i(q+1)}$ for all $i=0,1,2,...,p-1$ and also took subject number $p\cdot p+(q+1)$ Condition $(i)$ is obviously true Condition $(ii)$, If subject number is at least $p^2+1$, let it be $p^2+z$, then it taken by student number $p\cdot z+1,p\cdot z+2,...,p\cdot z+p$ If subject number is not more than $p^2$, let it be $p\cdot c+d$ where $0<d\leq p$, then it taken by student number $p\cdot i+j$ where $0\leq i,j\leq p,i\neq p,j\neq 0$ such that $T_{j+c(i+1)}=d$ which mean $j+c(i+1) \equiv_p d$ obviously have $p$ solutions Condition $(iii)$ let that $2$ student have number $p\cdot q_1+r_1,p\cdot q_2+r_2$ where $0<r_1,r_2\leq p$ If $q_1=q_2$ then that $2$ student have common subject number $p^2+q_1$ (Since $p\cdot i+T_{r_1+i(q_1+1)} \neq p\cdot i+T_{r_2+i(q_2+1)}$ for all $i=0,1,2,...,p-1$ and $r_1\neq r_2$) If $q_1\neq q_2$ then there exist exacly one number $i\in \{ 0,1,2,...,p-1\}$ such that $p\cdot i+T_{r_1+i(q_1+1)} = p\cdot i+T_{r_2+i(q_2+1)}$ $\centerline{Best regards,}$ $\centerline{ThE-dArK-lOrD}$