3. In the Cartesian coordinate system, with $(199,0)$ as the center and 199 as the radius, the number of integer points on the circumference of the circle is ( ) .
A. 2
B. 4
C. 8
D. 16
3. B Let $(x, y)$ be an integer point on a known circle, then $(x-199)^{2}+y^{2}=199^{2}$.

Obviously, $(x, y)=(0,0),(199,199),(199,-199),(398,0)$ are four solutions to the equation.
When $y \neq 0$ and $y \neq \pm 199$, $y$ is coprime with 199 (since 199 is a prime number), at this time $199, y,|199-x|$ form a Pythagorean triplet, so 199 can be expressed as the sum of the squares of two positive integers, i.e., $199=m^{2}+n^{2}$.
Since $199=4 \times 49+3$, we can set $m=2 k, n=2 l+1, k \in \mathbf{Z}, l \in \mathbf{Z}$.
Then $199=4 k^{2}+4 l^{2}+4 l+1=4\left(k^{2}+l^{2}+l\right)+1$.
This contradicts the fact that 199 is a prime number of the form $4 d+3$, hence the circle $O$ has only four integer points: $(0,0),(199, \pm 199)$, $(398,0)$.