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$F_6=2^3$ and $F_{12}=2^43^2$. Is there an $n>12$ such that $F_n=p^2k$ with $p$ prime and $k$ is $p$-smooth?

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For what it's worth, the first 1000 Fibonacci numbers are factored at mersennus.net/fibonacci/f1000.txt –  Gerry Myerson Feb 14 '12 at 1:20
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In general, integers that are divisible by the square of their largest prime factor are extremely rare. I suspect a probabilistic heuristic would predict that there are no other such Fibonacci numbers. - Also, despite what is more prevalent in the current literature, "friable" is better terminology than "smooth". –  Greg Martin Mar 21 '12 at 16:24
    
@GregMartin: I agree with your heuristic. But is there any hope of a proof? (If not, feel free to give "not known" as an answer...) –  Charles Mar 21 '12 at 20:14
    
oeis.org/A070003 –  Charles Mar 22 '12 at 12:36
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2 Answers

I think that this is an open question.

Note that if $~p~$ is a prime number then

$F_p \equiv \left(\frac{p}{5}\right) \pmod p ~~\text {and}~~ F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p$

It is not known whether there exists a prime $p$ such that:

$F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod {p^2}$

See Wikipedia article for more information .

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I don't see the relationship between my problem and Wall-Sun-Sun primes. I don't put any constraints on the indexes where the primes appear, so 8 and 144 are examples even though 6 and 12 are not Wall-Sun-Sun primes. –  Charles Feb 14 '12 at 14:38
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up vote 2 down vote accepted

There are $x\exp\left(-(1+o(1))\sqrt{\log x\log\log x}\right)$ numbers up to x which are divisible by the square of their largest prime factor. Interpreting this as a probability, there are heuristically about $$ \int^\infty_{1000}e^{-\sqrt{\log\varphi\cdot x\log x}}dx\approx2.8\cdot10^{-24} $$ such n. Insofar as there is no "conspiracy" among Fibonacci numbers, it seems likely that $F_{12}$ is the last example.

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