It follows that $X$ has exactly three distinct eigenvalues.

Note that the Cayley graphs having three distinct eigenvalues are strongly regular graphs. However, it is not easy to characterize the strongly regular orbit Cayley graphs by immediately calculating their spectra by applying Theorem \[thm-z-1\]. In the rest of this paper, we will characterize some strongly regular orbit Cayley graphs by analyzing their structures. We need some combination identities.

Combination identities
======================

In this part, we shall give some combination identities, which will be used to prove our main results. Throughout this paper, we always set ${s\choose t}=0$ if $t<0$ or $t>s$.

\[lem-z-5\] For a positive integer $n$, we have $$\sum_{i=0}^n{n\choose 2i}=\sum_{i=0}^n{n\choose 2i+1}=2^{n-1}.$$

It is clear that $$\left\{\begin{array}{l}
\displaystyle (1+1)^n=\sum_{i=0}^n{n\choose i}=\sum_{i=0}^n{n\choose 2i}+\sum_{i=0}^n{n\choose 2i+1}\\
\displaystyle (1-1)^n=\sum_{i=0}^n{n\choose i}(-1)^i=\sum_{i=0}^n{n\choose 2i}-\sum_{i=0}^n{n\choose 2i+1}
\end{array}\right.$$ By summation the two equations, we get $2\sum_{i=0}^n{n\choose 2i}=2^n$ and thus $\sum_{i=0}^n{n\choose 2i}=2^{n-1}$. Subtracting the second equation from the first one, we have $2\sum_{i=0}^n{n\choose 2i+1}=2^n$ and thus $\sum_{i=0}^n{n\choose 2i+1}=2^{n-1}$.

\[lem-z-6\] For a positive integer $m$, we have $$\left\{\begin{array}{l}
\displaystyle \sum_{j=0}^{m}{4m\choose 4j}=2^{4m-2}+(-1)^m2^{2m-1}\\
\displaystyle \sum_{j=0}^{m}{4m\choose 4j+2}=2^{4m-2}-(-1)^m2^{2m-1}\\
\displaystyle \sum_{j=0}^{m}{4m\choose 4j+1}=\sum_{j=0}^{m}{4m\choose 4j+3}=2^{4m-2}\\
\end{array}\right.$$

Let $i$ be the imaginary unit. Note that $1+i=\sqrt{2}e^{\pi i/4}$. We have $$\begin{aligned}
&&\displaystyle 2^{2m}\cos{m\pi}+(2^{2m}\sin{m\pi}) i=2^{2m}e^{m\pi i}=(\sqrt{2}e^{\pi i/4})^{4m}=(1+i)^{4m}\\
&=&\displaystyle \sum_{j=0}^m{4m\choose 4j}+\sum_{j=0}^m{4m\choose 4j+2}i^2+\sum_{j=0}^m{4m\choose 4j+1}i+\sum_{j=0}^m{4m\choose 