6. On the Cartesian plane, the number of integer points (i.e., points with both coordinates as integers) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$
4
6.【Analysis and Solution】Let $A(x, y)$ be a point on the circle, then it satisfies the equation $y^{2}+(x-199)^{2}=199^{2}$. Clearly, $(0,0),(199,199),(199,199),(389,0)$ are four sets of integer solutions. Their characteristic is that one of $y$ and $x-199$ is 0.
If there are other solutions such that $y$ and $x-199$ are both not 0, then there exist $m, n$ satisfying
$$
m^{2}+n^{2}=199^{2}, m \neq 0, n \neq 0
$$

By the general formula for Pythagorean triples, there exist $m_{1}, n_{1}$ satisfying
$$
m_{1}^{2}+n_{1}^{2}=199 \text{. }
$$

However, $199=4 \times 49+3$, which cannot be expressed in this form, leading to a contradiction. Therefore, there are only 4 sets of integer solutions.