Romanian Mathematical Olympiad

a) Since $400 = 1 \cdot 400 = 4 \cdot 100 = 16 \cdot 25$, we analyze three cases.
If $p(2m+1) = 1$, $p(2n+1) = 400$, we obtain $1 \le 2m+1 < 4$ and $400 \le 2n+1 < 441$, hence $m \in \{1, 2\}$ and $n \in \{200, 201, \dots, 219\}$, giving 40 pairs $(m, n)$. Similarly, in the second case we obtain 20 pairs, and in the last case, 24 pairs, leading to a total of 84 pairs.