A friend sent me this problem and I couldn't find a solution online. I'm wondering if my attempt is correct and whether if there is a more elegant solution?
The Problem: Suppose that $a_0, a_1, ... a_{n-1} $ are $n$ integers. Show that the following product $$ P(a_0,a_1,...,a_{n-1}) = \prod _{0\le i<j\le n-1}^{ }\frac{a_j-a_i}{j-i}\ \qquad(1)$$ is integer.
My Attempt: We will prove this with induction and an application of the pigeonhole principle
Base case: $n=1$
$$P(a_0,a_1) = \frac{a_1 - a_0}{1-0}$$ is clearly integer as $1-0 = 1$ and $a_1,a_0 \in \mathbb{N}$
We now assume $n=k$ Assume $P(a_0, ... , a_{k-1}) \in \mathbb{N}$
Note that $$P(a_0, ... , a_{k}) = P(a_0, ... , a_{k-1}) \cdot \prod _{0\le i \le k-1}^{ }\frac{a_k-a_i}{k-i}\ \qquad (2)$$
For any fixed $0 \le i \le k-1$, we see that it is possible that $k-i \not| \ a_k - a_i$, and then resulting in the product on the right of (2) being non-integer.
Specifically, we observe that the product on the right of (2) is non integer if the following occurs
$$a_k\ \text{mod}\ (k-i) \not\cong a_j \qquad (3)$$
For all $0 \le j \not= i \le k-1$
Note that as $0 \le i \le k-1$, then $1 \le k-i \le k$, meaning that the complete residual class of $k-i$ is at a maximum, $\{0, 1, 2, ..., k - 1\}$, $k$ possible distinct residuals.
Consider the $P(a_0, ... , a_{k-1})$ in (2)
By the definition of $P$ in (1), we see that for every unique permutation of the set $\{c,d\}$ corresponding to the term $\frac{a_c-a_d}{c-d}$ for $0 \le c<d \le k-i$ exists in $P(a_0, ... , a_{k-1})$.
Additionally, due to (3), we have that $a_c, a_d$ can have a possible $k-1$ distinct residuals at most, with one being the residual of $a_k\ \text{mod}\ (k-i)$
But note that $P(a_0, ... , a_{k-1})$ uses $k$ variables $a_0, a_1, ..., a_{k-1}$, so therefore by the pigeonhole principle, there must be one term in that product that results in $$k-i \ | \ a_c - a_d$$
Otherwise similar stated by saying that $a_c - a_d \ \text{mod} \ k - i\cong 0$
Thus, we show that for any $0\le i \le k-1$, the product on the right of $(2)$ will still evaluate to be an integer. Thus (2) is integral as we also assumed that $P(a_0, ..., a_{k-1})$ is integer.
Thus, by the induction principle, (1) must be true.
I'm not sure if this is a valid formal proof, also am wondering if there are better solutions.
