TITLE: Bound on the order of a finite group generated by elements $a$ and $b$ of order 2 and $n \geq 3$ such that the sum of the images of $a$, $b$ and $b^{-1}$ under any ordinary representation has only rational eigenvalues
QUESTION [5 upvotes]: Assume that $G = \langle a, b \rangle$ is a finite non-abelian group which is generated by an
involution $a$ and an element $b$ of order $n$ ($n\geq 3$) such that for every (complex) representation
$\varphi$ of $G$ the matrix $\varphi(a) + \varphi(b) + \varphi(b^{-1})$ has only rational eigenvalues.

Question: Is there an upper bound on the order of $G$?

REPLY [2 votes]: If you are looking for a more abstract proof based on representation and group theory, I suggest you to work on a more general question as follows:
Assume that $G=\langle S\rangle$  is a finite  group which is generated by $S=S^{-1}$ ($1\not\in S$) such that for every complex representation $\phi$  of G  the matrix $\sum_{s\in S} \phi(s)$  has only rational eigenvalues. Then is $|G|$  bounded above by a function of $|S|$?
The answer to the above question is positive and a crude known bound is
$\frac{|S|(|S| − 1)^{2|S|} − 2}{|S| − 2}$.
Note that, what you need by the representation theoretic assumption is that the  eigenvalues of the linear transformation $T=\sum_{s\in S}s$ on the vector space $\mathbb{C}(G)$ are all rational (and so integer). Note that $T$ is an element of the group ring $\mathbb{C}(G)$.
You may find some related graph theoretic results in the following paper:
Alireza Abdollahi and E. Vatandoost, Which Cayley graphs are integral?, The Electronic Journal of Combinatorics 16 (2009), #R122.
Sorry for the self-promotion!