1. Find all triples of integers $x, y, z$, for which

$$
\begin{aligned}
& x+y z=2005 \\
& y+x z=2006 .
\end{aligned}
$$
1. By subtracting the first equation from the second, we get

$$
(x-y)(z-1)=1 \text {, }
$$

from which it follows that either $x-y=z-1=1$, or $x-y=z-1=-1$.

In the first case, $z=2, y=x-1$ and by substituting into any of the original equations, we determine $x=669$, so $y=668$.

In the second case, $z=0, y=x+1$, so $x=2005$ and $y=2006$.

The solutions are two triples: $x=669, y=668, z=2$ and $x=2005, y=2006, z=0$.

Another solution. From the first equation, we express

$$
x=2005-y z
$$

and substitute this value into the second equation, which we then rearrange:

$$
\begin{gathered}
y+2005 z-y z^{2}=2006 \\
y\left(1-z^{2}\right)=2005(1-z)+1
\end{gathered}
$$

From the given system, it is clear that $z$ cannot be 1, so we can write

$$
y(1+z)=2005+\frac{1}{1-z} .
$$

The left side of the last equality is an integer, so the right side must also be an integer. This condition is satisfied only by $z=0$ and $z=2$.

As in the previous solution, by substituting into any of the original equations, we calculate $x=669, y=668$ for $z=2$ and $x=2005, y=2006$ for $z=0$.

For a complete solution, award 6 points. For each missing solution, deduct 2 points.