Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 166 (Diophantine Analysis)

For the general result, choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_s$ and algebraic coefficients ${\beta }_0,\dots ,{\beta }_s$, and putLetwhere $f_{{\alpha }_j}$ is the minimal polynomial and its $H$ is the naive polynomial height, and letThe general lower bound for a linear form in logarithms states that, if $\Lambda \ne 0$, thenwhere the effective constant $C$ depends only on $s$ and the degree of the number field generated by all the data.

For the improved homogeneous result, take $b_1,\dots ,b_s\in \mathbb{Z}$ andWith the same $A_j$, after ordering the terms setThe Baker lower bound for a homogeneous linear form in logarithms givesBoth constants are effective. The division by $\log A_s$ in $B^{\ast }$ is the improvement that matters when ${\alpha }_s$ has variable height.

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Put $\Lambda =n\log 2-m\log 3$. It is nonzero by unique prime factorization. If $3^m\le 2^{n-1}$ or $3^m\ge 2^{n+1}$, the claimed inequality follows immediately after increasing the effective constant. We may therefore assumewhich implies $m\asymp n$.

The Baker lower bound for a homogeneous linear form in logarithms, with the fixed algebraic numbers $2$ and $3$, givesfor an effective absolute constant $C_0$. If $|\Lambda |>1$, the desired conclusion is again immediate. Otherwise, the mean value theorem applied to the exponential function on $[-1,1]$ gives $|e^u-1|\ge c|u|$. HenceAbsorbing the fixed factor into a larger exponent proves $|2^n-3^m|\ge 2^n/n^C$.

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LetSuppose $F_n=x^k$ is a perfect power with $x\ge 2$. The finitely many small $n$ can be absorbed into the final effective constant. The Binet formula gives $x^k\asymp {\varphi }^n$ andThus, forthe local Lipschitz equivalence of $u$ and $e^u-1$ at zero yieldsThe form cannot vanish: applying the nontrivial field automorphism of $\mathbb{Q}(\sqrt{5})$ to ${\varphi }^n=\sqrt{5}{x}^k$ would give ${\psi }^n=-\sqrt{5}{x}^k$, whose absolute values are incompatible.

Apply the Baker lower bound for a homogeneous linear form in logarithms with the variable-height number $x$ placed last. The parameters belonging to $\varphi$ and $\sqrt{5}$ are absolute constants, while $\log A_x\asymp \log x$. Moreover,so $n/\log A_x\ll k$ and therefore $B^{\ast }\ll k$. The refined lower bound becomesComparison with the exponential upper bound gives $k\le C_3\log k$. Since $k/\log k$ tends to infinity, this bounds $k$ by an effective absolute constant. Enlarging it to cover the discarded small indices proves the claim.

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Let $K$ be a number field. Its places consist of its real embeddings, conjugate pairs of complex embeddings, and the finite places associated with nonzero prime ideals $\mathfrak{p}$ of its ring of integers of a number field. At a real or complex place use the usual absolute value. If $\mathfrak{p}$ lies over the prime number $p$ with ramification index $e_{\mathfrak{p}}$, normalize its absolute value byThese normalizations extend the standard absolute values on $\mathbb{Q}$.

Write $d_v=[K_v:{\mathbb{Q}}_v]$ for the local degree of a place. Thus $d_v$ is $1$ at a real place, $2$ at a complex place, and $e_v{f}_v$ at a finite place. The Absolute multiplicative Weil height isThe product formula shows that this is unchanged when $K$ is replaced by a larger number field containing $\alpha$.

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The product formula says that every $x\in K^{\times }$ satisfiesFirst suppose that $x$ is an algebraic integer. Its principal ideal has the prime ideal factorizationTaking the ideal norm givesOn the other hand, the field norm is the product over embeddings, soEquating these expressions proves the formula for algebraic integers. Every nonzero element of $K$ is a quotient of two algebraic integers, and multiplicativity completes the proof.

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For every $k\in \mathbb{Z}$,For $k\ge 0$, this follows directly fromat each place of a number field. The product formula gives $H({\alpha }^{-1})=H(\alpha )$, becauseand this handles negative $k$ as well.

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If $\alpha \ne 0$ has degree $d$, then at every Archimedean embeddingIndeed, the factor of the defining product belonging to that Archimedean place shows $\max (1,|\alpha |{)}^{1/d}\le H(\alpha )$; using the exact local degree only improves this estimate. This proves the upper bound. Apply it to ${\alpha }^{-1}$ and use $H({\alpha }^{-1})=H(\alpha )$ to obtain the lower bound. This is the Liouville height inequality.

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Write $\mathcal{L}(R)$ for the polynomial length. The height bound for a polynomial evaluation isprovided the denominator is nonzero. At non-Archimedean places the integral coefficients and ultrametric inequality give the local estimate without an extra constant; at Archimedean places the triangle inequality gives the polynomial length. Multiplication over every place of a number field and the product formula produce the displayed bound.

Taking $P(X,Y)=X+Y$ and $Q=1$ givesTaking $P(X)=X+1$ gives

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For coprime integers $p,q$ with $q>0$,This follows either directly from the real and p-adic absolute values, or from the height-Mahler measure formula applied to the primitive minimal polynomial $qX-p$.

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Choose distinct prime numbers $p,q$ so large that $pq>C(p+q)$, and setThe fraction $(p+q)/(pq)$ is reduced, because neither $p$ nor $q$ divides $p+q$. The height of a rational number therefore givesThe choice of $p,q$ proves the required strict inequality.

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For an integer $m>1$, let ${\alpha }_m$ be a root ofIts roots areThe polynomial is irreducible over $\mathbb{Q}$: its discriminant is $4m(m-1)$, and the product of the coprime consecutive integers $m$ and $m-1$ cannot be a square unless both are squares, which is impossible for consecutive positive squares beyond $0,1$.

The height-Mahler measure formula givesBoth algebraic conjugates of ${\alpha }_m+1$ exceed $1$, soConsequentlyFix, for example, $\delta =1/10$. For all sufficiently large $m$, the ratio is larger than $1+\delta$ by a fixed margin, while $H({\alpha }_m)\to \infty$. Hence, for every $C>0$, some sufficiently large $m$ satisfies

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For a linear form $L=a_1{X}_1+\cdots +a_N{X}_N$ over a number field $K$, define its height to be the projective height of its coefficient vector:The product formula makes this independent of multiplying $L$ by a nonzero scalar.

The Siegel lemma says the following. Let $D=[K:\mathbb{Q}]$, let $L_1,\dots ,L_M$ be linear forms in $N$ variables with $DM<N$, and suppose $H_p(L_i)\le \mathcal{H}$ for every $i$, where $\mathcal{H}\ge 1$. Then there is a nonzero $\mathbf{x}\in {\mathbb{Z}}^N$ annihilated by all the forms and satisfying

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Writeso there are $N=2(n+1)$ unknown integer coefficients. Let $t$ be the number of integers $r$ withFor each such $r$, impose the linear equation over $K=\mathbb{Q}(\alpha )$where $D_r$ is the normalized derivative of a polynomial. The coefficient of $p_j$ is $\left(\genfrac{}{}{0ex}{}{j}{r}\right){\alpha }^{j-r}$ and that of $q_j$ is $\left(\genfrac{}{}{0ex}{}{j}{r}\right){\alpha }^{j-r+1}$. The local definition of projective height, together with $\left(\genfrac{}{}{0ex}{}{j}{r}\right)\le 2^n$, givesfor a constant $C_0$ depending only on $\alpha$.

There are $M=t$ forms over the degree-$d$ field $K$, andMoreover $N-dM\ge \kappa n+2$. Applying Siegel lemma gives a nonzero integral coefficient vector withbecause the exponent is bounded in terms of $\kappa$ and every fixed power of $n$ is at most exponential in $n$. These $P,Q$ have all the required vanishing normalized derivatives.

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Suppose $P$ and $Q$ were linearly dependent over $\mathbb{Q}$. Then $P=cQ$ for some $c\in \mathbb{Q}$, with the zero-polynomial cases included. Since $d\ge 3$, $c+\alpha \ne 0$, and thereforeIf this has multiplicity $t$ at $\alpha$, then the minimal polynomial of $\alpha$ to the power $t$ divides $Q$ in $\mathbb{Q}[X]$. Hence $dt\le \mathrm{deg}Q\le n$.

The construction in part (b) givesWhen $\kappa <1$, this exceeds $n/d$ for all sufficiently large $n$ depending only on $\kappa$ and $d$, contradicting $dt\le n$. Thus $P,Q$ are linearly independent.

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The determinant in the hint is independent of $Y$ and equals the WronskianIt is nonzero because $P,Q$ are linearly independent. Also $\mathrm{deg}W<2n$, and coefficient convolution givesfor a constant $C_0$ depending only on $C$.

Fix $y$, and suppose $F(X,y)$ has multiplicity $r$ at $a=p/q$. Inthe two terms vanish to orders at least $r$ and $r-1$, so $W$ has multiplicity at least $r-1$ at $a$. The primitive polynomial $(qX-p{)}^{r-1}$ therefore divides $W$ in $\mathbb{Z}[X]$ by Gauss lemma for polynomials. Comparing leading coefficients givesIf $r>\delta n+1$, this implies $q^{\delta n}<C_0^n$ and hence $q<C_0^{1/\delta }$. Choosing $q$ larger than this bound proves that $r\le \delta n+1$, uniformly in $y$.

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It is enough to prove the result for $0<\kappa <1$, since a construction for a smaller positive value gives the weaker vanishing requirement for any larger one. Apply part (b) with $\kappa /2$. For large $n$, part (c) gives linearly independent polynomials $P_0,Q_0$, andsatisfies $H(G)\le C_1^n$ and has order at leastat $X=\alpha$ after setting $Y=\alpha$.

Set $\delta =\kappa /(4d)$. By the rational multiplicity bound for a linear auxiliary polynomial, once $q_1$ is sufficiently large, the one-variable polynomial $G(X,y)$ has multiplicity at most $\delta n+1$ at $p_1/q_1$. Consequently there is some integer $j\le \delta n+1$ such thatPut $F=D_j^{X}G$. The normalized derivative of a polynomial preserves integral coefficients and multiplies height by at most $2^n$, so $H(F)\le C_2^n$. Differentiation lowers the vanishing order at $(\alpha ,\alpha )$ by at most $j$; for sufficiently large $n$,Finally write $F=P-YQ$ by replacing the coefficient of $Y$ by its negative. Then $F(p_1/q_1,y)\ne 0$ and all the claimed bounds hold.

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If $\alpha$ is nonreal, its positive distance from $\mathbb{R}$ makes the assertion immediate, so assume $\alpha \in \mathbb{R}$. Fix $\epsilon >0$ and putChoose $\kappa >0$ so small thatSuppose for a contradiction that infinitely many reduced fractions satisfy $|\alpha -p/q|<q^{-\mu }$. Their denominators are unbounded.

Choose one such $p_1/q_1$ with $q_1$ arbitrarily large, and then a later one $p_2/q_2$ with $q_2$ arbitrarily large. Choose the integer $n$ so thatThen $n$ can be made sufficiently large for part (e). Apply it with $y=p_2/q_2$. Since $F$ has integral coefficients, degree at most $n$ in $X$, and degree at most one in $Y$, its nonzero rational value satisfies the denominator boundThe supplied upper estimate and the two approximation inequalities give, with $a=(2-\kappa )n/d$,Using the upper bound $q_2<q_1^{a+(2-\kappa )/d}$ in the denominator estimate and comparing yieldsAfter taking $n$th roots and letting the choice of $q_2$ make $n$ large, this bounds $q_1^{\eta }$ by a constant depending only on $\alpha ,\kappa$. That contradicts the ability to choose $q_1$ arbitrarily large. Hence only finitely many such rational approximations exist. This is the Thue-Siegel rational approximation bound.

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