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Let $a_1,a_2,\ldots,a_n$ be $n$ distinct integers. Show that the product of all the fractions of the form $\dfrac{a_k-a_l}{k-l}$, where $n \geq k > l$, is an integer.

I thought about first determining how many differences $a_k-a_l$ are divisible by $b$.

Assume that $n_0$ of the integers $a_1,a_2,\ldots,a_n$ are divisible by $b$, that $n_1$ of them yield the remainder $1$ upon division by $b$, that $n_2$ of them yield the remainder $2$, and so on up to $n_{b-1}$ integers the yield a remainder of $b-1$ when divided by $n$. It then follows that $$n_0+n_1+n_2+\cdots+n_{b-1} = n.$$ The difference $a_k-a_l$ is divisible by $b$ if and only if $a_k \equiv a_l \pmod{b}$. Now note that the number of differences $a_k-a_l$ divisible by $b$ such that $a_k \equiv a_l \equiv r \pmod{b}$ is $C^2_{n_r} = \dfrac{n_r(n_r-1)}{2}$. It follows that the number of differences divisible by $b$ is exactly \begin{align*}N &= \dfrac{n_0(n_0-1)}{2}+\dfrac{n_1(n_1-1)}{2}+\cdots+\dfrac{n_{b-1}(n_{b-1}-1)}{2}\\&= \dfrac{n_0^2+n_1^2+n_2^2+\cdots+n_{b-1}^2}{2}-\dfrac{n_0+n_1+n_2+\cdots+n_{b-1}}{2}\\&= \dfrac{n_0^2+n_1^2+n_2^2+\cdots+n_{b-1}^2}{2}-\dfrac{n}{2}.\end{align*}

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The following lemma solves the problem:

Lemma: Let $n$ be a positive integer and $p$ a prime. Given positive integers $a_1<a_2<\dots < a_n$ let $f_p(a_1,a_2,\dots a_n)=v_p(\prod\limits_{1\leq i < j\leq n} a_i-a_j)$. Then $f_p(a_1,a_2,\dots,a_n)\geq f_p(1,2,\dots,n)$.

Proof: We can calculate $v_p(a_1,a_2,\dots a_n)$ in the following way:

For every positive integer $k$ and $0\leq j < p^k$ we let $b^j_k$ be the number of terms among $a_1,a_2,\dots,a_n$ that are congruent to $j\bmod p^k$.

Let $c_k=\sum\limits_{j=0}^{p^k}\binom{b_k^j}{2}$.

Then we have $f(a_1,a_2,\dots a_n)=\sum\limits_{k=0}^\infty c_k$.

Notice that by Jensen's inequality for binomial coefficients we have that $c_k$ is minimized if and only if for every $j\neq j'$ we have that $|b_k^j-b_k^{j'}|\leq1$. Since this is clearly the case when $a_1,a_2,\dots,a_n=1,2,3,\dots,n$ the lemma follows.

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  • $\begingroup$ The idea to the formula is a bit similar to de Polignac's formula. If you are having trouble seeing why the formula holds I can give some help. $\endgroup$ – Jorge Fernández Hidalgo Dec 10 '16 at 3:47
  • $\begingroup$ Can you explain the Jensen's Inequality part? $\endgroup$ – user19405892 Dec 10 '16 at 3:55
  • $\begingroup$ Sure, the theorem is as follows: Let $x_1,x_2,\dots x_n$ be non-negative integers that add up to a constant $A$. Then the sum $\binom{x_1}{2}+\binom {x_2}{2}+\dots + \binom{x_n}{2}$ is minimized when $|x_i-x_j|\leq 1|$ for all $1|\leq i < j\leq n$. The proof is easy by contradiction. If $x_i<x_j-1$ then the sum becomes even smaller after changing $x_i$ for $x_i+1$ and $x_j$ for $x_j-1$. $\endgroup$ – Jorge Fernández Hidalgo Dec 10 '16 at 3:58

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