= Solution
The <eigenvalue> equation $D_rf=\lambda f$ is
$$
f(x+r)=(1+\lambda)f(x).
$$
If $\lambda=-1$, this forces $f(x+r)=0$ for every $x$, hence $f=0$, so $-1$ is not an <eigenvalue>. If $c=1+\lambda\ne0$, choose arbitrary values on representatives of the cosets of $r\mathbb Z$ and extend by
$$
f(t+kr)=c^kf(t),
\qquad k\in\mathbb Z.
$$
This produces nonzero eigenfunctions. Choosing <functions> supported on distinct cosets gives infinitely many linearly independent eigenfunctions. Therefore
$$
\boxed{\operatorname{spec}_{\rm point}(D_r)=\mathbb R\setminus\{-1\}},
$$
and every eigenspace is infinite-dimensional.
Direct expansion gives
$$
D_rD_sf(x)=f(x+r+s)-f(x+r)-f(x+s)+f(x),
$$
which is symmetric in $r,s$. Hence
$$
\boxed{D_rD_s=D_sD_r}.
$$
Suppose a degree-$n$ <polynomial> $p$ were a sum of $n$ periodic <functions>,
$$
p=f_1+\cdots+f_n,
\qquad D_{r_i}f_i=0,
\quad r_i\ne0.
$$
Apply the commuting product $D_{r_1}\cdots D_{r_n}$. Every term on the right is killed by its corresponding factor, whereas for leading coefficient $a_n$ the <mixed finite difference of a polynomial> gives
$$
D_{r_1}\cdots D_{r_n}p
=n!a_n\prod_{i=1}^nr_i\ne0.
$$
This contradiction proves
$$
\boxed{p\text{ cannot be a sum of }n\text{ periodic functions}}.
$$
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