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For each $n$, How can we find $n$ consecutive composite Fibonacci numbers?

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  • $\begingroup$ Know any sufficient conditions for $F_n$ to be composite? $\endgroup$ – Erick Wong May 31 '14 at 10:15
  • $\begingroup$ @ErickWong No, I don't know. $\endgroup$ – user118535 May 31 '14 at 10:18
  • $\begingroup$ Now you do, after DenisMath's answer :). $\endgroup$ – Erick Wong May 31 '14 at 10:19
  • $\begingroup$ For $F_n$ if $n$ is composite, $F_n$ is composite. Except $n = 4$ and $n = 1$ $\endgroup$ – Dane Bouchie Jun 21 '14 at 2:13
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$F_{kn}$ is divisible by $F_n$, so, apart from $F_4 = 3$, any Fibonacci prime must have a prime index.

http://en.wikipedia.org/wiki/Fibonacci_number

So you should simply take n consecutive composite indexes.

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