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Are there infinitely many Fibonacci numbers that are also powers of 2? If not, which is the largest?

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3 Answers 3

up vote 30 down vote accepted

Fibonacci numbers have just about the greatest divisibility rule you could expect. Fibonacci numbers share common divisors exactly when their corresponding indices share common divisors, $\gcd(F[m],F[n])$ = $F_{\gcd(m,n)}$.

This result means that the Fibonacci index of any power of $2$ greater than $8$ must be divisible by $6$ as $F_6 = 8$ and this means that the index of power of $2$ Fibonacci number greater that $8$ must be a power of 6 and therefore must be divisible by $F_{36}$.

However $F_{36}$ is also divisible by $F_{9}$ since $9$ divides $36$ and given that $F_9 = 34, F_{36}$ is therefore divisible by $34$ and cannot be a power of $2$.

Since any candidate powers of $2$ greater than $8$ must be divisible by $34$ there can be no Fibonacci numbers greater than $8$ which are powers of $2$.

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I got that the index must be divisible by 6, but didn't understand why it must be a power of 6. –  Anant May 15 at 12:55
I don't think it does; it just has to be divisible by 6 and have prime factorisation only containing 2s and 3s. But the same proof (using $F_4$ as well as $F_9$) seems to work. –  user73985 May 15 at 13:23
I'm a bit confused right now. Why is is gcd in the topmost equation? Shouldn't it be something like just cd? –  Max Ried May 15 at 15:53

There are three, and the biggest is 8.

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For posterity, please avoid link-only answers and also replicate the information here. Websites go down and information on them changes over time. If quoting from a source, also provide attribution. –  Kyle Falconer May 15 at 20:47

As far as I can tell, as a corollary of Carmichael's theorem, it follows that no Fibonacci numbers other than $1$, $2$ and $8$ can be powers of $2$. Thus, $8$ is the largest.

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