Problems & Puzzles:
Problems
Problem 14. N and N^21 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^21 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^21)?
(Ref. 2, B16, p. 71)
Polly T. Wang wrote
(23/11/2002):
Since x^21 is a powerful number, it can be written
in the form y^n. Thus x^2y^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^21) 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^21 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^21 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 Z1 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. 457460
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 OnLine
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.
***
