Determine the gcd of all numbers of the form

$$
(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)
$$

where $a, b, c, d$ are integers.
We will show that the gcd sought is 12. Indeed, by the pigeonhole principle, among $a, b, c, d$ there are two numbers congruent modulo 3, and thus their difference is divisible by 3. Now if we look modulo 4, there are two possibilities: either there are two numbers among $a, b, c, d$ congruent modulo 4, and at that moment their difference is divisible by 4, or $a, b, c, d$ are distinct modulo 4. But in the latter case, two of them are even and two are odd, so the product of interest has two even factors. Thus, in all cases, this product is divisible by 3 and by 4, hence by 12.

On the other hand, taking $a=0, b=1, c=2, d=3$, the product obtained is -12, so the divisor we have found is indeed the gcd.