Show that there is a number between 1 and 1000 such that there is a perfect square

Question

Show that there exists an integer $n \in S = \{1,2, \ldots, 1000\}$ such that

$$\prod_{i\in S-\{n\}}i!$$

is a perfect square.

I was thinking in trying to prove it by contradiction using the Pigeonhole Principle.

Answer 1

First observe that

$$\prod_{n\in S}n!=\prod_{n=1}^{1000}\prod_{k=1}^nk=\prod_{k=1}^{1000}\prod_{n=k}^{1000}k=\prod_{k=1}^{1000}k^{1001-k}\;.$$

For odd $k$ the exponent $1001-k$ is even, so

$$\prod_{k=1}^{500}(2k-1)^{1001-(2k-1)}$$

is a square, and we need only consider

$$\prod_{k=1}^{500}(2k)^{1001-2k}=\prod_{k=1}^{500}2^{1001-2k}\prod_{k=1}^{500}k^{1001-2k}\;.$$

Now $\sum\limits_{k=1}^{500}2k=500\cdot501$, so

$$\prod_{k=1}^{500}2^{1001-2k}=2^{500(1001-501)}=2^{500^2}$$

is a square, and we’ve reduced the problem to considering

$$\prod_{k=1}^{500}k^{1001-2k}=500!\cdot\prod_{k=1}^{500}k^{1000-2k}\;.$$

Each of the exponents $1000-2k$ is even, so $\prod\limits_{k=1}^{500}k^{1000-2k}$ is a square. Thus

$$\prod_{n\in S\setminus\{500\}}n!$$

is a square.