Problems & Puzzles: Puzzles

Puzzle 512. b^c=a(mod d)

JM Bergot poses the following nice puzzle:

Let a,b,c,d  be four consecutive primes,a<b<c<d such that b^c=a(mod d)

 Here are two solutions:

5^7=3(mod 11)
11^13=7(mod 17).

Are these rare solutions?



Wow!!!... contributions came from Farid Lian, Farideh Firoozbakht, Seiji Tomita, Jacques Tramu, Torbjörn Alm, Jim Howell, Fred Schneider, Jan van Delden, Luca Poletti & J. K. Andersen.

All of them found only two more solutions:

61^67=59 (mod 71)
77617^77621=77611 (mod 77641)

Fred & Andersen argued that there must be infinite solutions.




Records   |  Conjectures  |  Problems  |  Puzzles