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

Ifits Mahler measure isIf $f_{\alpha }\in \mathbb{Z}[X]$ is the primitive minimal polynomial of an algebraic number $\alpha$ and $d=\mathrm{deg}{f}_{\alpha }$, the height-Mahler measure formula is

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Choose a number field $L$ containing ${\alpha }_1,\dots ,{\alpha }_k$. At every place $v$ of $L$, putThe triangle inequality giveswhere $c_v=1$ at every non-Archimedean place because the coefficients are integers, while at an Archimedean place one may takeRaise these inequalities to the local weights and multiply over all places. The definition of the Absolute multiplicative Weil height and the product formula then giveThis is the height bound for a polynomial evaluation.

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Take distinct $x,y\in A_n$. Their difference has the formwhere $P\in \mathbb{Z}[X]$ has degree at most $n-1$ and polynomial length at most $nh$. By the height bound for a polynomial evaluation,The algebraic number $x-y$ is nonzero and has degree at most $d$, so the Liouville height inequality gives the separation

All elements of $A_n$ lie in an interval of length at mostSince $H(1-\alpha )\le 2H(\alpha )$, another application of the Liouville height inequality givesThe number of points in an interval is at most one plus its length divided by their minimum separation. ConsequentlyThus the requested statement holds, for example, with the absolute constant $C=2$.

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Suppose, for a contradiction, that suitable nonzero polynomials vanish at both ${\alpha }^k$ and ${\alpha }^{k+1}$. If $P(\gamma )=0$, then the primitive minimal polynomial $f_{\gamma }$ divides $P$. The multiplicativity of Mahler measure and the Mahler measure bounded by polynomial length give

If $[\mathbb{Q}({\alpha }^k):\mathbb{Q}]=d$, then, using $H({\alpha }^k)=H(\alpha {)}^k$, this inequality contradicts $H(\alpha {)}^{kd}>nh$. Hence $\mathbb{Q}({\alpha }^k)$ is a proper intermediate field of $\mathbb{Q}(\alpha )/\mathbb{Q}$. Its degree divides the prime $d$ by the tower law, so ${\alpha }^k\in \mathbb{Q}$. Applying the same argument to ${\alpha }^{k+1}$ is even stronger and gives ${\alpha }^{k+1}\in \mathbb{Q}$. Since $\alpha \ne 0$,contrary to $d\ge 2$. At least one of the two proposed values of $\beta$ therefore has no such polynomial relation.

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Write $H=H(\alpha )$. Since $\alpha$ has degree at least two and lies in $(0,1)$, $H>1$. PutThen $H^{kd}>nh$ andLetPart (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 $h^m$ sumsare distinct. Sincethey form a subset of $A_n$. Henceas required.

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