My problem! Here is the original problem statement and solution.

:crab: NT IS DEAD Let $S$ be the set of all functions $f: \mathbb{Z}/a\mathbb{Z}\to \mathbb{Z}/p^e\mathbb{Z}$. The problem asks us to determine the number of elements in cycles in the directed graph $G$ on $S$ defined by $f\to \Delta f$. Let $T$ be the set of all $f\in S$ such that $\Delta^k f\equiv 0$ for some $k$. Note that $S$ and $T$ are both additive groups, so we can quotient $S$ by $T$ to obtain $\frac{p^{ea}}{|T|}$ cosets. Claim: Each coset contains exactly one function in a cycle. The proof consists of two steps. Step one: Each coset contains at least one function in a cycle. Let $f$ be an arbitrary function in $S$ with coset $f + T$. Since every vertex in $G$ has outdegree one, $f$ eventually leads into a cycle $C$. Let $g_1 = \Delta^k f$ be an arbitrary function in the cycle, and let $g_2$ be another function in the cycle so that $g_1 = \Delta^k g_2$. Then $\Delta^k (f - g_2) \equiv 0$, so $g_2\in f + T$. Hence, every coset contains at least one function in a cycle. Step two: Each coset contains at most one function in a cycle. Suppose $f_1$ and $f_2$ are in the same coset and are both in cycles. Since $f_1$ and $f_2$ differ by an element of $T$, $\Delta^k f_1 = \Delta^k f_2$ for large enough $k$. In particular, $f_1$ and $f_2$ are in the same cycle. But then if $n$ is the length of the cycle, we have (for sufficiently large k) $f_1 = \Delta^{kn} f_1 = \Delta^{kn} f_2 = f_2$, as desired. The claim is proved. We now only need to find $|T|$. Claim: For any nonnegative integer $i$ and function $f: \mathbb{Z}/p^i\mathbb{Z}\to \mathbb{Z}/p^e\mathbb{Z}$, we have $\Delta^k f\equiv 0$ for large enough $k$. We will begin by proving the $i=1$ case. For each $j$, let $g_j:\mathbb{Z}/p^i\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ be the function defined by $g_j(j) = 1$ and $g_j(x) = 0$ everywhere else. By induction, for each $1\le k\le p^j - 1$, $\Delta^k g_j$ is a cyclic shift of the function $g(x) = (-1)^x\binom{k}{x}$. Then by applying $\Delta$ to $\Delta^{p^i-1}g_j$ we get that the outputs of $\Delta^{p^i}g_j$ are $-\binom{p^i}{1}$, $\binom{p^i}{2}$, $\dots$, $(-1)^{p^i-1}\binom{p^i}{p^i-1}$, $1 - (-1)^{p^i-1}$. But these are all multiples of $p$ (even if $p=2$!), so $\Delta^{p^i}g_j\equiv 0$ for each $j$. But every function $f: \mathbb{Z}/p^i\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ is a linear combination $$f(x) = \sum_{d=0}^{p^i-1}f(d)g_d(x)$$ and $\Delta$ is additive, so we also have $\Delta^{p^i} f\equiv 0$. To finish, we induct on $e$, with the base case of $e=1$ being what we just proved. For the inductive step, for any $e$ and $f:\mathbb{Z}/p^i\mathbb{Z}\to \mathbb{Z}/p^e\mathbb{Z}$, we know that $p$ divides every output of $\Delta^{p^i}f$. But then if $f' = \frac{1}{p}\Delta^{p^i}f$, by the inductive hypothesis $p^{e-1}$ divides every output of $\Delta^{k}f'$ for some $k$. Hence, $p^e$ divides every output of $$p\Delta^k f' = \Delta^{k+p^i}f$$ and the claim is proved. Finally, one last claim. Claim: write $a = p^ib$ where $p\nmid b$. Then $T$ is exactly the set of functions $f$ for which $f(x+dp^i) = f(x)$ for all $x,d$. Once more, the proof consists of two steps. Step one: all such functions are in $T$. Note that all such functions are really just a function $f':\mathbb{Z}/p^i\mathbb{Z}\to \mathbb{Z}/p^e\mathbb{Z}$ repeated $b$ times, so we're done by the previous claim. Step two: all functions in $T$ are of the described form. Suppose otherwise, and let $f\in T$ be a function not of the described form. Let $k$ be maximal such that $g = \Delta^k f$ is not of the described form, so that $g(x+dp^i)-g(x)=c\neq 0$ for some $x,d$. Since $k$ is maximal, we have $g(x+dp^i+1)-g(x+dp^i) = g(x+1)-g(x)$, $g(x+dp^i+2)-g(x+dp^i+1)=f(x+2)-f(x+1)$, etc, and this eventually gives $g(x+2dp^i)-g(x+dp^i)=g(x+dp^i)-g(x) = c$ and $g(x+(t+1)dp^i) - g(x+tdp^i) = c$ for each $t$. Summing, we get $g(x) = g(x+bp^i) = g(x) + bc$. But this means $p^e | bc$, implying $p^e | c$ and $c = 0$ in $\mathbb{Z}/p^e\mathbb{Z}$, a contradiction. The claim is proved. By the claim we have $|T| = p^{ep^i}$, so the number of functions in cycles is $\frac{p^{ea}}{|T|} = \boxed{p^{e(a-p^i)}}$.

NT is dead. The answer is $p^{e(a-p^{\nu_p(a)})}$. First observe that we can explicitly write $$\Delta^k f(n)=\sum_{i=0}^k \binom{n}{i} f(n+k-i)(-1)^i.$$Now, let $$S=\{f \mid \exists k\geq 1\text{ s.t. } \Delta^k f \equiv 0\}.$$We first characterize $S$ entirely. Claim: $S$ is precisely the set of functions $f$ such that $f(n)=f(n+p^{\nu_p(a)})$ for all $n$. Proof: First suppose that $f(n) \not \equiv f(n+p^{\nu_p(a)})$ but $f \in S$. Then there exists some $N$ such that $\Delta^k f \equiv 0$ for all $k \geq N$. Then, picking $m$ massive, we have by Kummer's theorem (or Legendre) that $p \mid \tbinom{p^m}{i}$ for all $0<i<p^m$. Therefore, $$\Delta^k f(n)\equiv f(n+p^m)-f(n) \pmod{p},$$but this must also be $0$, hence $f(n+p^m) \equiv f(n) \pmod{p}$. By repeating this fact $c$ times so that $ap^m \equiv p^{\nu_p(a)} \pmod{a}$, it follows that $f(n+p^{\nu_p(a)})\equiv f(n) \pmod{p}$. Now we consider $\Delta^k$ modulo $p^2$, so now $\tbinom{p^m}{i}$ vanishes unless $\nu_p(i) \geq m-1$. By picking $m$ large such that $m-1\gg \nu_p(a)$, we get the sum to be $$f(n+p^m)-f(n)+\sum_{i=1}^{p-1} \binom{p^m}{ip^{m-1}}f(n+p^m-ip^{m-1})(-1)^{i} \pmod{p^2}.$$We can pair up the terms involving $\tbinom{p^m}{ip^{m-1}}$ with $\tbinom{p^m}{(p-i)p^{m-1}}$ and make them vanish, since $p$ will divide both of these "coefficients" and the values of $f$ multiplying them are also congruent modulo $p$ (note: if $p=2$ we get an issue since the pairing isn't possible anymore, but this can be resolved by considering the fact that now $p^2$ will divide the central coefficient, so this stronger bound essentially lets us do the same thing). Thus the entire sum vanishes, so $f(n+p^m) \equiv f(n) \pmod{p^2} \implies f(n+p^{\nu_p(a)}) \equiv f(n) \pmod{p^2}$. We can then repeat this argument with $i$ such that $\nu_p(i) \geq m-2$, and then $m-3$, and so on until $m-e$ (which can be made large by picking $m$ large), so that $f(n+p^{\nu_p(a)}) \equiv f(n) \pmod{p^e}$: contradiction. On the other hand, if we do have $f(n)=f(n+p^{\nu_p(a)})$ for all $n$, then we have $\Delta^k f(n) =\Delta^k f(n+p^{\nu_p(a)})$ for all $n$. Additionally, we similarly get $$\Delta^{p^{\nu_p(a)}} f(n) \equiv f(n+p^m)-f(n) \equiv 0 \pmod{p},$$so if we define $f' \equiv \tfrac{\Delta^{p^{\nu_p(a)}} f}{p}$ from $\mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/p^{e-1}\mathbb{Z}$, which is legal since $p \mid \Delta^{p^{\nu_p(a)}} f(n)$ for all $n$, and $f'(n)=f'(n+p^{\nu_p(a)})$ as well, so we apply the same reasoning and induct down to finish (the base case of $e=0$ is vacuous). $\blacksquare$ The direct corollary of this claim is that $|S|=p^{ep^{\nu_p(a)}}$, since we have $p^e$ choices for the first $p^{\nu_p(a)}$ function outputs and the rest are uniquely determined. Now, separate the functions from $\mathbb{Z}/a\mathbb{Z}$ to $\mathbb{Z}/p^e\mathbb{Z}$ into $p^{ea-ep^{\nu_p(a)}}$ "equivalence families" such that $f$ and $g$ are in the same family iff $f-g \in S$. Obviously, there can be at most one function in each family that is sent to itself after some number of finite difference iterations. Indeed, if there exist $f \neq g$ both in the same family such that there exist $k_1,k_2 \geq 1$ with $\Delta^{k_1} f = f$ and $\Delta^{k_2} g = g$. Then pick $N$ large, so that $\Delta^N (f-g) \equiv 0$. Then $\Delta^{Nk_1k_2} f = f$ and $\Delta^{Nk_1k_2} g = g$, but $\Delta^{Nk_1k_2} (f-g) \equiv 0$: contradiction. Thus there exist at most $p^{e(a-p^{\nu_p(a)})}$ working functions. On the other hand, we can generate a unique function which works for each family: first, for any function $f$, consider the sequence $\Delta^0 f, \Delta^1 f, \ldots$. Since there are exactly $p^{ea}$ functions in total from $\mathbb{Z}/a\mathbb{Z}$ to $\mathbb{Z}/p^e\mathbb{Z}$, by Pigeonhole some function among $\Delta^0 f,\ldots,\Delta^{p^{ea}} f$ is repeated, hence by $\Delta^{p^{ea}} f$ the sequence is periodic (possibly before that as well), i.e. there exists some $k \geq 1$ such that $\Delta^{k} (\Delta^{p^{ea}} f) = \Delta^{p^{ea}} f$. Now, the idea is to pick a function $f$ in some family and look at $\Delta^{p^{ea}} f$. Specifically, if $f-g \not \in S$ but we have $\Delta^{p^{ea}} f = \Delta^{p^{ea}} g$, then $\Delta^{p^{ea}} (f-g) \equiv 0$: contradiction. Therefore, each family must produce at least one of these functions that is not shared by any other family, so there are at least $p^{e(a-p^{\nu_p(a)})}$ working functions. Therefore, there are exactly $p^{e(a-p^{\nu_p(a)})}$ working functions, so we are done. $\blacksquare$ Remark: The last part of the solution feels somewhat unsatisfying. In fact, the $\Delta^{p^{ea}}$ of any two functions in the same family is the same, since by that point the difference goes to zero: another corollary of the claim is that if $f$ goes to zero, it does so in at most $ep^{\nu_p(a)} \leq ea$ finite differences (simply note that we take $p^{\nu_p(a)}$ finite differences $e$ times in our induction), which is certainly less than $p^{ea}$. Therefore there actually exists a bijection between a family $\mathcal{F}$ and the function that results after $p^{ea}$ finite differences.

Source tells me this is the easiest problem on the TSTST!? We solve this one: The_Turtle wrote: For every pair of integers $a, b \in \mathbb{Z}^+$ and function $f: \mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/b\mathbb{Z}$, let $\Delta f \colon \mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/b\mathbb{Z}$ denote the function $n \mapsto f(n+1)-f(n)$. In terms of $a$ and $b$, determine the number of functions $f \colon \mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/b\mathbb{Z}$ for which $\Delta^N f = f$ for some $N \in \mathbb{Z}^+$. Write $a = p_1^{e_1} \dots p_n^{e_n}$ and $b = p_1^{f_1} \dots p_n^{f_n}$ for primes $p_1,\dots,p_n$ such that $\operatorname{rad}(ab) = p_1 \dots p_n$ and nonnegative integers $e_1,\dots,e_n$ and $f_1,\dots,f_n$. To begin, it's clear that $\Delta$ is a linear operator. Let $\mathcal F$ denote the set of functions $\mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/b\mathbb{Z}$. Call a function nilpotent if $\Delta^N f$ is identically zero for some $N$. Let $\mathcal N$ denote the set of functions in $\mathcal F$ that are nilpotent, and let $\mathcal P$ denote the set of functions in $\mathcal F$ such that $\Delta^N f = f$ for some $N$. We want to find $|\mathcal P|$. Note that $\mathcal F$, $\mathcal N$, and $\mathcal P$ are abelian groups under pointwise addition, since if $\Delta^N f = 0$ and $\Delta^M g = 0$, then $\Delta^{N + M} (f + g) = 0$, and if $\Delta^N f = f$ and $\Delta^M g = g$, then $\Delta^{NM} (f + g) = f + g$. Now by pigeonhole, $(\Delta^n f)_{n \ge 0}$ is always eventually periodic, so we can find some $n$ such that $\Delta^n f = \Delta^{2n} f$ (see e.g. Floyd cycle detection). Then, $\Delta^n (f - \Delta^n f) = 0$, so $f = (f - \Delta^n f) + \Delta^n f$, where $f - \Delta^n f \in \mathcal N$ and $\Delta^n f \in \mathcal P$. Finally, clearly $\mathcal N \cap \mathcal P = \{ 0 \}$, so $\mathcal F = \mathcal N \oplus \mathcal P$. It suffices to find $|\mathcal N|$. To do this, first identify \[ \mathbb{Z}/a\mathbb{Z} \cong \bigoplus_{i=1}^n \mathbb{Z}/p_i^{e_i}\mathbb{Z} \quad \text{and} \quad \mathbb{Z}/b\mathbb{Z} \cong \bigoplus_{i=1}^n \mathbb{Z}/p_i^{f_i}\mathbb{Z},\]and write $f$ in the obvious way, sending $n$ tuples to $n$ tuples. The key claim is as follows: Claim: Suppose $f$ is nilpotent. Then, there exist functions $g_1,\dots,g_n$ with $g_i \colon \mathbb{Z}/p_i^{e_i}\mathbb{Z} \to \mathbb{Z}/p_i^{f_i}\mathbb{Z}$ such that $f(a_1,\dots,a_n) = (g_1(a_1), \dots, g_n(a_n))$ for all $a_i \in p_i^{e_i}\mathbb{Z}$ for all $1 \le i \le n$. Proof: This follows from backwards induction on the sequence $(\Delta^n f)_{n \ge 0}$, where the base case is the $N$ such that $\Delta^N f = 0$. In particular, $\Delta^{n-1}f(t) = \Delta^n f(t-1)+\cdots+\Delta^n f(1)+\Delta^n f(0)+\Delta^{n-1}f(0)$, each of which have the projections of their output onto $\mathbb{Z}/p_i^{f_i}\mathbb{Z}$ determined by their input's projections onto $\mathbb{Z}/p_i^{e_i}\mathbb{Z}$ for each $i$. $\square$ Finally, if we WLOG identify $1 \in \mathbb{Z}/a\mathbb{Z}$ with $(1, \dots, 1) \in \textstyle\bigoplus_{i=1}^n \mathbb{Z}/p_i^{e_i}\mathbb{Z}$, then \[ \Delta f (a_1,\dots,a_n) = (g_1(a_1+1)-g_1(a_1), \dots, g_n(a_n+1)-g_n(a_n)) = (\Delta g_1 (a_1), \dots, \Delta g_n (a_n)),\]so $f$ being nilpotent implies that each of the $g_i$ are as well. Now, we claim that every $g \colon \mathbb{Z}/p^{e}\mathbb{Z} \to \mathbb{Z}/p^{f}\mathbb{Z}$ is nilpotent, or equivalently $\Delta$ is a nilpotent linear endomorphism on the free $\mathbb{Z}/p^{f}\mathbb{Z}$-module of such functions. But it's not hard to check that $\Delta$ has characteristic polynomial $(x + 1)^{p^e} - 1$ because $\Delta + \operatorname{id}$ is simply the shift operator with characteristic polynomial $x^{p^e}$. Now, it is well known that a matrix is nilpotent if and only if the coefficients of its characteristic polynomial (except the leading $1$) are nilpotent in the base ring. But every nonleading coefficient of $(x + 1)^{p^e} - 1$ is divisible by $p$ by Lucas's theorem, so we are done. Now to answer extract, there are $(p^f)^{p^e}$ functions $g \colon \mathbb{Z}/p^{e}\mathbb{Z} \to \mathbb{Z}/p^{f}\mathbb{Z}$, so \[ |\mathcal P| = \frac{|\mathcal F|}{|\mathcal N|} = b^a \left(\prod_{i=1}^n (p_i^{f_i})^{p_i^{e_i}} \right)^{-1}. \]We are done. $\blacksquare$ Remark: The motivation for looking at nilpotent functions is essentially orthogonal complements but on an $R$-module. Indeed, if we define the ``inner product'' $\langle \cdot, \cdot \rangle$ on $\mathcal F$ given by \[ \langle f, g \rangle = \sum_{x \in \mathbb{Z}/a\mathbb{Z}} f(x) g(x),\]then $\mathcal N$ is very nearly (or maybe exactly) the ``orthogonal complement'' $\mathcal P^\perp = \{ n \in \mathcal F \mid \langle p, n \rangle = 0, \forall p \in \mathcal P \}$. Indeed, $\mathcal N \subseteq \mathcal P^\perp$. This comes from the fact that $\Delta$'s adjoint $\Delta'$ is the operator mapping $f$ to $n \mapsto f(n-1) - f(n)$, which satisfies $(\Delta')^n f = 0 \iff \Delta^n f = 0$. Thus, if $n \in \mathcal N$ and $(\Delta')^k n = 0$, then for all $p \in \mathcal P$, \[ 0 = \langle p, (\Delta')^k n \rangle = \langle \Delta^k p, n \rangle.\]But $\Delta^k p$ traverses every element in $\mathcal P$, as desired. This proof may or may not be reversible because this inner product is super fake.

Here is a solution requiring minimal theory. Lemma 1: Let $V$ be a finite set and $F:V\to V$ be a function. Suppose $S\subseteq V$ satisfies both of the following: For all $v\in V$, there exists a nonnegative integer $j$ such that $F^j(v)\in S$. $F(S)=S$ Then $S$ is the set of elements, $s$, of $V$ such that there exists a positive integer $k$ with $F^k(s)=s$.

Main Claim: Set $V$ to be the set of all functions $\mathbb Z/a\mathbb Z\to\mathbb Z/p^e\mathbb Z$ and $F$ to be the function $\Delta^1$. Let $b=p^{v_p(a)}$, $c=\frac ab$, and $S$ be the set of functions $f\in V$ such that for all $m\in\mathbb Z/a\mathbb Z$, \[\sum_{i=0}^{c-1}f(m+bi)=0.\]Then $S$ satisfies both conditions of Claim 1.

Thus, $S$ is the set of desired functions, and the answer is $|S|=p^{e(a-p^{v_p(a)})}$.

A solution sketch using commutative algebra: given any finite abelian group $M$ with endomorphism $T \colon M \to M$, the number of elements in $m \in M$ such that $T^k(m) = m$ for some $k$ is exactly the size of the localization $M_t$, where we interpret $M$ as a $\mathbb{Z}[t]$ module in which multiplication by $t$ acts by $T$. To see how this helps us solve the problem, let $M_{a, e}$ be the abelian group of functions $\mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/p^e\mathbb{Z}$. Then, since localization is exact, we can use the exact sequences \[0 \to M_{a,e} \to M_{a,e+e'} \to M_{a,e'} \to 0\]to get that the answer is the answer for $e=1$ raised to the power $e$. To finish the $e=1$ case, we have to compute the size of \[\mathbb{F}_p[\Delta, \Delta^{-1}]/((\Delta + 1)^a - 1).\]However, in general, for any field $k$ and $P(x) \in k[x, x^{-1}]$, the $k$-dimension of the space $k[x, x^{-1}]/(P(x))$ is just the difference of the maximum and minimum degrees of the nonzero terms in $P$. In this case, the maximum and minimum degrees are $a$ and $p^{v_p(a)}$, respectively, yielding a final answer of $p^{e(a - p^{v_p(a)})}$.

Answer: $p^{e(a - p^{\nu_p(a)})}$. I think it suffices to consider ring theory over the ring $R[X]/\langle X^a - 1 \rangle$, where $R = \mathbb{Z}/p^e \mathbb{Z}$. Interpret a function $f : \mathbb{Z}/a\mathbb{Z} \to R$ as the polynomial $\displaystyle \sum_{j = 0}^{a - 1} f(j) X^j$. Modulo $X^a - 1$, finite difference just corresponds to multiplication by $X - 1$. Thus, we are just counting the number of elements $f \in R[X]/\langle X^a - 1\rangle$ such that $(X - 1)^c f = f$ for some $c \geq 1$. Let $a = p^m u$, where $m = p^{\nu_p(a)}$ and $u$ is a positive integer not divisible by $p$. Then we can factorize $$ X^a - 1 = (X^{p^m} - 1) \cdot h(X), \quad \text{ where } \quad h(X) = \sum_{j = 0}^{u - 1} X^{p^m j}. $$The two factors are co-prime since over $R = \mathbb{Z}/p^e \mathbb{Z}$, clearly $h(X) \equiv u \pmod{X^{p^m - 1}}$ and $p \nmid u$. In particular, by Chinese Remainder Theorem, $$ R[X]/\langle X^a - 1 \rangle = R[X]/\langle X^{p^m} - 1 \rangle \times R[X]/\langle h(X) \rangle. $$We now work in the latter ring. The main claim is that for any $(f, g) \in R[X]/\langle X^{p^m} - 1 \rangle \times R[X]/\langle h(X) \rangle$, $(X - 1)^c (f, g) = (f, g)$ for some $c \geq 1$ if and only if $f = 0$. If the claim is true, then the answer is just the cardinality of $R[X]/\langle h(X) \rangle$; since $h$ is monic, this cardinality is just $$ |R|^{\deg(h)} = {p^e}^{p^m(u - 1)} = p^{e (p^m u - p^m)} = p^{e(a - p^{\nu_p(a)})}. $$ It now remains to prove the claim. First, suppose that $(X - 1)^c (f, g) = (f, g)$ for some $c \geq 1$, where $f \in R[X]/\langle X^{p^m - 1} \rangle$ and $g \in R[X]/\langle h(X) \rangle$. Clearly, this means $(X - 1)^c f = f$. On the other hand... $(X - 1)^{p^m} = X^{p^m} - 1 + p g(X)$ for some polynomial $g \in R[X]$. So, over $R[X]/\langle X^{p^m - 1} \rangle$ (since $R = \mathbb{Z}/p^e \mathbb{Z}$), we have $(X - 1)^{p^m e} = 0$. (I think so, but all that's important is that $(X - 1)^{p^m c} = 0$ for $c$ large enough.) This suffices to prove that $f = 0$. For the converse, it suffices to prove that there exists $m \geq 1$ such that $(X - 1)^m = 1$ in $R[X]/\langle h(X) \rangle$. In fact, since $R[X]/\langle h(X) \rangle$ is a finite commutative ring, it suffices to show that $X - 1$. But $X - 1 \mid X^{p^m - 1}$, and we have seen that $X^{p^m - 1}$ and $h(X)$ are co-prime. So certainly, $X - 1$ and $h(X)$ are co-prime. Thus $X - 1$ must be a unit in $R[X]/\langle h(X) \rangle$.

Replace $f$ with an $a$ dimensional vector over the integers mod $p^e$. Let $\Delta$ be the linear map that operates on $a$-dimensional vectors and maps them to their common difference. The answer is $(p^e)^d$, where $d$ is the number of eigenvalues (with multiplicty) which are not divisible by $p$. The characteristic polynomial of $\Delta$ can be easily computed to be $(1-\lambda)^a-1=0$. However, by Vieta's, the lowest degree term in this polynomial not divisible by $p$ will be the term where we are summing over all products of $d$ roots, which has degree $a-d$. However, we also know by expanding the characteristic polynomial and binomial theorem that the lowest degree term not divisible by $p$ has degree $p^{v_p(a)}$. This implies that $d=a-p^{v_p(a)}$, so our answer is $p^{e(a-p^{v_p(a)})}$

\[ \mathbb{Z}/a\mathbb{Z} \cong \bigoplus_{i=1}^n \mathbb{Z}/p_i^{e_i}\mathbb{Z} \quad \text{and} \quad \mathbb{Z}/b\mathbb{Z} \cong \bigoplus_{i=1}^n \mathbb{Z}/p_i^{f_i}\mathbb{Z},\]What does $\bigoplus$ mean??

Any answer??

Why do people reply NT is dead in this section? Is there a reference I don’t get?

Sorry but the relation $\Delta^kf=f$ Is true for all $k$???

Joider wrote: \[ \mathbb{Z}/a\mathbb{Z} \cong \bigoplus_{i=1}^n \mathbb{Z}/p_i^{e_i}\mathbb{Z} \quad \text{and} \quad \mathbb{Z}/b\mathbb{Z} \cong \bigoplus_{i=1}^n \mathbb{Z}/p_i^{f_i}\mathbb{Z},\]What does $\bigoplus$ mean?? This is the direct sum of $\mathbb{Z}$-modules. The congruences follow by the Chinese remainder theorem. By the way, if you already know the latex is \bigoplus, you can type ``\bigoplus" in google.

Oky thank you

Doing the case of $e=1$ really helps a lot to this problem. Note that all of the $f$ forms a linear space $V$, we only need to calculate its dimension. Let $a=p^{t}b$ where $p\nmid b.$ Define $g(x):=\sum_{k=0}^{b-1}f(x+kp^t).$ We prove that $f$ is good iff $g\equiv 0.$ This gives $\dim V=\dim W-p^t=a-p^t.$ Thus the total number is $p^{e(a-p^t)}.$ On one hand if $g\equiv 0$ but for all $k\ge 1,$ $\Delta^k f\neq f.$ Then there exists $k>j\ge 1$ such that $\Delta^j f=\Delta ^k f.$ This gives $\Delta^{k-1}f=\Delta^{j-1}f+C.$ However sice $g\equiv 0,\Delta^k g\equiv 0.$ We have $p^e\mid C\cdot b$ which leads to contradiction since $C=0.$ On the other hand note that $$\Delta^{p^t}f(m)=\sum_{j=0}^{p^t}(-1)^{p^t-j}\binom {p^t}jf(m+j)\equiv f(m+p^t)-f(m)\quad[p].$$$$\Delta^{p^t}g(m)=\sum_{k=0}^{b-1}\Delta^{p^t}f(m+kp^t)\equiv 0\quad [p].$$By induction we may prove that $\Delta^{ep^t}g\equiv 0[p^e].$ but now if $\Delta^k f=f,$ $$g=g(\Delta^{kep^t}f)=\Delta^{kep^t}g=0,$$done.$\Box$