Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 166 / 1

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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