This is related to question Matrix-valued periodic Fibonacci polynomials. I want to find integer-valued matrices $x$ such that the Fibonacci polynomials $f_n(x)$, defined by the recursion $f_n(x)=xf_{n-1}(x)-f_{n-2}(x)$ with initial values $f_0(x)=0$ and $f_1(x)=1$ are periodic.
Special cases seem to be the matrices $x = G_k^{(m)} = \left( {{g_m}(i,j)} \right)_{i,j = 0}^{k - 1}$ with $g_m(i,j)=1$ if $|i-j|=m$ or $i+j=m-1$ and $g_m(i,j)=0$ else.
As observed by Lev Soukhanov it suffices to show that all eigenvalues are of the form $\pm2\cos(\frac{2\pi p}{q}).$
This is true for $m=1:$ The characteristic polynomial $a_k(x)$ of $ G_k^{(1)}$ satisfies
$a_k(x)=-x a_{k-1}(x)- a_{k-2}(x)$ with initial values $ a_{0}(x)=1$ and $ a_{1}(x)=1-x.$
This implies that $a_k(x+1/x)(-x)^k=\frac{1+x^{2k+1}}{1+x}.$ Therefore the eigenvalues are $-2\cos(\frac{2\pi j}{2k+1})$ for $1\le{j}\le{k}.$
Let now $b_k(x)=L_k(x)$ denote the Lucas polynomials which satisfy the recursion $b_k(x)=xb_{k-1}(x)-b_{k-2}(x)$ with initial values $b_0(x)=2$ and $b_1(x)=x$ and $c_k(x)=F_k(x)$ the Fibonacci polynomials which satisfy the same recursion with initial values $c_0(x)=1$ and $c_1(x)=x.$
These satisfy $x^n F_n(x+1/x)=\frac{1-x^{2n+2}}{1-x^2}$ and $x^n L_n(x+1/x)=1+x^{2n}.$
For arbitrary $m$ periodicity would follow if the following conjectured formula for the characteristic polynomials $c_{ G_k^{(m)}}(x)$ is true.
Conjecture: $$c_{ G_{2mk+j}^{(2m)}}(x)=\pm (a_k(x)a_k(-x))^{m-j}(b_{k+1}(x)c_k(x))^j$$, $$c_{ G_{2mk+m+j}^{(2m)}}(x)=\pm (a_{k+1}(x)a_{k+1}(-x))^{j}(b_{k+1}(x)c_k(x))^{m-j}$$ for $0\le{j}\le{m}$ and $$c_{ G_{(2m+1)k+j}^{(2m+1)}}(x)=\pm a_k(x) (a_k(x)a_k(-x))^{m-j}(b_{k+1}(x)c_k(x))^j$$ for $0\le{j}\le{m}$ and $$c_{ G_{(2m+1)k+m+j}^{(2m+1)}}(x)=\pm a_{k+1}(x) (a_{k+1}(x)a_{k+1}(-x))^{j-1}(b_{k+1}(x)c_k(x))^{m+1-j}$$ for $1\le{j}\le{m}.$
Edit: It should be noted that $ a_k(x)a_k(-x)=(-1)^k F_{2k}(x)$ and $ b_{k+1}(x)c_k(x) =F_{2k+1}(x).$
New Edit:
A simpler formulation of the conjecture: $$c_{ G_{2mk+j}^{(2m)}}(x)=(-1)^j F_{2k}(x)^{m-j}F_{2k+1}(x)^j, $$ $$c_{ G_{2mk-j}^{(2m)}}(x)=(-1)^j F_{2k}(x)^{m-j}F_{2k-1}(x)^j ,$$ $$c_{ G_{(2m+1)k+j}^{(2m+1)}}(x)=(-1)^j a_{k}(x) F_{2k}(x)^{m-j}F_{2k+1}(x)^j, $$ $$c_{ G_{(2m+1)k-j}^{(2m+1)}}(x)=(-1)^j a_{k}(x) F_{2k}(x)^{m-j}F_{2k-1}(x)^j $$ for $0\le{ j} \le{m}.$
Is there a simple way to prove this conjecture?
