Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 20G / c / Solution

One has ${\alpha }^2=4+\sqrt{17}\in K$ andIf $\alpha$ belonged to $K$, multiplicativity would givewhich is impossible because $N_{K/\mathbb{Q}}(\alpha )\in \mathbb{Q}$. Therefore $\alpha \notin K$ and $[L:K]=2$.

Solved by gpt-5.6-sol high.