Vtyjkyuk

It is known that for every $n$ consecutive integers, their product is divisible by $n!$, since $$mchoosen = fracm!n!(m-n)!$$ is also an integer.

So is it true that for every distinct integer $a_1, a_2, ..., a_n$, the number $$S = prod_1leq i< jleq n fraca_i - a_ji-j$$ is also an integer?

(Sorry for my grammar mistakes, English is my second language)

We observe that $prod_1 leq i < j leq n (i-j) = prod_r=1^n r!$ and the numerator is just the Vandermonde Determinant so that by performing row operations we obtain
$$prod_1 leq i < j leq n fraca_i-a_ji-j =
beginvmatrix
1 & 1 & cdots & 1 \
binoma_11 & binoma_21 & cdots & binoma_n1 \
binoma_12 & binoma_22 & cdots & binoma_n2 \
cdots & cdots & cdots & cdots \
binoma_1n-1 & binoma_2n-1 & cdots & binoma_nn-1 \
endvmatrix$$

which is obviously an integer.

Let $q = p^k$ for some prime $p$ and integer $k>0$. If you let $n_q$ = the number of pairs $i<j$ for which $a_i - a_j$ is divisible by $q$, then notice that $n_q$ is minimized for $a_1,ldots,a_n = 1,2,ldots,n$ (it is a bit of work to show this but it is not deep) (basically, you consider the set $mathbbZ/(q)$ and now you look the function $f: mathbbZ/(q) rightarrow mathbbN$ for which $f(a)$ is the number of $i$ for which $a_i$ mod $q$ is $a$. Then observe that you can compute $n_q$ from $f$, and that $n_q$ is minimized if all values of $f$ are inside $m,m+1$ where $m$ is $n/q$ rounded down).

Now apply this for $q = p, p^2, p^3, ldots$ and you see that the $p$-adic valuation of the numerator of $S$ is at least that of the denominator. That means $S$ is an integer.