Problems & Puzzles: Puzzles
Puzzle 186. Squareful Fibonacci numbers
A number is said to be "Squareful" if it contains at least one square in its prime factorization. A worse name for these numbers is the awkward "nonsquarefree".
Anyway, the first of these numbers are 4, 8, 9, 12, 16, 18, etc.
Regarding the Fibonacci numbers Fn (1, 1, 2, 3, 5, 8, 13, 21, 34, ... for n = 1, 2, ...) the first Squareful are for n = 6, 12, 18, 24, 25, 30, 36, 42, 48,... (See A037917 ).
It has been noticed that no one Fibonacci Squareful number has been found such that the index n is prime.
The observation seems to be interesting because no reason seems to exists for such notorious absence.
Would you like to find the first Squareful Fibonacci number whose index n is prime, or to demonstrate that such numbers are impossible?
Jim Fougeron wrote (14/7/2002):
There will not be any Fibonacci numbers with prime n's which are Squareful.
I looked at a quick factoring all Fibonacci's n>=1000. What I found was that for all Squareful numbers, the square was always made up of some factorization
of the n value. Thus F(25) has it's square in the 5's
f(25) trivially factors as: 5^2*3001 (25 is 5^2)
f(153) has factors: 2*17^2*1597 (153 is 3^2*17)
f(252) has factors: 2^4*3^3*13*17*... (252 is 2^2*3^2*7)
This "property" appears to hold for all Fibonacci's less than 1000, and I
see no reason why it would not continue to hold true.
If the above property is true, then for a f(p) to be Squareful, then the Fibonacci number would have to have a square of p^2 in it. That being the case, I quickly tested f($a) % $a within PFGW using a simply ABC2 file type:
a: primes from 3 to 100000
I was looking for a zero value, but did not run this "test" very long. It
quickly became apparent that f(p)%p is always 1 or p-1. This is why all
Fibonacci's with prime n values are squarefree (i.e. not Squareful).
This is by no means a proof. This is simply observations made in about a 10 minute time frame.