Define:A set $\mathcal{C}(t)$, a positive integer $n$ is in the $\mathcal{C}(t)$ if $x^t \pmod{n}$ describes a bijection from the set $\{0,1,...,n-1\}$ to itself.
Example table: \begin{array}{|c|c|} \hline & \text{Set} \\ \hline \mathcal{C}(2) & \{1, 2\} \\ \mathcal{C}(3) & \{1, 2, 3, 5, 6, 10, 11, 15, 17, 22, 23, 29, 30, 33, 34, 41, 46, 47, 51, 53, 55, 58, 59, 66, 69, 71, 82, 83, 85, 87, 89, 94\} \\ \mathcal{C}(4) & \{1, 2\} \\ \mathcal{C}(5) & \{1, 2, 3, 5, 6, 7, 10, 13, 14, 15, 17, 19, 21, 23, 26, 29, 30, 34, 35, 37, 38, 39, 42, 43, 46, 47, 51, 53, 57, 58, 59, 65, 67, 69, 70, 73, 74, 78, 79, 83, 85, 86, 87, 89, 91, 94, 95, 97\} \\ \mathcal{C}(6) & \{1, 2\} \\ \mathcal{C}(7) & \{1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 46, 47, 51, 53, 55, 57, 59, 61, 62, 65, 66, 67, 69, 70, 73, 74, 77, 78, 79, 82, 83, 85, 89, 91, 93, 94, 95, 97\} \\ \mathcal{C}(8) & \{1, 2\} \\ \mathcal{C}(9) & \{1, 2, 3, 5, 6, 10, 11, 15, 17, 22, 23, 29, 30, 33, 34, 41, 46, 47, 51, 53, 55, 58, 59, 66, 69, 71, 82, 83, 85, 87, 89, 94\} \\ \mathcal{C}(10) & \{1, 2\} \\ \mathcal{C}(11) & \{1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 47, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 91, 93, 94, 95, 97\} \\ \mathcal{C}(12) & \{1, 2\} \\ \mathcal{C}(13) & \{1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 51, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97\} \\ \mathcal{C}(14) & \{1, 2\} \\ \mathcal{C}(15) & \{1, 2, 3, 5, 6, 10, 15, 17, 23, 29, 30, 34, 46, 47, 51, 53, 58, 59, 69, 83, 85, 87, 89, 94\} \\ \mathcal{C}(16) & \{1, 2\} \\ \mathcal{C}(17) & \{1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97\} \\ \mathcal{C}(18) & \{1, 2\} \\ \hline \end{array}
and set $\mathcal{A}_b(t)=\{0,\pm 1^t, \ldots, \pm(b-1)^t\}$
We want to express integer $n$ in terms of base $b$ as $$n=b^t_kb^k+b^t_{k-1}b^{k-1} \cdots+ b^t_0\overset{say}{=}[b_k,b_{k-1},\cdots, b_0]^t_b$$ where $b_i\in \{0,\pm 1,\pm 2,\ldots,\pm(b-1)\}$ or $b^t_i\in\mathcal{A}_b(t)$
Example : $2= (1)^3\times10+(-2)^3 = [1,-2]^3_{10}$
Theorem 1 : Foe every $n\in \mathbb{Z}$ can be express in base $b$ with digits have power $t$ if and only if $b\in\mathcal{C}(t)$ (or There exist $b^t_i\in\mathcal{A}_b(t)$ to express $b=\sum_{i\ge 0} b^t_ib^i$ iff $b\in\mathcal{C}(t)$).
Proof to Theorem 1: Proof by strong induction:
Base case: 1 can be written in base $b$ as 1.
Assume that $P(n)$ is true i.e. for all $m$ such that $ 0 \leq m \leq n$, we can represent $m$ in base $b$ and digits have power $t$.
Now consider an integer $n+1$. We need to prove that we can represent $n+1$ in base $b$. We can write $n+1$ as $bm+i^t$ for some integer $m$ and fix power $t$ where $m < n$ and $0\le i\le b-1$, this will possible if $i^t \pmod{b}$ describes a bijection from the set $\{0,1,...,b-1\}$ to itself. By strong induction, we know $m$ has a base $b$ representation $[b_m,b_{m-1}, \cdots ,b_1,b_0]_b^t$ and so $bm$ has representation $[b_m,b_{m-1}, \cdots, b_1,b_0,0]_b^t$ and we can add $i^t$ to this depending on whether $n+1 = bm+i^t = [b_m,b_{m-1}, \cdots, b_1,b_0,i]_b^t$.
Thus if we can represent all integers less than $n+1$, we can also represent $n+1$.
How to prove following statement
Question: $\mathcal{C}(t)$ have following properties $$\text{If $t$ is prime, then set}\quad \mathcal{C}(t)=\left\{ N; \quad N=\prod_{\substack{p \text{ is prime}\\ 2t\ \nmid\ p-1}}p ,\text{ and $N$ is squarfree}\right\}$$
$$\text{If $t$ is composite, say $t=q_1\times q_2\times\ldots\times q_s$ for some $s$ then set} \quad\mathcal{C}(t)=\mathcal{C}(q_1)\cap\mathcal{C}(q_2)\cap\ldots\cap\mathcal{C}(q_s)$$ Where $q_i$ are distinct prime factor of $t$.
here are some more illustration for some satisfying $t$ (power) and $b$ (base), in the following table,
| t=1 | t = 3 | t= 5 | t = 7 | t = 9 | t = 11 | |
|---|---|---|---|---|---|---|
| b=2 | 2=[1, 0] | 2=[1, 0] | 2 =[1, 0] | 2=[1, 0] | 2=[1, 0] | 2=[1, 0] |
| b=3 | 2=[2] | 2=[1, -1] | 2=[1, -1] | 2=[1, -1] | 2=[1, -1] | 2=[1, -1] |
| b=5 | 2=[2] | 2=[-1, 0, 3] | 2=[-1, -1, 2] | 2=[1, 0, 1, -2] | 2=[-1, 0, -1, 0, 2, 1, -2, 3, 2] | 2=[-1, 0, 0, 1, 2, -1, -2, -3, 2, 3, 3, 0, -2] |
| b=6 | 2=[2] | 2=[-1, 2] | 2=[-1, 1, 2] | 2=[-1, -1, 2, 0, 1, -4, -4, 4, 4, 3, 2] | 2=[1, 0, -1, -2, 0, 2, 3, 4, -1, -4] | 2=[-1, 1, 2, 2, 2, 3, -4] |
| b=10 | 2=[2] | 2=[1, -2] | 2=[2, 2, -3, -8] | 2=[-1, 0, 2, 0, -1, -2, -6, 1, -5, 8] | 2=[-1, 1, -1, 0, -2, 4, -5, -5, 6, 9, 2] | 2=[-1, -2, -3, 1, 0, -2, 7, -2, -4, 2, 5, -2] |
The above claim came from my research work if you have any confusion or suggestion please share, thanks.
