Let $g\geq 1$. I would like to write down (for all $g$) a smooth projective geometrically connected curve $X$ over $\mathbf{Q}$ of genus $g$ with precisely one rational point.
Is this possible?
For which $g$ is this possible?
I think for $g=1$ this is possible. I just don't know an explicit equation, but I should be able to find it. (We just write down an elliptic curve without torsion of rank zero over $\mathbf{Q}$.)
For $g\geq 2$ things get more complicated for me.
I would really like the curve to be of gonality at least $4$, but I'll think about that later.
The smooth plane projective curve $E$ defined over $\mathbb Q$ by the equation $y^2z=x^3+2z^3$ has its point at infinity $[0:1:0]$ as its only rational point: $E(\mathbb Q)=\lbrace [0:1:0]\rbrace $. Indeed:
a) The torsion group of the curve $y^2z=x^3+az^3 $ is zero as soon as $a$ is a sixth-power free integer which is neither a square nor a cube nor equal to $-432$.
(Despite appearences I'm not making this crazy theorem up, but I am quoting theorem (3.3) of Chapter 1 in Husemöller's Elliptic Curves !).
b) On the other hand the curve $E$ has rank $0$, which means that its group of rational points is torsion (this is stated in the table following the theorem I just quoted).
The two results a) and b) prove the assertion in my introductory sentence.