An algebraic curve (in this question) is the zero set $C = f^{-1}(X\ Y)$ of any polynomial $f\in\mathbb R[X\ Y]$; we say then that $f$ represents $C$. An algebraic curve $C$ is non-rational $\ \Leftrightarrow\ $ there does not exist any polynomial $f\in \mathbb Q[X\ Y]$ which represents $C$. An algebraic curve $C$ is irreducible $\ \Leftrightarrow\ $ it is not a union of any two curves, different from $C$. The following problem is open to me:

**Question:** does there exist a non-rational irreducible curve which contains infinitely many rational points (i.e. when $C\cap\mathbb Q^2$ is infinite)?

**COMMENT**: Curve $(X-\frac 1{\sqrt 2})^2 + (Y-\frac 1{\sqrt 2})^2 = 1$ has exactly one rational point. This promises a taste for trying the rational points of non-rational curves, and of the geometric-combinatorial considerations related to them (other fields and dimensions are possible too).