Find the maximum the value $P_{1}\cdot P_{n}$ 
Suppose that $n \ge 2$ students attend a test of $m \ge 2$ problems. The scoring rule for each problem is: if $x$ students answer a problem incorrectly, then a correct answer is worth $x$ points and an incorrect answer is worth none.  The total score of a student is the sum of scores of all $m$ problems. The total score of all students will be arranged from high to low as $$P_{1}\ge P_{2}\ge P_{3}\ge \cdots\ge P_{n}$$ Find the maximum of the value $P_{1}\cdot P_{n}$.


Let $a_{k}$ students answer the $k$-th problem
then $$p_{1}\le\sum_{k=1}^{m}(n-a_{k}),p_{1}+p_{2}+\cdots+p_{n}=\sum_{k=1}^{m}a_{k}(n-a_{k})$$
so
$$p_{1}\cdot p_{n}\le p_{1}\cdot \dfrac{p_{2}+p_{3}+\cdots+p_{n}}{n-1}$$
I use this idea,at last,I can't solve this problem,(But I fell this problem can use this idea)
PS: Somedays ago, I have solve this problem :
How find this maximum of $P_{1}+P_{n}$

Let $x_i$ be the score of the $i$-th problem, then
\begin{equation}
\begin{split}
P_1 &\le \sum_{i=1}^{m} x_i \\
P_n &\le \frac{1}{n-1} \Big( \sum_{i=1}^m x_i (n-x_i) - P_1 \Big) 
\end{split}
\end{equation}
denote $S = \sum_{i=1}^{m} x_i, T = \sum_{i=1}^m x_i (n-x_i)$, $P_1 P_n \le \frac{1}{n-1} P_1 ( T - P_1 )$.
In case $T \ge 2S$,
\begin{equation}
\begin{split}
P_1 P_n &\le \frac{1}{n-1} S(T-S) \\
& =  \frac{1}{n-1} S\Big((n-1)S - \sum x_i^2 \Big) \\
& \le \frac{1}{n-1} S\Big((n-1)S - \frac{S^2}{m} \Big) \\
& = 4m^2(n-1)^2 \cdot \frac{S}{2m(n-1)} \cdot \frac{S}{2m(n-1)} \Big( 1-\frac{S}{m(n-1)} \Big) \\
& \le \frac{4}{27} m^2(n-1)^2.
\end{split}
\end{equation}
The strategy to this bound is: $P_1$ get every problem correct, others try to have equal total score, and every problem has score $\frac{2}{3}(n-1)$. If $n \equiv 1$ (mod $3$), this bound can be achieved, slightly less for other $n$. But to make $T \ge 2S$ possible, we need $n \ge 4$.
In case $T \le 2S$,
\begin{equation}
\begin{split}
P_1 P_n &\le \frac{1}{n-1} \cdot \frac{T^2}{4} \le \frac{1}{n-1} S^2 \le m^2(n-1).
\end{split}
\end{equation}
This is just a rough estimation we can't take equality, but proves for $n \ge 8$, previous strategy is the best.
Actually in case $T \le 2S$, we have $T \le 2m(n-2)$. (for $n \ge 3$)
Rewrite the condition $T \le 2S$ as
\begin{equation}
\sum_{i=2}^{n-1} (i-2)(n-i) \#_i \le (n-1) \#_1
\end{equation}
where $\#_i$ is number of $x_j$ equal to $n-i$. One can check
\begin{equation}
\frac{(i-2)(n-i)}{n-1} \ge \frac{i(n-i)-2(n-2)}{n-3}.
\end{equation}
Hence
\begin{equation}
\sum_{i=3}^{n-1} i(n-i)-2(n-2) \#_i \le (n-3) \#_1, \\
\sum_{i=1}^{n-1} i(n-i) \#_i \le \sum_i 2(n-2) \#_i,
\end{equation}
this is exactly what we want.
This gives a new bound $\frac{m^2(n-2)^2}{n-1}$, can be achieved by taking $x_i = n-2$ for any $i$. This is the previous bound for $n=4$, for $n=5,6,7$, previous bound is larger, for large $m$ previous bound is better, for small $m$, you probably need to compare. For $n=3$, result is $\frac{m^2}{2}$.
For n = 2, result is $\frac{m^2}{4}$.
