= (-1)^{\theta}$. Then equation implies that $c = \pm i e^{i2t}$. Since $e^{it}$ is a $4{\theta}$-th root of unity, so is $ce^{i(-{\theta}-1)t} = e^{it(1-{\theta})}$. On the other hand, from equation , $$e^{it(1-{\theta})} = \tfrac{{\theta}}{\sqrt{n}}(x-1) = \tfrac{-\sqrt{n} \pm i}{2{\theta}},$$ which has algebraic degree exactly four since $n$ is not a perfect square. This implies that $e^{it(1-{\theta})}$ is a primitive $m$-th root of unity for some $m$ such that $\phi(m) = 4$, namely $m \in \{5,8,10,12\}$. (Here $\phi$ is the Euler totient function.) But then it is not difficult to check that none of these primitive $m$-th roots of unity have imaginary part $\pm1/2{\theta}$, for any ${\theta}> 1$. Thus there are no solutions.

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[[**Case**]{} $(x,y) = (-1, \tfrac{2{\theta}^2-3 \pm i\sqrt{4{\theta}^2-5}}{2({\theta}^2-1)})$:]{} The characterizing equations reduce to $$\begin{aligned}
ce^{i2{\theta}^2t} \sqrt{n} & = -2 \pm i\sqrt{4{\theta}^2-5}  \label{eqn:flatsrg2.1} \\
ce^{i{\theta}t} \sqrt{n} & = 1 - {\theta}+ y(-{\theta}-1) \label{eqn:flatsrg2.2} \\
ce^{-i{\theta}t} \sqrt{n} & = 1 + {\theta}+ y({\theta}-1) \label{eqn:flatsrg2.3}.\end{aligned}$$ Simplifying plus we have $\sqrt{n} c{\mathrm{Re}}(e^{i{\theta}t}) = 1 - y$. Taking the absolute value, we get ${\mathrm{Re}}(e^{i{\theta}t})^2 = 1/n({\theta}^2-1)$. Similarly, from minus we get ${\mathrm{Im}}(e^{i{\theta}t})^2 = {\theta}^2(4{\theta}^2 - 5)/n({\theta}^2-1)$. Thus we can solve: $$\begin{aligned}
e^{i{\theta}t} & = \frac{1}{\sqrt{n({\theta}^2-1)}}[\pm 1 \pm {\theta}i\sqrt{4{\theta}^2-5}] \label{eqn:eitht}, \\
c & = \frac{1}{2\sqrt{{\theta}^2-1}}[\pm 