<s>[INST] Degree of splitting field of $X^4+2X^2+2$ over $\mathbf{Q}$ 
Find the degree of splitting field of $f=X^4+2X^2+2$ over $\mathbf{Q}$.

By Eisenstein, $f$ is irreducible. By setting $Y=X^2$, we can solve for the roots: $Y=-1\pm i \iff X=\sqrt[4]{2}e^{a\pi i/8}$, $a\in\{3,5,11,13\}$. Clearly $f$ splits in $\mathbf{Q}(\sqrt[4]{2},\zeta_{16})$. $\zeta_{16}$ is a zero of $X^8+1$, which is irreducible over $\mathbf{Q}$ by Eisenstein applied to $(X+1)^8+1$. I was not able to go further. 
Could someone help me to proceed?
 [/INST] The OP asked in a comment

is there some easy way to show that $−1+i$ is not a square in $\mathbb Q(\zeta_8)$?

Let $a,b,c,d \in \mathbb Q$ and $\zeta_8 = e^{\tfrac14\pi i}$.
Consider the number
$$\tag 1 \nu = a + b e^{\tfrac14\pi i}+ c e^{\tfrac12\pi i}  + d e^{\tfrac34\pi i}$$
Then
$\quad \nu^2 = a^2 + c^2 e^{i \pi} + 2 b d e^{i \pi} +$
$\quad\quad\quad\quad 2 a b e^{(i/4) \pi} + 2 c d e^{(5 i/4) \pi} +$
$\quad\quad\quad\quad b^2 e^{(i/2) \pi} + 2 a c e^{(i/2) \pi}  + d^2 e^{(3 i/2) \pi}+$
$\quad\quad\quad\quad 2 ad e^{(3 i/4) \pi} + 2 bc e^{(3 i/4) \pi} $
$\quad\quad\,=$
$\quad\quad\quad\quad a^2 - c^2- 2 b d +$
$\quad\quad\quad\quad (2 a b - 2 c d) e^{(1 i/4) \pi} +$
$\quad\quad\quad\quad (b^2  + 2 a c - d^2) e^{(1 i/2) \pi}+$
$\quad\quad\quad\quad (2 ad  + 2bc) e^{(3 i/4) \pi} $
Assume that $\nu^2 = i -1$. Then $a b - c d = 0$ and $ad  + bc = 0$. Multiplying the first equation by $c$ and the second equation by $a$ and then subtracting the first from the second and simplifying we get
$\quad (a^2 + c^2)d = 0$
Case 1: If both $a$ and $c$ are zero then $\nu^2 = -2bd + (b^2 -d^2)i$. So $b \ne 0$ and $d = \frac{1}{2b}$ and  
$\quad 4b^4 -4b^2 - 1 = 0$
But that would imply that $\sqrt 2$ is a rational number.
Case 2: If $d = 0$ then  $\nu^2 =
a^2  - c^2 + 2 abe^{πi/4}  + (b^2 +2ac)i + 2bc  e^{3πi/4}$. If we assume that $b \ne 0$ we must have both $a = 0$ and $c = 0$, which is impossible. But then $a^2  - c^2 = -1$ and $2ac = 1$. But then $a =\frac{1}{2c}$ and we wind up with, 
$\quad 4c^4 -4c^2 - 1 = 0$
But that would imply that $\sqrt 2$ is a rational number.
It is interesting to note that $f(x) = 4x^4 -4x^2 - 1 $ has the same splitting field as the initial polynomial the OP was examining, and it has two real roots,
$$ x = \pm\sqrt{\frac{1+\sqrt2}2}$$
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