Real part of a primitive root of unity is not an algebraic integer for n>4. I'm trying to prove the following.

Suppose that $\alpha$ is an $n$th root of unity whose real part is an algebraic integer. Then $\alpha^4 = 1$.

I've thought about this for a while, and have the following ideas/questions.
If I can show that the algebraic norm $\frac{1}{2}(\alpha + \alpha^{-1})$ is not an integer for $n> 4$, then I will be done. Since $x^2-(\alpha +\alpha^{-1})x +1$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\alpha+\alpha^{-1})$, we have that
$$
N\left(\frac{1}{2}(\alpha + \alpha^{-1})\right) = \frac{1}{2^{\phi(n)/2}}N(\alpha+\alpha^{-1})
$$
where $\phi(n)$ is the Euler $\phi$ function. So we just need to consider $N(\alpha+\alpha^{-1})$. Is there a way that we can bound $N(\alpha + \alpha^{-1})$? If we can show that this norm is less than $2^{\phi(n)/2}$ then we're done.
Another thought is that $\frac{1}{2}(\alpha+\alpha^{-1})$ is a root of $T_n(x)-1$ where $T_n$ is the $n$th Chebyshev polynomial. This however is not generally irreducible, so the fact that $T_n$ is not monic does not tell us that $\frac{1}{2}(\alpha+\alpha^{-1})$ is not an algebraic integer.
Thanks for any thoughts or hints.
 A: Here is a more direct argument that shows that the norm of $(\alpha + \alpha^{-1})/2$ is less than one unless $\alpha^2 = 1$. The norm is the product of the conjugates. Note that the conjugates of this are all of the form $(\beta + \beta)/2$ for some other root of unity $\beta = \alpha^k$. But note by the triangle inequality that
$$\left| \frac{\beta + \beta^{-1}}{2} \right| \le \frac{1}{2} + \frac{1}{2} = 1.$$
In particular, either the norm has absolute value strictly less than one, or equality holds for all conjugates and particular for $\beta = \alpha$. But the triangle inequality $|x + y| \le |x| + |y|$ for complex numbers is strict unless $\arg(x) = \arg(y)$. Since $\alpha$ and $\alpha^{-1}$ both have absolute value one, that means the triangle inequality is strict unless $\alpha = \alpha^{-1}$, which implies that $\alpha^2 = 1$.
To finish the original problem, if $(\alpha + \alpha^{-1})/2$ is an algebraic integer then its norm is an algebraic integer. Since the norm is less than $1$ unless $\alpha^2 = 1$, the only remaining possibility is that the norm is $0$, in which case $\alpha + \alpha^{-1} = 0$ and $\alpha^2 = -1$, or $\alpha^4 = 1$.
A: Every root of unity is an algebraic integer. The "real part" (not really so well defined) of a root of unity is (however we interpret it) (EDIT)! half the sum of it and one of its conjugates (also an algebraic integer...), so is (EDIT) at worst half an algebraic integer.
Or... are you asking the question you mean to?
(Thanks to comments for the correction about "half"...)
