How do I write down a curve with exactly one rational point 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.
Harry
  When x is rational, the y coordinates are not rational for the majority of the x values. 
   If you look at (x^2 + 5) mod (6 - x) the value is only 0, at x = 5; similarly for the other example (x^2 + 203) mod (114 -x) only as 0 at x = 47 and x = 113.  It is the same for larger numbers(100 of digits).