Is divisible by 12?

Any integer number has one of the following expressions:

$3k-1$ , $3k$ or $3k+1$ $\quad k \in \mathbb{Z}$ .

In a first approach, let´s assume that:

$a=3k_a-1$ , $b=3k_b$ and $c=3k_c+1$ , $\quad k_a, k_b, k_c \in \mathbb{Z}$ .

So, $d$ must have one of that expressions:

$d=3k_d-1$ or $d=3k_d$ or $d=3k_d+1$ , $\quad k_d \in \mathbb{Z}$ .

Then, one of $(a-d)$ , $(b-d)$ or $(c-d)$ , will have the expression:

$\ 3(k_i-k_d) \quad i \in \{a, b, c\}$

By the exposed, two of them, $a$ , $b$ , $c$ or $d$ , will be such that,

$i-j=3(k_i-k_j) \quad i,j \in \{a,b,c,d\} \quad i\neq j$

By other hand, any integer number is even or odd. So, three of them, or two pairs of them, $a$ , $b$ , $c$ or $d$ , will be such that two differences will have the expression:

$i-j=2(K_i-K_j) \quad i,j \in \{a,b,c,d\} \quad i\neq j \quad K_i,K_j \in \mathbb{Z}$

Then,

$\Large \overline{\prod_{i,j \in \{a,b,c,d\}}^{i \neq j}{(i-j)}}=2.2.3 \prod{\small residual\left\{{i-j}|{k_i-k_j}|{K_i-K_j}\right\}}$

$\quad$

Note:

$\small \overline{\prod{(i-j)}}$ defines a product such that if $(i-j)$ is a factor, $(j-i)$ is not.