Problems & Puzzles: Puzzles Puzzle 133.
P^{a}
 Q^{b} = K Here we deal with solutions to the above equation for any K=>1 value being:
Some solutions are: 3
^ 2  2 ^ 3 = 9  8 = 1 But  in a slight search  I haven't found solutions for the following K values below 50: 6, 14, 20, 25, 26, 27, 29, 31, 34, 35, 36, 42, 43 & 50 (*) Questions: a) Can you find
solutions for these unsolved K values As for sure you know it's conjectured that 3 ^ 2  2 ^ 3 = 9  8 = 1 is the only solution for K=1 (this is exactly the socalled Catalan Conjecture). c) But what about
the solution for K=2: 3 ^ 3  5 ^ 2 = 27  25 = 2, is this unique (**)
also, or can you find another solution (releasing P & Q from the prime
condition)? ______ (**) Other form of expressing this question goes like this: is 26 the only number 'squeezed' between two powers? Solution Luis Rodríguez extended the search for 50< K <=510 and found no solution  even releasing the prime condition  for the following K values: 58, 62, 66, 70, 78, 82, 86, 90, 102, 110, 114, 130, 134,,158,,178,,182, 202, 206, 210, 226, 230, 238, 246, 254, 258, 266, 274, 278, 290, 302, 306, 310, 314, 322, 330, 358, 374, 378, 390, 394, 398, 402, 410, 418, 422, 426, 430, 434, 438, 442, 446, 450, 454, 458, 462, 466, 470, 474, 478, 482, 494 y 510. You should notice the all these K values without solutions are even values. Does anybody knows why this happens? *** Jean Brette sent (10/4/01) the following for out question c):
He also sent the following reference:
*** Thanks to Brette we know that  at least  for K=2 and cubic & square powers there is one and only one solution. Is this true for any other pair of powers? *** Paul Jobling sent this follow up (10/4/01):
*** Carlos Rivera found two triplets as the asked by Paul but releasing the prime condition: 7^2  5^2 = 5=2 1^2 = 24 Maintaining the prime condition here are 3 more triplets solutions: 7^3  2^8 = 2^8  13^2 = 87 Can you find a quadruplet? *** Jean Brette & JeanCharles Meyrignac solved the question about the solubility for k odds & releasing the prime condition for P & Q:
*** Luis Rodríguez notices the following facts about the equation of this puzzle: 1) If no solution if found for a k value then: a) k mod 4 <>0, b) k
is not a perfect square. *** Jean Charles Mayrignac found (12/4/019 a solution missed by Luis Rodríguez: 74=99^3985^2 ***





