Each of the numbers $x_{1}, x_{2}, \ldots, x_{2004}$ can be equal to $\sqrt{2}-1$ or $\sqrt{2}+1$. How many distinct integer values can the sum

$$
\sum_{k=1}^{2004} x_{2 k-1} x_{2 k}=x_{1} x_{2}+x_{3} x_{4}+x_{5} x_{6}+\cdots+x_{2003} x_{2004}
$$

assume?

## List 10
We have that the possible products $x_{2 k-1} x_{2 k}$ where $k \in\{1,2, \ldots, 2004\}$ are $(\sqrt{2}-1)(\sqrt{2}-1)=3-2 \sqrt{2},(\sqrt{2}+1)(\sqrt{2}+1)=3+2 \sqrt{2}$ and $(\sqrt{2}-1)(\sqrt{2}+1)=1$.

Suppose that $a$ products are equal to $3-2 \sqrt{2}$, $b$ products are equal to $3+2 \sqrt{2}$ and $1002-a-b$ products are equal to 1.

The sum is equal to

$$
a(3-2 \sqrt{2})+b(3+2 \sqrt{2})+1002-a-b=1002+2 a+2 b+2(b-a) \sqrt{2}
$$

Thus, for the sum to be an integer, we must have $a=b$. Therefore, the sum is equal to $1002+4 a$. Since $a$ varies from 0 to 501 (because $a+b$ cannot be greater than 1002), the sum can take 502 distinct integer values.

## List 10