Extension of $\zeta_5^2 + \zeta_5^3$ over $ℚ$ is quadratic. Consider, for $\zeta_5$ the fifth root of unity, $\alpha := \zeta_5^2 + \zeta_5^3 ∈ ℂ$. 
To demonstrate: $[ℚ(α): ℚ] = 2$.
We know that $\zeta_5$ is a root of the cyclotomic polynomial $f := \frac{X^5-1}{X-1} = X^4 + X^3 + X^2 + X +1 ∈ ℤ[X]$. As it is monic and irreducible, we must in fact have that $f = f^{ζ_5}_ℚ$ (the minimum polynomial of $ζ_5$ over ℚ) and hence $[ℚ(ζ_5): ℚ] = \deg f = 4$.
How then can  $[ℚ(α): ℚ]$ be 2? Or have I already concluded something wrongly?
Edit
Without resorting to trigonometric identities (because the end game here is to derive them), does this allow one to prove that $ℚ(\alpha) = ℚ(√{5})$?
 A: As you note $[\Bbb{Q}(\zeta_5):\Bbb{Q}]=4$ and clearly $\alpha\in\Bbb{Q}(\zeta_5)$, so you have a tower of fields
$$\Bbb{Q}\subset\Bbb{Q}(\alpha)\subset\Bbb{Q}(\zeta_5).$$
The degree is multiplicative over towers of fields so $[\Bbb{Q}(\alpha):\Bbb{Q}]$ must divide $[\Bbb{Q}(\zeta_5):\Bbb{Q}]=4$. 
As you know, the first four powers of $\zeta_5$ form a basis for $\Bbb{Q}(\zeta_5)$ as a vector space over $\Bbb{Q}$. From this it is immediate that $\alpha\notin\Bbb{Q}$, which shows that $[\Bbb{Q}(\alpha):\Bbb{Q}]\neq1$. To determine whether the degree equals $2$ or $4$ you can compute a few powers of of $\alpha$: If the degree is $2$ then $\alpha^0,\alpha^1,\alpha^2\in\Bbb{Q}(\alpha)$ must be linearly dependent over $\Bbb{Q}$. A few simple computations show that
\begin{eqnarray*}
\alpha^0&=&1=\zeta_5^0,\\
\alpha^1&=&\zeta_5^2+\zeta_5^3,\\
\alpha^2&=&(\zeta_5^2+\zeta_5^3)^2=\zeta_5^4+2\zeta_5^5+\zeta_5^6=2+\zeta_5+\zeta_5^4,
\end{eqnarray*}
where we used the fact that $\zeta_5^5=1$. Now keep in mind that $1+\zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4=0$, so
$$\alpha^2+\alpha-1=0.$$
This shows that $\Bbb{Q}(\alpha)$ is a quadratic extension of $\Bbb{Q}$.
A: Fast method: (even though not very intuitive)
$$H(X):=X^2+X-1\\
Q(X):=H(X^2+X^3)=(X^2+X^3)^2+X^2+X^3-1=X^6+2X^5+X^4+X^3+X^2-1\\
Q(\zeta_5)=\zeta_5+2+\zeta^4+\zeta^3+\zeta^2=\zeta^4+\zeta^3+\zeta^2+\zeta+1=0\\
H(\alpha)=0$$
Actually, with a little bit more computations, one is able to prove
$$\alpha=-\frac{1}{2}-\frac{\sqrt{5}}{2}$$
Intuitive method:
First note that, since $\alpha\in \mathbb{R}$ (because $\overline{\zeta_5^2}=\frac{1}{\zeta_5^2}=\zeta_5^3$), $\mathbb{Q}(\alpha)\subsetneq\mathbb{Q}(\zeta_5)$
We then have
$$4=[\mathbb{Q}(\zeta_5):\mathbb{Q}]=[\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]\\
P(X):=X^3+X^2-\alpha\in \mathbb{Q}(\alpha)[X]$$
So 
\begin{cases}1<[\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]\le \text{deg}(P)=3\\
[\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]|4\end{cases}
And thus
$$[\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]=2\\
[\mathbb{Q}(\alpha):\mathbb{Q}]=\frac{[\mathbb{Q}(\zeta):\mathbb{Q}]}{[\mathbb{Q}(\alpha):\mathbb{Q}]}=2$$
A: A lot of nice solutions have already been posted, but I thought I'd add one that was a bit more elementary. The solution below exploits that palindromic property of the coefficients of $x^4+x^3+x^2+x+1$.
First, let $\omega=\zeta_5^2$, and note that
$$\alpha=\zeta_5^2+\zeta_5^3=\omega+\frac{1}{\omega}.$$
Since the minimal polynomial of $\omega$ over $\Bbb{Q}$ is $x^4+x^3+x^2+x+1$, we have that
$$\omega^4+\omega^3+\omega^2+\omega+1=0.$$
And since $\omega\ne0$, we can divide by $\omega^2$ to obtain
$$\omega^2+\omega+1+\frac{1}{\omega}+\frac{1}{\omega^2}=0,$$
which we can rearrange to obtain
$$\left(\omega^2+\frac{1}{\omega^2}\right)+\left(\omega+\frac{1}{\omega}\right)+1=0.$$
We can then complete the square:
$$\left(\omega^2+2+\frac{1}{\omega^2}\right)+\left(\omega+\frac{1}{\omega}\right)-1=0.$$
Hence:
$$\left(\omega+\frac{1}{\omega}\right)^2+\left(\omega+\frac{1}{\omega}\right)-1=0.$$
So $\alpha=\omega+\dfrac{1}{\omega}$ is a root of $x^2+x-1$. Since this is a quadratic polynomial with no rational roots, it is irreducible over $\Bbb{Q}$, and is therefore the minimal polynomial of $\alpha$ over $\Bbb{Q}$. Thus, $\left[\Bbb{Q}(\alpha):\Bbb{Q}\right]=2$.
