Existence of a prime with some additional properties

Question

Does there exist a prime $p$ such that $p-1$ is squarefree, divisible by at least three primes, and $$ \{1^{\sigma(1)},\ldots,(2p)^{\sigma(2p)}\}=\{1,\ldots,2p\} $$ in $\mathbf{Z}/(2p)\mathbf{Z}$ for some permutation $\sigma$ of $\{1,\ldots,2p\}$?

I am asking only about numerical evidences: I conjecture the answer is negative, but I would be happy with a (counter)example as well.. For instance, does $p=31$ work?

[The question comes from a characterization of special type of primes which have a number of applications, e.g., in cryptography and primality testing. The motivation for this question is related to this thread and this problem.]

Answer 1

$p = 31$

octave/matlab format:

sigma = [30   48   41   12   50    5   14    4   62   58   19   17   49   46   57   54   29   56   52   26   11    7   27   47   20   35   51   34   39   15   60   42   24    1   18   40   25   32   36   22  38   53   43   23   44   33    8   31   16    2   28   13   59   21   37   10   55    3   61    9   45    6];

sigma = [ 30   48   41   12   50    5   14    4   62   58 ...
          19   17   49   46   57   54   29   56   52   26 ...
          11    7   27   47   20   35   51   34   39   15 ...
          60   42   24    1   18   40   25   32   36   22 ...
          38   53   43   23   44   33    8   31   16    2 ...
          28   13   59   21   37   10   55    3   61    9 ...
          45    6];

Octave was used to generate Boolean equations in CNF format.

zChaff was used to solve Boolean selection variables.

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More details:

A power residue matrix was created $R(r,c) = r^c \pmod {2p}$ for $r,c = 1 \dots 2p$

A Boolean decision matrix can be imagined where $D(r,c) = \delta_{r,c}$ where $\delta_{r,c} \in \mathbb{B}$ Boolean so that $\sigma = [1:2p]D^T$.

The conditions on $\delta_{r,c}$ are that exactly one is selected from each row and each column.

For each residue $0 \dots (2p-1)$ only one of the $\delta_{r,c}$ corresponding to the positions of that residue in $R$ will be true.

i.e. create Boolean variables to select only one of each residue from $R$.

Answer 2

This is not an answer, just an idea. Let o(x) denote the multiplicative order of x in Z*/(31)and let s be sigma. Note that if o(i^s(i)) = 30, then o(i) =30. Consider i: 55 53 11 17 43 21 13 3 and i^s(i): 3 55 53 11 17 43 21 13 where i+1 is the j for which j^s(j)=i The rows are just permutations of the phi(30)=8 members of Z*/(31) with o(x)=30. To find a permutation for n=2(43) maybe you can first determine what works for the 12=phi(42) members of Z*/(43) with order 42. I'm GUESSING 2(43) will be nice but 2(71) will be naughty. Please let me know if that guess turns out to be right.