Determine all non-zero natural numbers $a, b$ and $c$ for which there exist non-zero natural numbers $x, y$ and $z$ such that $x!=a b+1, y!=b c+1$ and $z!=c a+1$.
Let $(a, b, c)$ be a potential solution triplet, and $(x, y, z)$ be given as in the statement. Without loss of generality, and since $x, y,$ and $z$ play symmetric roles, we assume that $x \leqslant y \leqslant z$.

Consider modulo 3. Suppose that $x \geqslant 3$. In this case, $a b \equiv -1 \equiv b c \equiv c a \pmod{3}$. In particular, 3 does not divide $a$, $b$, or $c$: $a, b, c$ are 1 or -1 modulo 3. By the pigeonhole principle, two elements are then equal modulo 3. By symmetry, we can assume that $a$ and $b$ are equal modulo 3. Therefore, $a b \equiv (\pm 1)^2 \equiv 1 \equiv -1 \pmod{3}$, which is absurd.

Thus, $x$ is 1 or 2, so $a b + 1$ is 1 or 2, and $a b$ is 0 or 1. According to the statement, we deduce that $a = b = 1$ and $x = 1$, and $c = z! - 1$. Therefore, up to permutation, $(a, b, c) = (1, 1, z! - 1)$.

Conversely, if $(a, b, c)$ is a triplet of the form $(1, 1, t! - 1)$ up to permutation, then $(a b + 1, b c + 1, a c + 1)$ is $(2!, t!, t!)$ up to permutation.