Fibonacci numbers that are powers of 2

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

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 $2$.

• I got that the index must be divisible by 6, but didn't understand why it must be a power of 6. May 15, 2014 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. May 15, 2014 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? May 15, 2014 at 15:53
• Not sure if it helps anyone, was just hovering over this question. @Christopher I think in the case above, $n$ must be divisible by 36. Substitute $m$ = 36, $F_n$ = $2^k$. Then in the gcd equation, LHS is a power of 2 (>8 because $F_{6}$ is a common factor). RHS is $F(6*gcd(6, n/6))$. By using same gcd equation (if $d | n$ then $F_{d} | F_{n}$), we know $F_{gcd(6, n/6)}$ must divide LHS which is a power of 2. This means $6 | gcd(6, n/6)$, which means n/6 is a multiple of 6, or in other words $36 | n$ Sep 8 at 13:53

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.

There are three, and the biggest is 8.

• 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. May 15, 2014 at 20:47
• There are 4. Note that one is repeated. So they are 1, 1, 2, 8. Jul 16, 2015 at 15:02