Define the product $P_n=1! \cdot 2!\cdot 3!\cdots (n-1)!\cdot n!$
a) Find all positive integers $m$, such that $\frac {P_{2020}}{m!}$ is a perfect square.
b) Prove that there are infinite many value(s) of $n$, such that $\frac {P_{n}}{m!}$ is a perfect square, for at least two positive integers $m$.
$a)$ We can rearrange the product the following way
$P_{2020}= 1!2!3!\cdots2020!$
$=1!(2\cdot1!)3!(4\cdot3!)\cdots2019!(2020\cdot2019!)$
$=2\cdot4\cdot6\cdots2020(1!)^2(3!)^2\cdots(2019!)^2$
$=2^{1010}1010!(1!)^2(3!)^2\cdots(2019!)^2$
So it suffices to take $m=1010$. Notice that the argument can me modified for any $n$ such that $4|n$ and taking $m=n/2$.
$b)$ Will notice that it suffices to take $n=8k^2$ for some $k\in\Bbb{Z}^+$, repeating the same argument as in $(a)$ we get
$$P_n=2^{4k^2}(1!)^2(3!)^2\cdots[(8k^2-1)!]^2(4k^2)!$$
Then it follows that $m=4k^2$ and $m=4k^2-1$ work for every $k\in\Bbb{Z}^+$
Andradessssss wrote:
$a)$ We can rearrange the product the following way
$P_{2020}= 1!2!3!\cdots2020!$
$=1!(2\cdot1!)3!(4\cdot3!)\cdots2019!(2020\cdot2019!)$
$=2\cdot4\cdot6\cdots2020(1!)^2(3!)^2\cdots(2019!)^2$
$=2^{1010}1010!(1!)^2(3!)^2\cdots(2019!)^2$
So it suffices to take $m=1010$. Notice that the argument can me modified for any $n$ such that $4|n$ and taking $m=n/2$.
says find all $m$