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Fix $k \geq 2$ and let $[n]$ denote the set $\{0, 1, \ldots, n-1\}$. A polynomial $p(x) = \sum_{i=0}^d a_i x^i$ with integer coefficients in $[2^k]$ is a permutation polynomial modulo $2^k$ if $p(x) \mod 2^k$ permutes the elements of $[2^k]$. It is known (due to Rivest) that $p(x)$ is a permutation polynomial modulo $2^k$ if and only if $a_1$ is odd and $\Delta_1$ and $\Delta_2$ are even, where $\Delta_1 = a_2 + a_4 + \ldots$ and $\Delta_2 = a_3 + a_5 + \ldots$.

I am interested in the opposite question. In particular, given a permutation $\pi: [2^k] \to [2^k]$, under what conditions does there exist a permutation polynomial $p$ modulo $2^k$ such that $p$ and $\pi$ produce the same permutation? I suspect that there are permutations that have no such polynomial, but I haven't been able to find or construct a class of examples.

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A class of permutations that cannot be constructed is where $\pi(0)$ and $\pi(1)$ are both even based on Rivest's criterion.

First note that $x \bmod 2^k$ is even iff $x$ is even.

If $\pi(0)$ is even we find that $a_0$ must also be even as $p(0) = a_0$.

If $\pi(1)$ is also even we find that $\sum_i a_i$ must be even as $p(1) = \sum_i a_i$.

Rivest's criterion tells us that $\Delta_1$ and $\Delta_2$ are even, thus $\sum_{i\geq2}a_i$ is even.

But this leads to an impossibility. $a_0$ and $\sum_{i\geq2}a_i$ are even, but $a_1$ must be odd due to Rivest's criterion, thus $p(1) = \sum_i a_i$ can't be even.

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  • $\begingroup$ This is clever! I didn't think of that. I am hoping that there are criteria that more characterizing, so the speak. $\endgroup$ – user340082710 Aug 10 at 2:29
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This answer is partial.

Fix any natural $k$. Let $S_k$ be the group of all permutations of the set $[2^k]$. A permutation $\pi\in S_k$ is representable, if there exists a polynomial $p\in \Bbb Z[x]$ which represents $\pi$, that is $p(x)\equiv \pi(x)\pmod {2^k}$ for each $x\in [2^k]$. Let $G_k$ be a set of all representable permutations of $S_k$. Let $\pi,\sigma\in G_k$ be permutations represented by polynomials $p,q\in \Bbb Z[x]$, respectively. Then $p(q(x))\equiv \pi(\sigma(x))\pmod {2^k}$ for each $x\in [2^k]$, thus $\pi\sigma\in G_k$. Therefore $G_k$ is a semigroup of a finite group $S_k$, so it is a group.

Put $G^0_k=\{\pi\in G_k:\pi(0)=0\}$. Let $\pi_1\in G_k$ be a permutation of $[2^k]$ such that $\pi_1(x)\equiv x+1\pmod {2^k}$ for each $x\in [2^k]$ and $C$ be a cyclic group of order $2^k$ generated by the map $\pi$. Since for each permutation $\pi\in G_k$ we have $\pi_1^{-\pi(0)}\pi(0)=0$, we have $\pi_1^{-\pi(0)}\pi\in G^0_k$, $G_k=CG^0_k$, so it suffices to describe the group $G^0_k$.

For each natural $l\le k$ put $X_l=\{0,2^l, 2\cdot 2^l,\dots, 2^k-2^l\}\subset [2^k]$. Let $\pi\in G^0_k$ be any permutation. It is easy to check that $\pi(X_l)\subset X_l$ for each $l$. Since $\pi$ is a bijection, we have $\pi(X_l)=X_l$ for each $l$.

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