Past exam of the number theory course of the University of Oxford / 2007 / A / 3 / Solution

Put $\theta =\sqrt{p+\sqrt{p}}$. Then $({\theta }^2-p{)}^2=p$, so $\theta$ is a root ofThis polynomial satisfies Eisenstein's criterion at $p$: every nonleading coefficient is divisible by $p$, while the constant coefficient $p(p-1)$ is not divisible by $p^2$. ThereforeNow suppose $p=n^2+1$, taking $n>0$ without loss of generality. Let $\beta =\sqrt{p-\sqrt{p}}$. Since $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$,Consequently all four distinct roots $\pm \theta ,\pm \beta$ lie in this field. This gives a cyclic quartic field from a nested quadratic radical. It is a splitting field in characteristic zero, hence a normal extension and a Galois extension.

There is a Galois automorphism $\sigma$ sending $\theta$ to $\beta$: a root image defines an embedding, and normality makes it an automorphism of the splitting field. It sends $\sqrt{p}={\theta }^2-p$ to $-\sqrt{p}$, soThus $\theta \mapsto \beta \mapsto -\theta \mapsto -\beta \mapsto \theta$ is a four-cycle. The Galois group has order four and contains an element of order four, givingFor the requested lower degrees, $(q,r)=(3,4)$ gives $\sqrt{q+\sqrt{r}}=\sqrt{5}$ and degree two; $(q,r)=(3,1)$ gives $2$ and degree one. Both pairs use positive integers.