## Exercise 3

## Statement.

Find all integers $x, y, z$ that satisfy: $1<x<y<z$ and $x+y+z+xy+yz+zx+xyz=2009$.
## Solution.

The idea is to factorize the left-hand side of the equation so as to obtain a divisibility relation between the left-hand side and the right-hand side (which will be a fixed integer whose divisors are known).

We notice that $x+y+z+xy+xz+yz+xyz=(1+x)(1+y)(1+z)-1$. Therefore, the equation can be rewritten as: $(1+x)(1+y)(1+z)=2010=2 \times 3 \times 5 \times 67$ (prime factorization of 2010). Since $2<1+x<1+y<1+z$, we obtain: $(1+x=3,1+y=2 \times 5,1+z=67)$ or $(1+x=3,1+y=5,1+z=2 \times 67)$ or $(1+x=5,1+y=2 \times 3,1+z=67)$. The solution triples $(x, y, z)$ are therefore: $(2,9,66),(2,4,133)$ and $(4,5,66)$.