# Squares in a Cyclotomic Extension

Is there an algorithm that can tell me whether some element of $\mathbb{Q}(\omega_n)$ is a square (i.e. in $(\mathbb{Q}(\omega_n)^{\times})^2$) ?

An element $a = \sum_{i=0}^{\phi(n) - 1} a_i \omega_n^i$ is a square iff the equation $\sum_{i=0}^{\phi(n) - 1} a_i \omega_n^i = (\sum_{i=0}^{\phi(n)-1} b_i \omega_n^i)^2$ is solvable in rationals. So by expanding, expressing $\omega_n^j$ as linear combination of $\{ \omega_n^i \}_{i=0}^{\phi(n) - 1}$ and equating coefficients we get that $a \in (\mathbb{Q}(\omega_n)^{\times})^2$ iff some quadratic polynomial equations (depending on a) have a common rational solutions. But this system can be complicated and unapproachable.

Is there a nice way to solve this system, or some reciprocity law allowing to simplify the calculations?

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Have you tried solving this problem in a simple case first? Say, $\mathbf Q(i)$ or even $\mathbf Q$? –  Bruno Joyal Nov 6 '13 at 4:23