20190905, 16:03  #1 
Sep 2009
3^{2}·227 Posts 
Factoring near Fibonacci numbers.
Hello,
Do Fibonacci numbers 1 have algebraic factors? I've noticed some PRPs in factordb like (10*I(5545)1)/9 where N1 has a lot of small factors which makes me think I would be able to prove it prime if I knew how to factor it. There are several other PRPs where N1 or N+1 simplify to Fibonacci(whatever) +or a small integer. Are they likely to have algebraic factors? Chris 
20190905, 16:14  #2  
Nov 2003
2^{2}·5·373 Posts 
No.
Quote:
What is the point of this, anyway? We already have enough factoring projects underway. We don't need another one. 

20190905, 16:25  #3 
Nov 2016
2819_{10} Posts 
Fibonacci number (>8) 1 cannot be prime
Fibonacci number (>2) +1 cannot be prime Conjecture: For every k>1, there are infinitely many k*F(n)1 primes and infinitely many k*F(n)+1 primes 
20190905, 16:51  #4 
Sep 2009
3^{2}·227 Posts 

20190905, 17:41  #5 
Jun 2003
11552_{8} Posts 
Sort of, yes. I don't know why, but fibonacci(n)1 is divisible by certain smaller fibonacci numbers. Perhaps the same reason cyclotomic polynomials1 have factors.
I've gone ahead and added those for your number  now it is provable. 
20190905, 17:46  #6 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}×367 Posts 
I've worked out, (it was new for me), there is a factorization:
Code:
fibonacci(2*n+1)+(1)^n=fibonacci(n+1)*(fibonacci(n+1)+fibonacci(n1)) 
20190905, 18:15  #7  
Jun 2003
2·5·7·71 Posts 
Quote:
BTW, since (fibonacci(n+1)+fibonacci(n1)) = lucas(n), it got me to https://en.wikipedia.org/wiki/Lucas_...onacci_numbers which gives F(n+k) + (1)^k F(nk) = L(k)F(n). I bet that can explain these. 

20190906, 03:38  #8  
Feb 2017
Nowhere
29×157 Posts 
Quote:
The factorizations are related to the trigonometric identity 2*sin(A+B)2*sin(AB) = (2*sin(A))(2*cos(B)) and similar identities involving sums and differences of sines or of cosines. Algebraically, "Fibonacci numbers act like sines, Lucas numbers act like cosines." Note that the difference in arguments on the right side is the second argument on the left side. But there's a twist. The quadratic x^2  x  1 has constant term 1, not 1, so the two roots are negativereciprocal, not reciprocal. As a result, the form of the factorization "wigwags," or alternates. I will illustrate with F_{2n+1}  2, where, note that 2 = F_{3}, and in each identity the indexes of the factors differ by 3. F_{5}  F_{3} = 3 = L_{1}F_{4} F_{7}  F_{3} = 11 = L_{5}F_{2} F_{9}  F_{3} = 32 = L_{3}F_{6} F_{11}  F_{3} = 87 = L_{7}F_{4} [etc] Also, F_{5} + F_{3} = 7 = L_{4}F_{1} F_{7} + F_{3} = 15 = L_{2}F_{5} F_{9} + F_{3} = 36 = L_{6}F_{3} F_{11} + F_{3} = 91 = L_{4}F_{7} [etc] There are similar identities involving F_{m+2k} + or  F_{m}, and slightly more complicated identities involving Lucas numbers. 

20190906, 15:45  #9 
Sep 2009
7FB_{16} Posts 
Thanks for that. Now I've got a suitable toolkit to deal with any similar PRPs that turn up.
Chris 
20190907, 13:05  #10 
Feb 2017
Nowhere
29×157 Posts 
The factorizations for sums and differences of Lucas numbers whose indexes differ by an even number are actually not that difficult to characterize. They may be described as follows (F indicates Fibonacci numbers, L indicates Lucas numbers):
L_{m+2k}  L_{m} = 5F_{k}F_{k+m} if k is even, and L_{k}L_{k+m} if k is odd L_{m+2k} + L_{m} = 5F_{k}F_{k+m} if k is odd, and L_{k}L_{k+m} if k is even 
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