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).