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?

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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.

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