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[i-1] + 27321362062956*d[i-1]
d[i] = 347082488429404*c[i-1] + 97379496466615*d[i-1]

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
14^2-21*3^2=7
and
55^2-21*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^2-21*b^2=7.

Then I selected solutions with a divisible by 49 and
b divisible by 3 to get
343*(a/49)^2-27*(b/3)^2=1

I think one could extend the above family
by taking:
(14-3*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 Erdos-Mollin-Walsh 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

***



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