6. The positive integer solutions of the equation $x^{2}+y^{2}=2000$ are $\left(x_{0}, y_{0}\right)$. Then the value of $x_{0}+y_{0}$ is $\qquad$
6.52 or 60.

Since $2=|1 \pm i|^{2}, 5=|2 \pm i|^{2}$, therefore, $2000=2^{4} \cdot 5^{3}=|1 \pm i|^{8} \cdot|2+i|^{6}$ or $|1 \pm i|^{8} \cdot|2-i|^{6}$ or $|1 \pm i|^{8} \cdot|2-i|^{4} \cdot|2+i|^{2}$ or $|1 \pm i|^{8} \cdot|2+i|^{2} \cdot|2-i|^{4}=|8+$ $44i|^{2}$ or $|40+20i|^{2}$ or $|40-20i|^{2}$ or $|40+20i|^{2}=8^{2}+44^{2}$ or $40^{2}+20^{2}$.
$\therefore x_{0}+y_{0}=8+44$ or $40+20=52$ or 60.