There are 2 well-known formulas involving Cyclotomic polynomials, which can be described roughly as writing $\phi_n$ as norms of elements in some quadratic extension. They appear in wikipedia and mathworld, but the references either don't contain a proof or are untraceable. I consulted several books, but I found a (partial) proof only on the internet and only of one identity.
Let $n > 3$ be an odd squarefree integer, and let $\phi_n$ be the n'th cyclotomic polynomial. The identities are:
$$ (Gauss) \exists R, S\in \mathbb{Z}[x] \text{ such that}: 4\cdot \phi_n(x) = R^2 - (-1)^{\frac{n-1}{2}} n S^2$$ $$ (Lucas) \exists R, S\in \mathbb{Z}[x] \text{ such that}: \phi_n(x) = R^2 - (-1)^{\frac{n-1}{2}} nx S^2$$
What I'm looking for is a proof of Lucas's identity (the proof is suppose to be in "Kraitchik, M. Recherches sue la théorie des nombres, tome I").
A reference for Gauss's identity will also be appreciated.
The proof of the 1st identity, which I've found long ago on the web, is very neat: The Galois group of $Q_n:=\mathbb{Q}(e^{\frac{2\pi i}{n}})$ over $\mathbb{Q}$ is isomorphic to $G_n := (\mathbb{Z}/n\mathbb{Z})^{\times}$. Let $Q'_n$ be the quadratic subfield which is obtained by taking the elements of $Q_n$ fixed by $\{ a \in G_n: \left( \frac{a}{n} \right) = 1 \}$. By considering the (generalized) Gauss sum $\sum_{a=0}^{n-1} \left( \frac{a}{n} \right) e^{a \cdot \frac{2 \pi i}{n}}$, which lies in $Q'_n$ and its square can be evaluated as $(-1)^{\frac{n-1}{2}}n$, which implies $Q'_n = \mathbb{Q}(\sqrt{(-1)^{\frac{n-1}{2}}n})$.
Now consider the polynomial $\psi_n(x) := \prod_{1\le a \le n, (a,n)=1, \left( \frac{a}{n} \right) = 1 } (x- e^{\frac{2\pi i}{n} a})$. It lies in $Q'_n[x]$. In fact, its coefficients are algebraic integers. Since the ring of integers of $Q'_n$ is $\mathbb{Z}[\frac{1+\sqrt{(-1)^{\frac{n-1}{2}}n}}{2}]$, we can write $\psi_n(x) = \frac{R(x) +\sqrt{(-1)^{\frac{n-1}{2}} n} S(x)}{2}$. Now, by taking the product of $\psi_n$ by its Gaolis conjugate, we find $\phi_n(x) = \frac{R^2 - (-1)^{\frac{n-1}{2}}n S^2}{4}$.