Finding the norm in the cyclotomic field $\mathbb{Q}(e^{2\pi i / 5})$

I agree completely with Qiaochu. If you understand what he wrote, then you can use this to give a much shorter and easier proof than the hint is suggesting.

In fact I didn't even read the hint. I saw the question, thought "Oh, that looks like the norm from $\mathbb{Q}(\sqrt{5})$". I also know that $\mathbb{Q}(\sqrt{5})$ is the unique quadratic subfield of $\mathbb{Q}(\zeta_5)$ (since $5 \equiv 1 \pmod 4$; there's a little number theory here), and that for any tower of finite field extensions $L/K/F$, we have

$N_{L/F}(x) = N_{K/F}(N_{L/K}(x))$.

Norms also carry algebraic integers to algebraic integers, so this shows that the norm of any element of $\mathbb{Z}[\zeta_5]$ is also the norm of some algebraic integer of $\mathbb{Q}(\sqrt{5})$, i.e., is of the form $(\frac{A+\sqrt{5}B}{2})(\frac{A-\sqrt{5}B}{2})$ for some $A,B \in \mathbb{Z}$. I think we're done.

[I have never read Stewart and Tall, so it may be that they have not assumed or developed this much theory about norm maps at the point they give that exercise. But if you know it, use it!]

You're almost there: set $A=2q-r-s$ and $B=r-s.$ Then your expression for $\textbf{N}(\alpha)$ reduces to the desired form. i.e. your

$$\textbf{N}(\alpha) = \frac14 \left \lbrace (2q-r-s)^2 - 5(r-s)^2 \right \rbrace = \frac14(A^2-5B^2).$$