Exercise 1. Find 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 $z!=c a+1$.

Note: For any non-zero natural number $n$, the integer $n!$ denotes the product $1 \times 2 \times \cdots \times n$.
Solution to Exercise 1 Let $(a, b, c)$ be a potential solution triplet. Without loss of generality, and since $a, b$, and $c$ play symmetric roles, we assume that $a \leqslant b \leqslant c$, so that $x \leqslant z \leqslant y$ and that $x!$ divides both $y!$ and $z!$.

Let $d$ be a divisor of $x!$, that is, a common divisor of $x!, y!$, and $z!$. Since

$$
a b \equiv b c \equiv c a \equiv-1 \quad(\bmod d)
$$

we know that

$$
-1 \equiv(a b)(b c)(c a) \equiv(a b c)^{2} \quad(\bmod d)
$$

is a square modulo $d$. However, -1 is not a square modulo 3. Thus, $d \neq 3$, which means that $x \leqslant 2$.

Since $x!=a b+1 \geqslant 2>1$ !, we conclude that $x=2$, so that $a b=x!-1=1$, and thus $a=b=1$. The triplet $(a, b, c)$ is therefore, up to permutation, equal to the triplet $(1,1, z!-1)$.

Conversely, any triplet of the form $(1,1, z!-1)$ clearly works.

Grader's Comment The exercise was poorly performed: indeed, many students incorrectly assumed that the number $n!-1$ is a prime number for all $n \geqslant 1$. This is false, for example, for $n=5$, since $5!-1=119=7 \times 17$. Some students tried to apply Wilson's theorem, with little success: here, this theorem did not help, as it did not allow determining $x, y$, and $z$ in terms of $a, b$, and $c$ or vice versa.