Take distinct $x,y\in A_n$. Their difference has the form
$$
x-y=P(\alpha),
$$
where $P\in\mathbb Z[X]$ has degree at most $n-1$ and <polynomial length> at most $nh$. By the <height bound for a polynomial evaluation>,
$$
H(x-y)\leq nh\,H(\alpha)^{n-1}.
$$
The algebraic number $x-y$ is nonzero and has degree at most $d$, so the <Liouville height inequality> gives the separation
$$
|x-y|
\geq(nh)^{-d}H(\alpha)^{-d(n-1)}.
$$

All elements of $A_n$ lie in an interval of length at most
$$
\frac h{1-\alpha}.
$$
Since $H(1-\alpha)\leq2H(\alpha)$, another application of the <Liouville height inequality> gives
$$
\frac1{1-\alpha}\leq(2H(\alpha))^d.
$$
The number of points in an interval is at most one plus its length divided by their minimum separation. Consequently
$$
|A_n|
\leq
2^dh(nh)^dH(\alpha)^{dn}+1
\leq
(2hn)^{d+1}H(\alpha)^{dn}+1.
$$
Thus the requested statement holds, for example, with the absolute constant $C=2$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 166 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-166/2/c/solution