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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.

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    Curve = smooth projective geometrically connected curve over $\mathbf{Q}$. The genus is the arithmetic genus (or geometric genus because of our definition of curves).2012-08-03

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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.

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    Hi Georges, I think your assertion b) is incorrect, as pointed out by Emiliano Ambrosi in [his answer](https://math.stackexchange.com/a/1783724/7719). According to Table 2 following Theorem (3.3) of Husemoller's "Elliptic Curves", the elliptic curve $y^{2} = x^{3} + 2$ has rank $1$. However perhaps you can take $a = 6$ instead.2018-08-01
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I can't comment, but I'm afraid that the answer of George is incorrect. Indeed the equation has at least 2 solutions, namely $[0:1:0]$ and $[-1:-1:1]$

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I believe that there is a class of curve that has a single rational point on it, it is an hyperbola given by (x^2 + A)/(B - x). The conditions to give a single rational point would be for B^2 + A = a prime number, e.g. B= 6 A = 5 the rational point would be at (5, 30). The conditions for 2 rational points would be that B^2 + A = N, where N = pq, e.g. B = 114 A = 203, the ratioanl points are at (47, 36) and (113, 12972).

Hope that this may help you.

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    I don't think this works. Maybe you're writing down curves with only one integer point? In fact, any rational value for $x$ will give a value for $y$ which is a rational number. I think that the projective curve induced by your equation is isomorphic to the projective line.2012-08-03
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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).