Problems & Puzzles: Problems

Problem 14.- N and N^2-1 powerful numbers

A powerful number N is such that if a prime p divides N, then p^i also divides to N, for some i=>2.

The smallest powerful numbers are 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, …

It’s conjectured (By Erdös?) that there are only finite many powerful numbers N such that N^2-1 is also powerful.

But it happens that I haven’t been successful to find any…. Can you find the five smaller powerful couples of numbers (N, N^2-1)?

(Ref. 2, B16, p. 71) Polly T. Wang wrote (23/11/2002):

Since x^2-1 is a powerful number, it can be written in the form y^n. Thus x^2-y^n=1. (1) Except the solution x=n=3(x is not powerful), it is proven that (2,n) must be a pair of Double Wieferich primes, so n must be 1093. It is checked that there isn't any pair of (x,y) satisfying (1), so there isn't any pair (x,x^2-1) which are both powerful numbers... See http://www.sciencenews.org/20001223/mathtrek.asp

The Wang's note is pertinent because the problem has been reduced to the Catalan's Conjecture and this Conjecture has been linked to the pairs of Double Wieferich primes, but I have read the article pointed out by Wang and the Preda's work (hunting these pairs) hasn't been finished yet, not yet...

***

T. D. Noe and Jean Brette observed something wrong in the argument of Wang that I didn't see:

T. D. Noe wrote:

'Polly Wang says "Since x^2-1 is a powerful number, it can be written in the form y^n." This is not true. For example, the number 72 is powerful, but not a power of some number'.

Jean Brette wrote:

1) I don't understand the first sentence of this contribution :

Since x^2-1 is a powerful number, it can be written in the form y^n. I don't see why it can be. A powerful number is not always a power of some y.

2) since x is a powerful number, so is Z = x^2, and Z et Z-1 are consecutive. Do you know if there is some solution without the condition : Z is a square?

In other words: Does it exist two consecutive powerful numbers (other than 8 and 9)?

3) If Polly T. Wang was right, one don't need the full Catalan conjecture. A recent article on this conjecture give a reference (but I have not read it): Ko Chao : On the Diophantine equation x^2 = y^n+1. Sc. Sinica 14 (1965) p. 457-460 where the author shows that there is no solution , except x=3, y=2, n=3

***

One additional interesting comment over this issue by T. D. Noe, is the following one (13/12/02):

In the paper "On Powerful Numbers," Mollin and Walsh show that there are an infinite number of pairs of consecutive powerful numbers. The smaller number of each pair is 8, 288, 675, 9800, 12167, 235224... (See sequence A060355 in the On-Line Encyclopedia of Integer Sequences.) They actually prove a much stronger result: for every integer n, there are an infinite number of pairs of powerful numbers whose difference is n.

*** Records   |  Conjectures  |  Problems  |  Puzzles  Home | Melancholia | Problems & Puzzles | References | News | Personal Page | Puzzlers | Search | Bulletin Board | Chat | Random Link Copyright © 1999-2012 primepuzzles.net. All rights reserved.