Find all integers $x$, $y$ and $z$ that satisfy $x+y+z+xy+yz+zx+xyz=2017$.
In arithmetic, one can't do much with sums; one must always try to factorize! Here, we notice that $x+y+z+xy+yz+zx+xyz=(x+1)(y+1)(z+1)-1$, so the equation to solve can be rewritten as:

$$
(x+1)(y+1)(z+1)=2018
$$

The prime factorization of 2018 is $2 \times 1009$, so there are few possibilities for 2018 to be written as the product of three integers:

- Case $1: x+1=1, y+1=2, z+1=1009$
- Case $2: x+1=1, y+1=1009, z+1=2$
- Case $3: x+1=2, y+1=1, z+1=1009$
- Case $4: x+1=2, y+1=1009, z+1=1$
- Case $5: x+1=1009, y+1=1, z+1=2$
- Case $6: x+1=1009, y+1=2, z+1=1$

The triplets $(x, y, z)$ that are solutions to the problem are therefore all 6 permutations of the triplet $(0,1,1008)$.