Some motivation: The matrix $M$ is Butson Hadamard if the entries of $M$ are $k^{\textrm{th}}$ roots of unity (for some $k$), and distinct pairs of rows are orthogonal under the usual Hermitian inner product. I am interested in classifying the Butson Hadamard matrices for which some power is a real scalar matrix. (This is equivalent to the corresponding unitary matrix having finite multiplicative order.)

In the case of $2\times 2$ matrices, after some reductions, the problem is equivalent to classifying the pairs of roots of unity $(\zeta_{1}, \zeta_{2})$ for which

$$\Re( \zeta_{1} ) = \sqrt{2} \Re(\zeta_{2}) $$

This can be restated as $2\Re(\zeta_{2})^{2} - \Re(\zeta_{1})^{2} = 0$. The solutions we have found, up to negation and complex conjugation, are $(i, i)$, $(1, \omega_{8})$ and $(\omega_{8}, \omega_{6})$. We ran computer searches which suggest that any further solution involves a $k^{\textrm{th}}$ root of unity for some $k \geq 33$.

Q1: Is this list of solutions to the displayed equation complete?

Q2: More generally, what techniques exist for finding all solutions to a polynomial equation in real parts of roots of unity?

  • $\begingroup$ I found no further solutions up to $k \le 10000$. $\endgroup$ – Yaakov Baruch Aug 4 '17 at 16:12

We have $$ \frac{\zeta_1+\frac{1}{\zeta_1}}{\zeta_2+\frac{1}{\zeta_2}}=\sqrt{2}. $$ By applying Galois group automorphisms we can assume that $\zeta_2=e^{2\pi i/n}$ for some $n$, where we can take either value of $\sqrt{2}$. Solving for $\zeta_1$ gives $$ \sqrt{2}\zeta_1=\zeta_2+\frac{1}{\zeta_2} \pm\sqrt{\zeta_2^2+\frac{1}{\zeta_2^2}}. $$ If $n>8$ then $\zeta_2^2+\frac{1}{\zeta_2^2}>0$. This implies that $\zeta_1$ is real, a contradiction. Thus there are no further solutions.


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