Functional Equation Season

Let us define the following operators $T$ (shift operator) and $I$ (identity operator) by the following formulas: $T f(x)=f(x+1)$ and $If(x)=f(x).$ Take into consideration that these two operators commute. That is going important for the formulas that we are going to introduce below. Then for any natural number $n$ the we can define $T^n$ and the composition of $T$ with itself $n$ times. Clearly, $T^nf(x)=f(x+n).$ We can also define operators that are going to be linear combinations of "powers" of $T$ and $I.$ For example, the functional equation $f(x+2)-2\cos{\frac{\pi}{2018}}f(x+1)+f(x)=0$ can be written in the form $(T^2-2\cos{\frac{\pi}{2018}}T+I)f(x),$ where the corresponding operator will be $(T^2-2\cos{\frac{\pi}{2018}}T+I).$ In general, such operators can be defined by the formula $P(T)=a_n T^n+a_{n-1}T^{n-1}+...+a_1T+a_0I,$ where $n$ is any whole number and $a_0, a_1,...,a_n$ are real numbers. We can "multiply" operators of this kind by following the rule of polynomial multiplication. It can be proved that if $P(T)$ and $Q(T)$ are two of this "polynomials" and $R(T)=P(T)Q(T),$ then $R(T)f(x)=P(T)(Q(T)f(x)).$ Now, since $(T^2-2\cos{\frac{\pi}{2018}}T+I)=(T-e^{i\pi/2018}I)(T-e^{-i\pi/2018}I)$ and $e^{i\pi/2018}$ and $e^{-i\pi/2018}$ are 4036th roots of the unity. Then, $(T^2-2\cos{\frac{\pi}{2018}}T+I)$ is a factor of $T^{4036}-I.$ Therefore, since $(T^2-2\cos{\frac{\pi}{2018}}T+I)f(x)=0,$ if we "multiply" both sides of this equation by a convenient "polynomial", we get that $(T^{4036}-I)f(x)=0.$ The latter means that $f(x+4036)=f(x).$ Therefore, $f(x)$ is periodic with period less than or equal to 4036.