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Prime numbers p such that there exists an integer k such that p-1 does not divide k-1 and x -> x + x^k is a bijection from Z/pZ to Z/pZ.
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%I #32 Jul 17 2019 06:55:41

%S 31,43,109,127,157,223,229,277,283,307,397,433,439,457,499,601,643,

%T 691,727,733,739,811,919,997,1021,1051,1069,1093,1327,1399,1423,1459,

%U 1471,1579,1597,1627,1657,1699,1723,1753,1777,1789,1801,1831,1933,1999,2017

%N Prime numbers p such that there exists an integer k such that p-1 does not divide k-1 and x -> x + x^k is a bijection from Z/pZ to Z/pZ.

%C If x -> x + x^k is a bijection from Z/pZ to Z/pZ then the following facts hold:

%C -v_2(k-1) >= v_2(p-1)

%C -gcd(k+1,p-1) = 2

%C -2^(k-1) = 1 (mod p).

%C The third fact is very important as it shows that for a given k there are a finite number of solutions p.

%C If p = 1 (mod 3) and 2^((p-1)/3) = 1 then either k = (p-1)/3+1 or k = 2*(p-1)/3+1 has the wanted property (see sequence A014752 for more information when this happens). It is a sufficient but not necessary condition since 3251 also appears in this sequence but 3 does not divide 3250.

%H Elias Caeiro, <a href="/A306787/b306787.txt">Table of n, a(n) for n = 1...212</a>

%H Problèmes du 9ème Tournoi Français des Jeunes Mathématiciennes et Mathématiciens, <a href="https://tfjm.org/wp-content/uploads/2019/01/Problemes-TFJM2019.pdf">Problem 7 question 7</a>, 2019 (in French).

%e For p = 31 and k = 21, x -> x + x^k is a bijection.

%Y Cf. A014752.

%K nonn

%O 1,1

%A _Elias Caeiro_, Apr 16 2019