= Solution
Put $\theta=\sqrt{p+\sqrt p}$. Then $(\theta^2-p)^2=p$, so $\theta$ is a root of
$$
F(X)=X^4-2pX^2+p^2-p.
$$
This <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 $p^2$. Therefore
$$
\boxed{m_\theta(X)=X^4-2pX^2+p^2-p,\qquad[\mathbb Q(\theta):\mathbb Q]=4.}
$$
Now 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 $\sqrt p=\theta^2-p$,
$$
\beta=\frac{n(\theta^2-p)}\theta\in\mathbb Q(\theta).
$$
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 $-\sqrt p$, so
$$
\sigma(\beta)=\frac{-n\sqrt p}{\beta}=-\theta.
$$
Thus $\theta\mapsto\beta\mapsto-\theta\mapsto-\beta\mapsto\theta$ is a four-cycle. The <Galois group> has order four and contains an element of order four, giving
$$
\boxed{\operatorname{Gal}(\mathbb Q(\theta)/\mathbb Q)\cong C_4.}
$$
For the requested lower degrees, $(q,r)=(3,4)$ gives $\sqrt{q+\sqrt r}=\sqrt5$ and degree two; $(q,r)=(3,1)$ gives $2$ and degree one. Both pairs use positive <integers>.
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