Example 5 On the Cartesian plane, the number of integer points on the circumference of a circle centered at $(199,0)$ with a radius of 199 is
Let $A(x, y)$ be any integer point on circle $O$, then its equation is $(x-199)^{2}+y^{2}=199^{2}$.
Obviously, (0,0), $(199,199),(199,-199),(398,0)$ are 4 solutions to the equation. However, when $y \neq 0, \pm 199$, $(y, 199)=1$ (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 $m=2 k, n=2 l+1$. Then
$$
199=4 k^{2}+4 l^{2}+4 l+1=4\left(k^{2}+l^{2}+l\right)+1 \text {, }
$$

This contradicts the fact that 199 is a prime number of the form $4 d+3$, hence there are only 4 integer points on circle $O$: $(0,0),(199,199)$, $(398,0),(199,-199)$.