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Fibonacci Primes

  1. Aug 8, 2011 #1
    I have read that we do not know if there are an infinite number of Fibonacci primes. So far, no one has produced a proof to show if they are infinite. I wanted to know why this seems to be so challenging? I'm sure it is, and maybe there is a subtle mathematical principle I am missing.

    I thought of using Cantor's set theory. I essentially would compare the set of Fibonacci primes with the natural numbers and since both have cardinality aleph null, that should imply that the Fibonacci primes are infinite. Right?

    If anyone has any insight into this open problem, please let me know. I would be very interested in your input.
  2. jcsd
  3. Aug 8, 2011 #2
    That is exactly what one would like to prove isn't it? But how do we?
  4. Aug 8, 2011 #3
    Maybe a step to prove or disprove might be to prove my conjecture: Fib(2n) = A(n)*Fib(n);A(3n) = B(n)*C(n); C(5n) = D(n)*C(n);D(7n)=E(n)*D(n);E(11n)=F(n)*E(n)... This continues undefinitly and with each increase in an alpha the constant numerical multiplyer in the index is the next prime in the sequence of primes. All values set forth are integers.
    Last edited: Aug 8, 2011
  5. Aug 8, 2011 #4
    Dedekind proposed that an infinite set can be set up with a one-to-one correspondence with a proper subset of itself. I think it would be easy to create a one-to-one correspondence between the natural numbers and every Fibonacci prime because Fibonacci primes are subsets of the natural numbers. Also all Fibonacci primes are unique and so are all the natural numbers. This correspondence would simply go on ad infinitum.

    The fact the no one has come forward with a proof makes me wonder how naive my method of attack is.
  6. Aug 9, 2011 #5

    Dedekind did indeed prove that an infinite set can be set up in a bijection with a proper subset of itself. But that just proves the existence of A proper subset (say in N) to which it is bijective (say even natural numbers). It doesn't mean that "Given any proper subset of an infinite set, there exists a bijection between them."

    For example,
    1. N is a proper subset of R, but there is no bijection. (or one to one correspondence from R to N for that matter)
    2. {1,2,3,4} is a proper subset of N, does that mean there exists a bijection?

    You seem to suggest that showing that a one one correspondence exists is a very easy job, I would like to highlight that that would be the end of the proof that there are infinitely many Fibonacci primes.
  7. Aug 9, 2011 #6
    In case this wasn't clear I meant we have to prove that cardinality of the set of Fibonacci primes is aleph null.

    How do you say that each new term is a next prime? It could be that it has a prime factor that is a new prime but may not be prime itself?
  8. Aug 9, 2011 #7
    I did not say that each new term is a next prime, I said that the index multiplyer is a next prime. That is I identified the Fibonacci series by Fib which is distinct from the A,B,C,D,E and F series. You determine the A series by dividing Fib(2n) by Fib(n) to get A(n). Then you divide A(3n) by A(n) to get B(n) etc. My last relation was that E(11n) = E(n)*F(n) for all n. So after determining the F series, you would use the relation F(13n) = F(n) * G(n) (i.e., F(13) = F(1)*G(1), F(26) = (F2)*G(2), F(39)=F(3)*G(3)...). Then after determining the G series by dividing F(13), F(26), F(39) ... respectively by F(1), F(2), F(3) ... , you would use the relation G(17n) = G(n)*H(n) to determine the H series (if your computer was powerful enough). I know it is a stretch to go from this conjecture to topic of the post but both involve a relationship between primes and the Fibonacci series, so there could be a connection. One thing is certain, only Fibonacci numbers with a odd index can be prime if the index is greater than 4.
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