It is a famous theorem of Faltings, previously a conjecture by Mordell, that any algebraic curve of genus at least $2$ defined over the rational numbers have at most finitely many rational points. A *hyperelliptic curve* is a special algebraic curve of the form

$$\displaystyle z^2 = f(x,y),$$

where $f(x,y) \in \mathbb{Z}[x,y]$ is a binary form of degree $2n+2$. The genus of the curve is equal to $n$. For genus $1$ and $2$, all algebraic curves are hyperelliptic.

What is unknown from Faltings' theorem is an effective bound for the height of rational points lying on a given curve. Therefore, even if one obtains an effective upper bound for the number of rational points lying on a given curve $C$, there is no way to check how close to being optimal the bound is.

A recent theorem of Bhargava asserts that most hyperelliptic curves, ordered by naive height of the coefficients of $f$, have no rational points as the genus $n$ tends to infinity. Indeed, there are $o(2^{-n})$ hyperelliptic curves of genus $n$ defined over $\mathbb{Q}$ with at least one rational point.

My question is, what is the quantity

$$\mathfrak{S}(n) = \displaystyle \sup_{\substack{C \text{ hyperelliptic} \\ \text{genus of } C = n}}\#\{(x,y,z) \in C: (x,y,z) \text{ rational; inequivalent}\}.$$

Here two points $(x, y,z), (x',y',z')$ are inequivalent if there does not exist a non-zero rational number $\lambda$ such that $(x,y,z) = (\lambda x', \lambda y', \lambda^{n+1} z')$. In other words, the points $(x,y,z), (x',y',z')$ are not in the same equivalence class in the weight projective space $\mathbb{P}(1,1,n+1)$.

Plainly, $\mathfrak{S}(n) \gg n$. Indeed, this can be seen by considering the curve

$$\displaystyle C: z^2 = x^{2n+2} + (y - x)(y - 2x) \cdots (y - (2n+2)x)$$

which has the pairwise inequivalent points $(1,1,1), (1,2,1), \cdots, (1,2n+2,1)$. It is not clear to me whether or not these are essentially all of the inequivalent rational points on $C$.

Is $\mathfrak{S}(n)$ bounded? If so, then can we give an upper bound in terms of $n$? If not, can one give an explicit family of curves $C_1, C_2, \cdots$ such that

$$\displaystyle \# C_1(\mathbb{Q}) < \#C_2(\mathbb{Q}) < \cdots ?$$

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