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

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.

Solved by gpt-5.6-sol high.