About "The product of the six numbers surrounding any interior number in Pascal’s triangle is a perfect square"

Question

The current Futility Closet has this statement: "The product of the six numbers surrounding any interior number in Pascal’s triangle is a perfect square."

Here is the link with a nice illustration: http://www.futilitycloset.com/2016/07/14/a-square-triangle/

Here is the proof I found:

At $c(n, m)$, the neighbors are $(n-1, m-1), (n-1, m), (n, m-1), (n, m+1), (n+1, m), (n+1, m+1) $.

The product is

$\begin{array}\\ p(n, m) &=c(n-1, m-1)c(n-1, m) c(n, m-1)c(n, m+1) c(n+1, m)c(n+1, m+1)\\ &=\frac{(n-1)!^2n!^2(n+1)!^2}{(m-1)!(n-m)!m!(n-1-m)!(m-1)!(n-m+1)!(m+1)!(n-m-1)!m!(n-m+1)!(m+1)!(n-m)!}\\ &=\frac{(n-1)!^2n!^2(n+1)!^2}{(m-1)!^2(n-m)!^2m!^2(n-1-m)!^2(n-m+1)!^2(m+1)!^2}\\ &=\left(\frac{(n-1)!n!(n+1)!}{(m-1)!m!(m+1)!(n-1-m)!(n-m)!(n-m+1)!}\right)^2\\ \end{array} $

I have two questions:

(1) Since the result is so nice, it seems to me that there should be a simpler, more intuitive proof. Is there?

(2) These are the product of the binomial coefficients adjacent to $(n, m)$, or at distance 1. What about the coefficients and distance 2, or, more generally, k. How would we state the product of such coefficients and what properties would it have (e.g., would it be a k-th power)?

Answer 1

Ad. (2): A generalisation of this property can be found in Generalized hidden hexagon squares by A.K.Gupta.

For positive integers $r_1,r_2$ the following product spaced around $\binom{n}{m}$ is a perfect square \begin{align*} \binom{n-r_1}{m-r_2}\binom{n-r_1}{m}\binom{n}{m-r_2} \binom{n+r_2}{m+r_1}\binom{n+r_2}{m}\binom{m}{m+r_2} \end{align*} If $r_1=r_2$ the product is equally spaced around $\binom{n}{m}$, if $r_1=r_2=1$ we get the point of origin.

Answer 2

For (1), noticing that
$$\binom{n-1}{m-1}\binom{n}{m+1}\binom{n+1}{m}=\binom{n-1}{m}\binom{n+1}{m+1}\binom{n}{m-1}\tag3$$ holds should make it simpler to prove the claim.

Writing $\binom ab$ as $\frac{a!}{b!(a-b)!}$, one sees that the numerators on both sides of $(3)$ are $(n-1)!n!(n+1)!$, and that the denominators on both sides are the product of $(m-1)!m!(m+1)!$ and $(n-m-1)!(n-m)!(n-m+1)!$.