of $C_1(\tau)$, $C_2(\tau)$ and $K(\tau)$ are equal to $\pm1/\sqrt{n}$. Since $U(\tau)$ is unitary, $$I=U(\tau)U(\tau)^* ={\begin{pmatrix} C_1^2 +KK^T& i(C_1K-KC_2)\\ i(K^TC_1-C_2K^T)& K^TK+C_2^2 \end{pmatrix}}$$ and thus all entries of $\sqrt{n}U(\tau)$ are fourth roots of unity. If $D$ is diagonal of order $n\times n$ and $D_{u,u}$ is 1 or $i$ according as $u$ is in the first or second colour class of $X$, then $$H = \sqrt{n} DU(\tau)D$$ is a real Hadamard matrix. In particular, we have $$H = \sqrt{n}{\begin{pmatrix} C_1& -K\\ -K^T&-C_2 \end{pmatrix}}$$ It is well known that if $n$ is a real Hadamard matrices of order $n$ such that $n > 2$, then $n$ is divisible by four.[  ]{}

Adamczak et al. [@adamczak07] show that if uniform mixing occurs on $C_{2m}$, then $m$ must be a sum of two squares. We generalize this result to all bipartite graphs.

If $X$ is a regular, bipartite graph with $n$ vertices, then $n$ is the sum of two integer squares.

[*Proof.* ]{}Suppose $X$ is regular (and bipartite). Then $$\sqrt{n}U(\tau){\textbf{1}}=(a+ib){\textbf{1}}$$ for some integers $a$ and $b$ and then $$n{\textbf{1}}= \sqrt{n}{{\mkern2mu\overline{\mkern-2muU(\tau)}}}\, \sqrt{n}U(\tau){\textbf{1}}= (a-ib)(a+ib){\textbf{1}}$$ and this shows that $n$ is the sum of two integer squares.[  ]{}

Cycles
======

It was conjectured by Ahmadi et al. [@ahmadi03] that 