Problem 1037

Answer: a) will not be found. Solution. a) Suppose, to the contrary, that there exist two rational points $M_{1}\left(x_{1}, y_{1}\right)$ and $M_{2}\left(x_{2}, y_{2}\right)$ on the circle of the given family. Then $\left(x_{1}-\sqrt{2}\right)^{2}+\left(y_{1}-\sqrt{3}\right)^{2}=\left(x_{2}-\sqrt{2}\right)^{2}+\left(y_{2}-\sqrt{3}\right)^{2}$. Therefore, $2\left(x_{1}-x_{2}\right) \sqrt{2}+2\left(y_{1}-y_{2}\right) \sqrt{3}=q-$ a rational number. If $\left(x_{1}-x_{2}\right) \cdot\left(y_{1}-y_{2}\right) \neq 0$, then squaring the last equality, we get that $\sqrt{6}$ is a rational number, which is false. If one of the differences, for example, $x_{1}-x_{2}$, is 0, then with $y_{1} \neq y_{2}$ we obtain a contradiction with the irrationality of $\sqrt{3}$. Therefore, two different rational points on the circle cannot exist. b) Notice that inside a small-radius circle, there are no integer points (we can take a radius less than the distance from $M$ to the nearest integer point $A(1 ; 2)$). On the other hand, if we take a sufficiently large radius (for example, greater than 3000), then inside the circle there will be more than 2014 points. Since, by part a), when the radius is gradually increased, the jump in the number of integer points occurs only by one, there must necessarily come a moment when there will be exactly 2014 points inside the circle.