# Talk:Fibonacci number

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- If is a prime number then is prime. (The converse is false.)

is not correct:

n | F(n) |
---|---|

0 | 0 |

1 | 1 |

2 | 1 |

3 | 2 |

4 | 3 |

5 | 5 |

The forth Fibonacci number is 3. 3 is a Primenumber, but 4 is not a Primenumber. --arbol01 07:30, 30 December 2007 (CST)

- True. This is the only exception, due to the fact that 4=2*2 while F_2=1 is not greater than 1. Thus for the composite number 4=n=a*b=2*2 we do not get a proper divisor of F_a | F_n, i.e. F_2 | F_4, because F_2 is not proper, it is equal 1. It was my fault to overlook this case; I was happy (too happy) to see that the general theorem works accidentally also for prime F_3=2, where the index is prime 3. Thank you, Wlodzimierz Holsztynski 16:23, 30 December 2007 (CST)