$n\in \mathbf{N}$ such that a solution of $X^4+nX^2 +1$ is a root of unit

Question

Consider $f_n(X)=X^4+nX^2 +1$ in $\mathbf{Q}[X]$. I found that for all natural $n$ such that $n\neq 2-m^2$ for a natural $m$, $f_n(X)$ is irreducible in $\mathbf{Q}$. Consider $K_n=\mathbf{Q}(x)= \mathbf{Q}[X]/(f_n(X))$. Using Dirichlet Unit theorem, it is easy to see that $\mathcal{O}^*=\mu(K_n)\times \mathbf{Z} $, since all the roots of $f_n(X)$ are complex. It also easy to see that $x\in \mathcal{O}^*$. So my question is how to determine the natural $n$ such that $\mathcal{O}^*/(x)$ is finite. Clearly since $x$ is a unit, we have that $x=(z,a)$, where $a\in \mathbf{Z}$. So, that quotient is finite for $a\neq 0$, i.e., for $x\notin \mu(K_n)$. For example, for the $12$th cyclotomic polynomial, i.e., $X^4-X^2+1$, $x$ is a solution for $n=-1$ and so $x\in \mu(K_{-1})$. But I do not know how to find the other $n$ such that it is not finite.

Answer 1

If some root $r$ of $X^4+nX^2+1$ is a root of unity, then $r^2$ is a root of unity and vanishes $X^2+nX+1$. Thus $-n$ is double the real part of $r^2$ (because $r^2$ and its complex conjugate are the roots of $X^2+nX+1$, since the product of roots is $1$), thus $|n| \leq 2$.

If $|n|=2$, your polynomial is $(X^2 \pm 1)^2$. If $n=0$, your polynomial is $\Phi_4$. If $|n|=1$, $j$ or $ij$ is a root.