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.
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