Every Fibonacci term $F(3n)$ is divisible by two

$F(3) = 2$

$F(6) = 8$

$F(9) = 34$


After seeking Fibonacci tables factorization until $F(10000)$, for every term $\frac{F(3n)}{2}$, it appears there is just one result which is prime

$\frac{F(9)}{2} = \frac{34}{2} = 17$

all other $\frac{F(3n)}{2}$ = composite numbers

I can't see some reason for this to happen, but since there is no more primes in so huge quantity of terms, I suppose there is not more any prime for $\frac{F(3n)}{2}$


Miguel Velilla


Recall the well-known fact: $n \mid m \iff F_n \mid F_m$.

Thus $F_{3n}$ is divisible by $F_3=2$ and $F_n$. So if $n > 3$ and $\displaystyle \frac{F_{3n}}{2}$ is equal to prime number $p$ then $F_n \mid F_{3n}=2p$ so we have $F_{n}\in\{p, 2p\}$ (because $F_n>F_3=2$). In particular $F_{3n} \le 2F_n$ which is a contradiction as $2F_n < F_n+F_{n+1}=F_{n+2}<F_{3n}$.


It is well known that if $a$ divides $b$ then $Fib(a)$ divides $Fib(b)$.

Ignoring for the moment the cases where $a=1,2,3$ (in which case Fib(a)=$1,1,2$), this means that $Fib(3a)$ is divisible by $Fib(a) \gt 2$ so $\frac{Fib(3a)}{2}$ is divisible by $\frac{Fib(a)}{2} \gt 1$ if $Fib(a)$ is even or by $Fib(a) \gt 1$ if it is odd. So $\frac{Fib(3a)}{2}$ cannot be prime.

That leaves the early cases $\frac{Fib(3)}{2},\frac{Fib(6)}{2},\frac{Fib(9)}{2}$ which are $1,4,17$, and only the last of these is prime.


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