Determine the gcd of all numbers of the form

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

where $a, b, c, d$ range over $\mathbb{Z}$.
If we calculate the product in question for $a=0, b=1, c=2, d=3$, we find -12, so the gcd in question must divide $12=3 \times 4$.

Conversely, we will show that this gcd is 12, by showing that all numbers of this form are divisible by 3 and by 4.

An integer $n$ divides the difference between two of the integers among $a, b, c, d$ if these integers have the same remainder in the Euclidean division by $n$. Since there are only 3 possible remainders modulo 3, there must be two integers among $a, b, c, d$ that have the same remainder. All the products considered in the statement are therefore divisible by 3.

Now let's look at the remainders modulo 4: there are then two cases to consider. If there exist two integers among $a, b, c, d$ that have the same remainder in the Euclidean division by 4, the difference between these two integers will be divisible by 4. If $a, b, c, d$ all have different remainders modulo 4, this means that the 4 possible remainders in the Euclidean division by 4 are represented. Thus, there are among $a, b, c, d$ two even integers and two odd integers. The difference between two integers of the same parity being even, we have shown that in this case too all the products considered are divisible by $2 \times 2=4$.