One of the NCTM's President Mike Shaughnessy's Problems to Ponder requested to find the GCD of the set of all Prod Difs of four integers. To remind, Prod Dif \{a_1, a_2, ..., a_n\} is given by

Prod Dif \{a_1, a_2, ..., a_n\} = \prod_{1\leq i<j\leq n}(a_{j}-a_{i}).

For example Prod Dif \{a_1, a_2, a_3, a_4\} = (a_4-a_3)(a_4-a_2)(a_4-a_1)(a_3-a_2)(a_3-a_1)(a_2-a_1). The answer to Mike Shaughnessy's question is 12. In other words, the product of all possible differences of four integers is always divisible by 12, and 12 is the largest number with this property. (Here the two differences a-b and b-a are considered the same.

Matt Hendersen has conjectured that Prod Dif of n integers is always divisible by the superfactorial of n-1, which is defined as (n-1)!(n-2)!\cdot\ldots\cdot2!1!. Since this is exactly Prod Dif\{1,2,\ldots,n\}, this is the largest number with this property.

David Radcliffe has pointed out that this is so due to a result of Robin Chapman. I shall reproduce the latter below.

It so happens that Prod Dif \{a_1, a_2, a_3, a_4\} = (a_4-a_3)(a_4-a_2)(a_4-a_1)(a_3-a_2)(a_3-a_1)(a_2-a_1) is exactly the value of the 4\times 4 Vandermonde determinant. More generally,

\left|\matrix{
1 & a_{1} & a_{1}^{2} & \ldots & a_{1}^{n-1} \\
1 & a_{2} & a_{2}^{2} & \ldots & a_{2}^{n-1} \\
1 & a_{3} & a_{3}^{2} & \ldots & a_{3}^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & a_{n} & a_{n}^{2} & \ldots & a_{n}^{n-1} \\
}\right| = \prod_{1\leq i<j\leq n}(a_{j}-a_{i})
.

Now, applying elementary column operations, we can obtain, for any collection of monic polynomials f_i, i=1,2,\ldots n-1

\left|\matrix{
1 & f_{1}(a_{1}) & f_{2}(a_{1}) & \ldots & f_{n-1}(a_{1}) \\
1 & f_{1}(a_{2}) & f_{2}(a_{2}) & \ldots & f_{n-1}(a_{2}) \\
1 & f_{1}(a_{3}) & f_{2}(a_{3}) & \ldots & f_{n-1}(a_{3}) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & f_{1}(a_{n}) & f_{2}(a_{n}) & \ldots & f_{n-1}(a_{n}) \\
}\right| = \prod_{1\leq i<j\leq n}(a_{j}-a_{i})
.

For our purposes, we choose f_{i}(x)= x(x-1)(x-2)\cdot\ldots\cdot(x-i+1). For a non-negative a, we then have f_{i}(a)=i!{{a}\choose{i}}. So, each entry of this determinant in column i is divisible by (i-1)!, implying that the determinant itself is divisible by \prod_{i=1}^{n-1}i!.

References

  1. R. Chapman, A Polynomial Taking Integer Values, Math Mag VOL. 69, NO. 2, APRIL 1996, p 121