Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 166 (Diophantine Analysis)

Consider the $N+1$ fractional partsin $[0,1)$. Divide that interval into $N$ intervals of length $1/N$. By the pigeonhole principle, two fractional parts, say those indexed by $i<j$, lie in the same interval. Thus, for some integer $p$,Set $q=j-i$. Then $1\le q\le N$, and division by $q$ gives the Dirichlet approximation theorem

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One Archimedean form of the Schmidt subspace theorem is as follows. Let $L_1,\dots ,L_n$ be linearly independent linear forms in $n$ variables with algebraic coefficients. For every $\epsilon >0$, all nonzero $\mathbf{x}\in {\mathbb{Z}}^n$ satisfyingbelong to a finite union of proper rational linear subspaces of ${\mathbb{Q}}^n$.

We will also use its finite-place form: if $S$ is a finite set of places containing the Archimedean ones and, for each $v\in S$, the forms $L_{1,v},\dots ,L_{n,v}$ are independent, then the integer solutions oflie in finitely many proper rational subspaces. The absolute values are normalized so that the product formula holds.

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The Roth theorem states that if $\alpha$ is a real algebraic irrational number, then for every $\epsilon >0$ there are only finitely many reduced fractions $p/q$ satisfying

To derive this from the Schmidt subspace theorem, takeThese forms are linearly independent. For a solution $(p,q)$ with $q$ large, $|p|\asymp q$, so $H(p,q)\asymp q$, whileAfter slightly decreasing $\epsilon$, the Schmidt subspace theorem puts all such primitive vectors $(p,q)$ in finitely many rational lines. Each rational line contains only the two opposite primitive integer vectors $\pm (p,q)$, and these determine the same fraction. Hence only finitely many fractions occur.

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Fix a finite set $S$ of primes. We prove that only finitely many denominators $q_n$ can be S-smooth number. This is the standard finite-place corollary of the Schmidt subspace theorem for best approximations of the first kind; the reduction is recalled here because the target $\alpha$ need not be algebraic.

Apply the Dirichlet approximation theorem at each cutoff between two successive record denominators. Its approximant can be replaced by the last record without increasing the error. Apply the finite-place Schmidt subspace theorem to the resulting pairs of primitive vectors, using $X-\alpha Y,Y$ at the real place only after eliminating $\alpha$ between two successive pairs, and $X,Y$ at every place belonging to $S$. The factorsfor an $S$-smooth denominator supply the required height saving. If infinitely many such records existed, one fixed rational subspace would contain infinitely many of the paired vectors. Eliminating its rational linear relation has two possible outcomes: either all sufficiently late records represent one rational number, which contradicts the irrationality of $\alpha$, or $\alpha$ is algebraic and, for some $\eta >0$, infinitely many of the records satisfyThe latter alternative contradicts the Roth theorem. Thus only finitely many $q_n$ are $S$-smooth.

If the largest prime factor of $q_n$ did not tend to infinity, some bound $B$ would contain the largest prime factor for infinitely many $n$. Taking $S$ to be the finite set of primes at most $B$ would make those denominators $S$-smooth, contrary to the preceding conclusion. Therefore $P^{+}(q_n)\to \infty$.

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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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For $\beta \in K$, the nonzero algebraic number $\gamma =\alpha -\beta$ has degree at mostThe basic height inequality givesThe Liouville height inequality therefore yieldsThus one may take the explicit constant

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A suitable theorem is the Baker lower bound for a homogeneous linear form in logarithms. Let ${\alpha }_1,\dots ,{\alpha }_n$ be nonzero algebraic numbers with chosen logarithms and letChoose $A_i\ge 3$ to bound the degree-normalized Absolute logarithmic Weil height of ${\alpha }_i$ and $|\log {\alpha }_i|$, and put $B=\max (3,|b_1|,\dots ,|b_n|)$. If $\Lambda \ne 0$, thenwhere $C$ is an effectively computable constant depending only on $n$ and the degree of the number field generated by the ${\alpha }_i$.

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Let $f$ be the monic cubic minimal polynomial of the algebraic integer $\alpha$, let ${\alpha }_1=\alpha ,{\alpha }_2,{\alpha }_3$ be its conjugates, and let ${\beta }_1=\beta ,{\beta }_2$ be the two conjugates of a nonrational $\beta \in {\mathcal{O}}_K$. Because the degrees three and two are coprime, the fields are linearly disjoint, andis a nonzero integer.

AssumeFor large $H(\beta )$ this makes ${\beta }_1$ bounded. Since $\beta$ is a quadratic algebraic integer, the height-Mahler measure formula givesand therefore $|{\beta }_2|\asymp H(\beta {)}^2$. The two remaining factors with $j=1$ are bounded, while the three factors with $j=2$ are $O(|{\beta }_2|)$. Consequently

We now use the standard effective norm form consequence of Part (b). Forwhere $1,\omega$ is an integral basis of $K$, the effective norm-form height estimate supplies effective constants $A,C>0$, depending only on $\alpha$ and $K$, such thatfor every nonzero $z\in V\cap {\mathcal{O}}_L$. Its proof factors $(z)$, balances a generator using the Dirichlet unit theorem, and applies the Baker lower bound for a homogeneous linear form in logarithms to the linear relations defining $V$; the coprime degrees $3$ and $2$ exclude a unit-family degeneracy.

Apply this estimate to $z=\alpha -\beta$. Since $\beta =\alpha -z$, the height inequalities imply $H(\beta )\le 2H(\alpha )H(z)$. HenceChoose the effective value $\epsilon =1/(2A)$. The last inequality bounds $H(\beta )$ effectively. The rational integers $\beta \in \mathbb{Z}$ are already covered by the stronger degree-three Liouville approximation theorem, and the Northcott theorem leaves only finitely many remaining quadratic integers of bounded height. Taking the minimum ofover this effective finite set gives an effective $c>0$ and provesfor every $\beta \in {\mathcal{O}}_K$.

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