Let
$$
\phi=\frac{1+\sqrt5}{2},
\qquad
\psi=\frac{1-\sqrt5}{2}=-\phi^{-1}.
$$
Suppose $F_n=x^k$ is a <perfect power> with $x\geq2$. The finitely many small $n$ can be absorbed into the final effective constant. The <Binet formula> gives $x^k\asymp\phi^n$ and
$$
\left|\frac{\phi^n}{\sqrt5x^k}-1\right|
=\frac{|\psi|^n}{\sqrt5x^k}
\ll\phi^{-2n}.
$$
Thus, for
$$
\Lambda=n\log\phi-\log\sqrt5-k\log x,
$$
the local Lipschitz equivalence of $u$ and $e^u-1$ at zero yields
$$
0<|\Lambda|\ll\phi^{-2n}.
$$
The form cannot vanish: applying the nontrivial <field automorphism> of $\mathbb Q(\sqrt5)$ to $\phi^n=\sqrt5x^k$ would give $\psi^n=-\sqrt5x^k$, 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 $\phi$ and $\sqrt5$ are absolute constants, while $\log A_x\asymp\log x$. Moreover,
$$
k\log x=n\log\phi+O(1),
$$
so $n/\log A_x\ll k$ and therefore $B^*\ll k$. The refined lower bound becomes
$$
|\Lambda|>\exp(-C_1\log x\log k)
>\exp\!\left(-C_2\frac nk\log k\right).
$$
Comparison with the exponential upper bound gives $k\leq C_3\log k$. 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-166/1/c/solution