Olimpiada Matemática del Cono Sur

a) First, note that
$$\begin{aligned} P_{2020} &= 1 \cdot (1 \cdot 2) \cdot (1 \cdot 2 \cdot 3) \cdots (1 \cdot 2 \cdot 3 \cdots 2020) = 1^{2020} \cdot 2^{2019} \cdot 3^{2018} \cdots 2019^2 \cdot 2020 \\ &= (1^{1010} \cdot 2^{1009} \cdot 3^{1009} \cdots 2018 \cdot 2019)^2 \cdot (2 \cdot 4 \cdot 6 \cdots 2020) \\ &= (1^{1010} \cdot 2^{1009} \cdot 3^{1009} \cdots 2018 \cdot 2019)^2 \cdot 2^{1010} \cdot 1010! \end{aligned}$$
which implies that $m = 1010$ is a solution.

Assume there is another solution $m$. Then, as $P_{2020}/1010!$ is a perfect square, we have that
$$\frac{P_{2020}/m!}{P_{2020}/1010!} = \frac{1010!}{m!}$$
is the square of a rational number.

If $m < 1009$, then $1010!/m!$ is an integer that is a multiple of $1009$, which is prime, but not a multiple of $1009^2$; therefore, $m$ is not a solution. It is clear that $m = 1009$ is not a solution either.

If $m \ge 1013$, then $m!/1010!$ is a multiple of $1013$, which is prime; so, in order that it is a multiple of $1013^2$, we should have $m \ge 2 \cdot 1013 = 2026$. But $2027$ is prime and $P_{2020}$ does not have $2027$ as a factor; then $m < 2027$. Thus, the only possibility is $m = 2026$, which is not a solution, since $2026!/1010!$ is a multiple of $1019$, which is prime, but not a multiple of $1019^2$.

The remaining cases are $m = 1011$ and $m = 1012$. It is immediate to verify that they are not solutions, since $1011$ and $1011 \cdot 1012$ are not perfect squares.

b) Similarly as in a), note that if $n = 4t$ and $m = 2t$ for a positive integer $t$, then $P_n/m!$ is a perfect square, since
$$P_n = 1^{4t} \cdot 2^{4t-1} \cdot 3^{4t-2} \cdots (4t-1)^2 \cdot 4t = (1^{2t} \cdot 2^{2t-1} \cdot 3^{2t-1} \cdots (4t-1))^2 \cdot 2^{2t} \cdot (2t)!$$

Consider $n = 8(k^2 + k)$. Then, as we have already shown, for $m = 4(k^2 + k)$ we have a solution. We will now show that $P_n/(m+1)!$ is also a perfect square. Note that $m+1 = 4k^2 + 4k + 1 = (2k+1)^2$, and
$$\begin{aligned} \frac{P_n}{(m+1)!} &= \frac{1}{m+1} \cdot \frac{P_n}{m!} = \frac{1}{(2k+1)^2} \cdot (1^m \cdot 2^{m-1} \cdots (2k+1)^{m-k} \cdots (n-1))^2 \cdot 2^m \\ &= (1^m \cdot 2^{m-1} \cdots (2k+1)^{m-k-1} \cdots (n-1))^2 \cdot 2^m \end{aligned}$$
which is an integer, since $m-k-1 = 4k^2 + 3k - 1 > 0$ for $k \ge 1$.