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Let us consider $\mathbb{Z}_N=\{0,1,\cdots N-1\}$.

Does there exist any permutation $\sigma$ on $\mathbb{Z}_N$ satisfying :

$$\exists p,q\in\mathbb{N}, ~~p,q\neq0 (mod~N) ~~\forall i,j\in \mathbb{Z}_N$$ the following formula holds $$q\sigma(i)(1+2\sigma(j))\equiv_{N} qi(1+2j)+2p$$

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  • $\begingroup$ Well, there is $q=p=0.$ There is an $i$ with $\sigma(i)=0$ Then for this $i$ and all $j$ one has $qi(1+2j)+2p=0$ $\endgroup$
    – Aaron Meyerowitz
    Commented Apr 15, 2022 at 7:22
  • $\begingroup$ Thanks for your comment. It needs a minor amendment. $\endgroup$
    – ABB
    Commented Apr 15, 2022 at 7:35
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    $\begingroup$ Take even $N$, $p=N/2$, $q=1$ and the identity permutation as $\sigma$. $\endgroup$
    – Max Alekseyev
    Commented Apr 15, 2022 at 13:27

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We must have $4p \equiv_N 0$ meaning $N=2p$ or $N=4p$ (since $p=0$ is forbidden): Let $i_0$ have $\sigma(i_0)=0.$ Then you require that for all $j,$ $$0 \equiv_N qi_0(1+2j)+2p$$ If $i_0=0$ then $0 \equiv_N 2p.$ There are still the other values of $i$ to consider but @MaxAlekseyev has given a solution.

Otherwise, putting $j=0$ and $1$, we have that $qi_0+2p\equiv 3qi_0+2p \equiv 0$ Hence $2qi_0 \equiv 0$ and also $4p \equiv 0.$

So another solution (after checking) is $N=4p=2q$ and $\sigma$ the identity permutation. In fact, $\sigma$ can be any permutation which preserves parity.

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