29° Olimpiada Matemática del Cono Sur

a) First, note that $$\begin{array}{rl}{P}_{2020}& =1\cdot (1\cdot 2)\cdot (1\cdot 2\cdot 3)\cdots (1\cdot 2\cdot 3\cdots 2020)=1^{2020}\cdot 2^{2019}\cdot 3^{2018}\cdots 2019^2\cdot 2020\\ & =(1^{1010}\cdot 2^{1009}\cdot 3^{1009}\cdots 2018\cdot 2019{)}^2\cdot (2\cdot 4\cdot 6\cdots 2020)\\ & =(1^{1010}\cdot 2^{1009}\cdot 3^{1009}\cdots 2018\cdot 2019{)}^2\cdot 2^{1010}\cdot 1010!\end{array}$$ 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 $$\begin{array}{rl}\frac{P_n}{(m+1)!}& =\frac{1}{m+1}\cdot \frac{P_n}{m!}=\frac{1}{(2k+1{)}^2}\cdot (1^m\cdot 2^{m-1}\cdots (2k+1{)}^{m-k}\cdots (n-1){)}^2\cdot 2^m\\ & =(1^m\cdot 2^{m-1}\cdots (2k+1{)}^{m-k-1}\cdots (n-1){)}^2\cdot 2^m\end{array}$$ which is an integer, since $m-k-1=4k^2+3k-1>0$ for $k\ge 1$.