I am having fun with some problems and am wondering if my solution to the following one is correct and well written.
Prove that a circle centered at the point $C=\left(\sqrt{2},\sqrt{3}\right)$ in the Cartesian plane passes at most through one point with integer coordinates.
I go by contradiction: assume there exists a circle centered at $C$ passing through two points $(x_1,y_1)\neq(x_2,y_2)$, with $x_i,y_i \in \mathbb{Z}, i = 1,2$. Let $R > 0$ be the radius of such a circle. Then \begin{align} \left(x_1-\sqrt{2}\right)^2+\left(y_1-\sqrt{3}\right)^2&=R^2,\\ \left(x_2-\sqrt{2} \right)^2+\left(y_2-\sqrt{3} \right)^2&=R^2. \end{align} Subtracting the two equations and simplifying I get $$\left(x_1-x_2\right)\left(x_1+x_2-2\sqrt{2}\right)+\left(y_1-y_2\right)\left(y_1+y_2-2\sqrt{3} \right)=0.$$ Define $m:=x_1-x_2, n:=x_1+x_2, r:=y_1-y_2, s:=y_1+y_2$. By assumption $m,n,r,s$ are integers, and I can rewrite the equation above as $$mn-2\sqrt{2}m+rs-2\sqrt{3}r=0$$ which reduces to saying that $$\sqrt{2}m'+\sqrt{3}r' \in \mathbb{Z}$$ for $m'=2m, r'=2r$. If I assume one between $m'$ and $r'$ is not zero, then squaring I would get that $\sqrt{2}, \sqrt{3}, \sqrt{6}$ are rational, which is false. So $m'=r'=0$, which means $m=r=0$, so $x_1=x_2,y_1=y_2$, contradiction.
If there were two points with rational coordinates the center’s irrational coordinates $c_1$, $c_2$ and $1$ are $\mathbb Q$-linearly dependent.
Consider the normal of the midpoint $(m_1,m_2)$ of these points.
The normal’s slope $q$ and the coordinates of the midpoint are rational. Now the center of the circle lies on the normal, hence we have $$q=\frac{m_2-c_2}{m_1-c_1}\Rightarrow qc_1-c_2\in\mathbb Q.$$
PS: If there are three points with rational coordinates, the center must have rational coordinates, too.