applies to the complement: uniform mixing occurs when ${\theta}$ is even, and no solutions occur when ${\theta}$ is odd.[  ]{}

There are infinite families of strongly regular graphs with these parameters [@haemers08].

Symplectic-type graphs
----------------------

We turn the graphs in part (iii) of Theorem \[thm:chan\]. These are the strongly regular graphs with parameters $(4{\theta}^2-1,2{\theta}^2, {\theta}^2,{\theta}^2)$; their eigenvalues satisfy ${\theta}+ \tau = 0$. Chan shows in this case that $(x,y)$ is one of $$\left\{\left(-1, \frac{2{\theta}^2-3 \pm i\sqrt{4{\theta}^2-5}}{2({\theta}^2-1)}\right),
        \left(\frac{-2{\theta}^2+1 \pm i\sqrt{4{\theta}^2-1}}{2{\theta}^2},1\right)\right\}.$$

[**Case**]{} $(x,y) = (\tfrac{-2{\theta}^2+1 \pm i\sqrt{4{\theta}^2-1}}{2{\theta}^2},1)$: The characterizing equations reduce to $$\begin{aligned}
ce^{i2{\theta}^2t} \sqrt{n} & = \pm i\sqrt{n} \label{eqn:flatsrg1.1} \\
ce^{i{\theta}t} \sqrt{n} & = {\theta}(x-1) \label{eqn:flatsrg1.2}  \\
ce^{-i{\theta}t} \sqrt{n} & = -{\theta}(x-1) \label{eqn:flatsrg1.3} .\end{aligned}$$ Equations and imply that $e^{2i{\theta}t} = -1$ and so $e^{i{\theta}t} = \pm i$ and $e^{i2{\theta}^2t} = (-1)^{\theta}$. Then Equation implies that $c = \pm i$. But then the LHS of equation is real while the RHS is not, a contradiction. So there are no solutions.

The characterizing equations for the complement of $X$ are: $$\begin{aligned}
ce^{i(2{\theta}^2-2)t} \sqrt{n} & = \pm i\sqrt{n} \label{eqn:flatsrgc1.1} \\
ce^{i(-{\theta}-1)t} \sqrt{n} & = {\theta}(x-1) \label{eqn:flatsrgc1.2}  \\
ce^{i({\theta}-1)t} \sqrt{n} & = -{\theta}(x-1) \label{eqn:flatsrgc1.3} .\end{aligned}$$ Equations and imply that $e^{i2{\theta}t} = -1$ and so $e^{i2{\theta}^2t} 