power of 2 involving floor function divides cyclic product if $S_n={a_1,a_2,a_3,...,a_{2n}}$} where $a_1,a_2,a_3,...,a_{2n}$ are all distinct integers.Denote by $T$ the product
$$T=\prod_{i,j\epsilon S_n,i<j}{(a_i-a_j})$$ Prove that $2^{(n^2-n+2[\frac{n}{3}])}\times \text{lcm}(1,3,5,...,(2n-1)) $ divides $T$.(where [] is the floor function)
I have tried many approaches.I tried using the fact that either number of odd or even integers$\ge(n+1)$ and since in $T$ every integer is paired with every other integer only once.Since even-even and odd-odd both give even numbers:$2^{\frac{n(n+1)}{2}}$ divides $T$.As for the $lcm(1,,3,5,....(2n-1))$ I have no Idea what to do.Please help.I am quite new to NT.Thank you.
 A: HINT:
You're not quite correct for the powers of $2$.  There are $2n$ numbers total, so there can in fact be exactly $n$ odds and $n$ evens (i.e. neither has to be $\ge n+1$.)
Instead: Let there be $k$ odds, and $2n-k$ evens.  This gives you ${k \choose 2}$ factors of $2$ from the (odd - odd) terms and ${2n-k \choose 2}$ factors of $2$ from the (even - even) terms.


*

*First you can show that their sum is minimized at $k=n$ and that sum is $n(n-1) = n^2 - n$.

*So now you're need another $2[{n \over 3}]$ more factors of $2$.  These come from  splitting evens further into $4k$ vs $4k+2$, and the odds into $4k+1$ vs $4k+3$, because some of the differences will provide a factor of $4$, i.e. an additional factor of $2$ in addition to the factor of $2$ you already counted.


*

*Proof sketch: e.g. suppose there are $n$ evens, and for simplicity of example lets say $n$ is itself even.  In the worst split exactly $n/2$ will be multiples of $4$, which gives another $\frac12 {\frac n 2}({\frac n 2}-1)$ factors of $2$ from these numbers of form $4k$, and a similar thing happens with the $4k+1$'s, the $4k+2$'s, and the $4k+3$'s.  This funny thing is that if you add up everything, $n({\frac n 2}-1) \ge 2[{\frac n 3}]$ (you can try it, and you will need to prove it). In other words $2[{\frac n 3}]$ is not a tight bound at all, but rather a short-hand that whoever wrote the question settled on just to make you think about the splits into $4k + j$.  In fact a really tight bound would involve thinking about splits into $8k+j, 16k+j$ etc.  


*As for $lcm(1, 3, 5, \dots, 2n-1)$, first note that $lcm$ divides $T$ just means each of the odd numbers divide $T$.  You can prove this by the pigeonhole principle.  


*

*Further explanation: E.g. consider $lcm(3, 9) = 9$, so this lcm divides $T$ if both odd numbers divide $T$.  In general, the lcm can be factorized into $3^{n_3} 5^{n_5} 7^{n_7} 11^{n_{11}} \dots$ and it divides $T$ if every term $p^{n_p}$ divides $T$.  but in your case $p^{n_p}$ itself must be one of the numbers in the list $(1, 3, 5, \dots, 2n-1)$ or else the lcm wouldn't contain that many factors of $p$.



Can you finish from here?
A: Let me give a sketch of the solution.
Well, there is a following fact:

Proposition. Given $n$ distinct positive integers $a_1, a_2, \ldots, a_n$. Then,
  $$
\prod_{i>j}\frac{a_i-a_j}{i-j}\in \mathbb{Z}.
$$
  In other words, $A_n$ divides $\prod\limits_{i>j}(a_i-a_j)$, where $A_n=\prod\limits_{i>j}(i-j)$.

This fact can be proved by showing that number of multiplicands of the product $\prod\limits_{i>j}(a_i-a_j)$ which are divisible by $p^s$ for every prime $p$ and positive integer is greater or equal the number of multiplicands of the product $\prod\limits_{i>j}(i-j)$ which are divisible by $p^s$.
It's clear that in the statement of the proposition above constant $A_n$ is sharp (because we can simply consider $a_k=k$ for $k=\overline{1, n}$).
Thus, it's sufficient to prove that $A_{2n}$ is divisible by $2^{(n^2-n+2[\frac{n}{3}])}\times\text{lcm}(1,3,5,...,(2n-1))$. Since $A_{2n}$ is divisible by $1, 3, \ldots, 2n-1$ we need to show that 
$$
2^{n^2-n+2[\frac{n}{3}]}|A_{2n}.
$$
This can be done in the way which is described in the previous answer.
