12. On the Cartesian plane, the number of lattice points (i.e., points with both integer coordinates) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ .
Moving the center of the circle to the origin $O(0,0)$, the equation of the circle becomes $x^{2}+y^{2}=199^{2}$, at this point the integer points on the circumference are $(199,0),(0,199),(-199,0),(0,-199)$. If there is an integer point $(x, y)$ on the circle with $x y \neq 0$, then $|x|,|y|, 199$ form a Pythagorean triple, thus there exist $a, b \in \mathbf{N}^{*}$ satisfying $a^{2}+b^{2}=199$. Assuming $a=2 k, b=2 l-1 \Rightarrow a^{2}+b^{2}=4\left(k^{2}+l^{2}-l\right)+1=4 \times 49+3$, which is a contradiction. Therefore, the number of integer points on the circumference is 4.