On one of our tests, the extra credit was to find which number you would take out from the set $\{1!,2!,3!,...(N-1)!,N!\}$ such that the product of the set is a perfect square. My answer was as follows:
Assume $N$ is even. First note that $(n!)=(n-1)!\cdot n$.
Apply this to the odd numbers to get the product: $$(2!)(2!)\cdot 3 \cdot(4!)(4!)\cdot 5\cdots ((N-2)!)((N-2)!)\cdot (N-1)\cdot N!$$
Let $ 2!4!6!...(N-2)!=E$. Then our equation is equal to: $$E^23\cdot 5\cdot 7\cdots (N-1)\cdot (N!)$$
Expand $N!$:
$$E^2\cdot 3\cdot 5\cdot 7\cdots (N-1)\cdot 2\cdot 3\cdot 4\cdots N$$
Group the odd terms together:
$$E^2\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7 \cdot 7\cdots (N-1)\cdot 2\cdot 4\cdot 6\cdots N$$
Let $O=1\cdot3\cdot5...\cdot (N-1)$:
$$E^2O^2 2\cdot 4\cdot 6\cdots N = E^2O^2\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot (2\cdot 4)\cdots (2\cdot (N/2))$$
Group together the $2$'s:
$$E^2O^22^{(N/2)}1\cdot2\cdot3...(N/2)=E^2O^22^{(N/2)}\cdot(N/2)!$$
So, if $N/2$ is even, it can be expressed as $2m$ for some $m$. So we have:
$$E^2O^22^{2m}(N/2)!=(EO2^m)^2(N/2)!$$ Therefore, if $N$ is even, the number missing is $(N/2)!$ if $N/2$ is even. For example, for $N=4$, $2!$ is missing, $N=6$ is impossible ($3$ is odd), and for $N=100$, $50!$ is missing.
I got partial credit - but my teacher said I missed a case, and that I should leave a sticky note on her desk with a number on it to show that I've fixed the proof.
I think the flaw might be in assuming $N$ is even, but I'm not sure how to deal with the case of $N$ odd.
And, I've found that my solution gives $4!$ for $N=8$, but $3!$ is also a solution. Past that, how can one number show that I've fixed the proof?
According to the paper "Square products of punctured sequences of factorials" by Rick Mabry and Laura McCormick. The given problem for $N\geq 2$ has at least a solution if ad only if $N$ is even and has one of the following forms:
1) $4k$ or $4k^2-2$ for any positive integer $k$ (take away $(2k)!$ and $(2k^2)!$ respectively);
2) $2k^2$ or $2k^2-4$ for any positive odd integer $k\geq 3$ such that $2(k^2-1)$ is a perfect square (take away $(k^2-2)!$ and $(k^2)!$ respectively).