Write $H=H(\alpha)$. Since $\alpha$ has degree at least two and lies in $(0,1)$, $H>1$. Put
$$
x=\frac{\log(nh)}{d\log H}>1,
\qquad
k=\lfloor x\rfloor+1.
$$
Then $H^{kd}>nh$ and
$$
k+1\leq x+2\leq3x.
$$
Let
$$
m=\left\lceil\frac n{k+1}\right\rceil.
$$
Part (d), applied with $m$ in place of its polynomial-degree parameter, supplies $\ell\in\{k,k+1\}$ such that no nonzero integer polynomial of degree at most $m-1$ and with coefficients of absolute value less than $h$ vanishes at $\beta=\alpha^\ell$.

It follows that the $h^m$ sums
$$
\sum_{j=0}^{m-1}a_j\beta^j,
\qquad 0\leq a_j<h,
$$
are distinct. Since
$$
(m-1)\ell\leq(m-1)(k+1)\leq n-1,
$$
they form a subset of $A_n$. Hence
$$
|A_n|\geq h^m
\geq h^{\,n/(k+1)}
\geq h^{\,dn\log H/(3\log(nh))}
=H^{\,dn\log h/(3\log(nh))},
$$
as required.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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