Problems & Puzzles: Puzzles

Puzzle 133.  Pa - Qb = K

Here we deal with solutions to the above equation for any K=>1 value being:

a) P & Q prime numbers ( Q may be greater equal or less than P)
b) and a & b greater than 1 (not necessarily primes)?

Some solutions are:

3 ^ 2 - 2 ^ 3 = 9 - 8 = 1
3 ^ 3 - 5 ^ 2 = 27 - 25 = 2
2 ^ 7 - 5 ^ 3 = 128 - 125 = 3
5 ^ 3 - 11 ^ 2 = 125 - 121 = 4
2 ^ 5 - 3 ^ 3 = 32 - 27 = 5

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
b) Are the cases with solution finite or infinite?

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 so-called 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)?

______
(*) Note: as a matter of fact for these K values in bold-red I have not found solutions even if P & Q are released of 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):

"... x^3 = y ^2 + 2 : is the solution x=3, y= 5 unique? YES. (Leonard Euler)

The best way is to use complex numbers of the form a + i sqrt(2) , where a and b are integers.

x^3 = (y+ i sqrt(2)) (y-i sqrt(2))

y+ i sqrt(2) and y-i sqrt(2) have no common factors , so each must be a cube and

y + i sqrt(2) = (A + i B sqrt(2)) ^3 = (A^3 - 6 A.B^2) + i sqrt(2) (3 A^2 . B - 2 B^3)

Identify : the coefficient of i sqrt(2) must be equal to 1 , and y = (A^3 - 6 A.B^2)

1) (3 A^2 . B- 2 B^3) = B ( 3 A^2- B^2) = 1, implies B = 1 and ( 3 A^2- B^2) = 1, i.e. A = ±1

2) B = 1 and A = ±1 give y = ± 5 and x = 3

He also sent the following reference:

"you can read on this problem in R. K. Guy, Unsolved problems in number theory, 2nd edition, problem D9, p. 155-157"

***

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):

"I noticed that you could find a triplet of prime powers: (3^4, 5^3, 13^2) = (81, 125, 169), and 169-125 = 125-81 = 44. Are there any other triplets?"

***

Carlos Rivera found two triplets as the asked by Paul but releasing the prime condition:

7^2 - 5^2 = 5=2 -1^2 = 24
8^2 - 6^2 = 6^2 - 2^3 = 28

Maintaining the prime condition here are 3 more triplets solutions:

7^3 - 2^8 = 2^8 - 13^2 = 87
19^2 - 3^5 = 3^5 - 5^3 = 118
17^2 - 13^2 = 13^2 - 7^2 = 120

***

Jean Brette & Jean-Charles Meyrignac solved the question about the solubility for k odds & releasing the prime condition for P & Q:

"For the odd number 2n+1 we have 2n+1 = (n+1)^2 - n^2 (I write : (n+1) square minus n square)" [Brette]

"Let's compute: S = (a+b)^2 - a^2 = b^2 + 2*a*b. If b=1, S = 1+2*a, thus all ODD values can be reached. Ex: 3=1+2*1 -> a = 1, and 2^2-1^2 = 3. If b=2, S = 4+4*a, thus all multiples of 4 can be reached" [Meyrignac]

***

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.
2) if k is prime then there is only one solution.

***

Jean Charles Mayrignac found (12/4/019 a solution missed by Luis Rodríguez: 74=99^3-985^2

***

 Records   |  Conjectures  |  Problems  |  Puzzles