Problems & Puzzles: Collection 20th

Coll.20th-020. Another puzzle on consecutive primes.

On Jun 17, 2018 Zak Seidov wrote:

Ten consecutive primes:

Mod[60962211883,60962211917,60962211929, 60962211941,60962211953,
60962212009,60962212021,60962212033, 60962212067,60962212123]11=
{1,2,3,4,5,6,7,8,9,10}

Later he added a second example:

Mod[{92728531127, 92728531139, 92728531151, 92728531229, 
92728531307, 92728531319, 92728531331, 92728531409, 
92728531421, 92728531433}]11 ={1,2,3,4,5,6,7,8,9,10}

Q. Find other sets of p-1 consecutive primes that are congruent 1, 2, ... p-1 mod p, for p>11.


 


 

Jan van Delden wrote on May 1, 2020

The given example suggests two different problems.

A) The residues q[k] modulo p are in a set, comprising the values {1,2,3,…,p-1}

B) The residues q[k] modulo p are part of this set and are consecutive q[k] mod p = k.

Needless to say that condition B) is much stricter than condition A). Fixing the order of appearance of the residues seriously decreases the probability of finding a solution.

 A few results

 A)

 A list of values [p,q[1]], smallest q[1]: [[2,3],[3,5],[5,11],[7,11],[11,41],[13,10267],[17,832361],[19,1565437],[23,52112257],[29,355021996291],[31,15203906741]]

 B)

 A list of values of [p,q[1]], smallest q[1]:

 [2,3],[3,7],[5,251],[7,61223],[11,23700022897]]

For p=13 I was not able to find a solution, actually I gave up. I searched until 142*10^12. The first partial solution where the residueset has size 11 instead of 12:

82076064730973 82076064730987 82076064731027 82076064731041 82076064731081 82076064731147 82076064731161 82076064731201 82076064731293 82076064731333 82076064731347

The routine is quite fast, but I would not expect a solution of size 12 to appear soon.

 

***


Records   |  Conjectures  |  Problems  |  Puzzles