Problems & Puzzles:
Problems
Problem 53.
Powerful numbers revisited
I will suppose that you know exactly
what a powerful
number is. So, I will pose directly two of the most popular unsolved
problems
Perhaps you know that all the known
pairs of consecutive powerful numbers (*) are such that one of the member of
the couple is a perfect square, with only one exception already known, and
pointed out by Golomb, 1970: 12167=23^3 & 12168=2^3*3^2*13^2
Q1. Can you find another pair like
(12167 & 12168)?
Perhaps you know that while it has been proved that there
are infinite cases of consecutive powerful numbers, it has been conjectured
by several mathematicians (Erdös, Golomb) that there can not be three consecutive powerful
numbers.
Q2. How far this search has been made
(*)? Would you like to try to find
the first triplet
of consecutive powerful numbers, or show that they are impossible?
____
(*) See Sloane
A060355
Contribution came from J. Wroblewski.
***
Jaroslaw wrote:
Regarding Q1, there are infinitely many
solutions.
Here is how I got an infinite family (I believe one can easily
construct more families like that).
Consider the equation
3^3*c^2 + 1 = 7^3*d^2
It has infinitely many solutions, hence there are infinitely
many pairs of consecutive powerful numbers, neither of which
is a square.
c[0] = 376766
d[0] = 1342879
3^3*d[0]^2 = 48689748233307
7^3*c[0]^2 = 48689748233308
c[i] = 97379496466615*c[i1] + 27321362062956*d[i1]
d[i] = 347082488429404*c[i1] + 97379496466615*d[i1]
c[1] = 73378566731480957414
d[1] = 261537761671184312049
c[2] = 14291135759507045763965540575090454
d[2] = 50936831077090977385276030140145391
c[3] = 2783327208393663274732558923371899994022551429006
d[3] = 9920405923784291913924768096855946930622570930881
c[4] = 542078004110408249263550623360891312259489327239068251468178926
d[4] = 1932088267205077938842953341158497093365330223626511071637930239
More details:
I took 3 and 7, primes congruent mod 4 and such that
7 is quadratic residue mod 3,
3 is quadratic residue mod 7.
I have found
14^221*3^2=7
and
55^221*12^2=1
where 21=3*7.
Then for any n I get
(14+3*Sqrt[21])*(55+12*Sqrt[21])^n=a+b*Sqrt[21]
with
a^221*b^2=7.
Then I selected solutions with a divisible by 49 and
b divisible by 3 to get
343*(a/49)^227*(b/3)^2=1
I think one could extend the above family
by taking:
(143*Sqrt[21])*(55+12*Sqrt[21])^n=a+b*Sqrt[21]
I believe that replacing 3 and 7 with other appropriate primes
should lead to other families of solutions.
Regarding Q2, I think there is no solution at
all, but I see no prospect of proving
that. The reason for such an opinion is the close connection of the
problem to the abc conjecture:
http://www.math.unicaen.fr/~nitaj/abc.html
The ErdosMollinWalsh conjecture on consecutive powerful numbers.
Therefore I consider the problem hopelessly hard and see no point
in putting any effort to it.
No point searching for a solution, I strongly believe does not exist.
No point trying for a proof, which is probably beyond reach for anyone
on this world.
***
Later (Dec. 08) he added:
I have done an exhaustive search up to 500^6
for pairs of consecutive powerful numbers, neither being a square.
There are 5 such pairs:
12167 = 23^3
12168 = 2^3 * 3^2 * 13^2
5425069447 = 7^3 * 41^2 * 97^2
5425069448 = 2^3 * 26041^2
11968683934831 = 13^2 * 79^3 * 379^2
11968683934832 = 2^4 * 23^3 * 7841^2
28821995554247 = 7^2 * 17^2 * 23^3 * 409^2
28821995554248 = 2^3 * 3^4 * 13^2 * 16223^2
48689748233307 = 3^3 * 139^2 * 9661^2
48689748233308 = 2^2 * 7^3 * 13^2 * 43^2 * 337^2
15625000000000000  search limit
***
