Fibonacci terms with primal distances

Question

Given any prime $p$, are there fibonacci numbers $F_k $ and $F_n$ such that $|F_n - F_k|=p^i , i \in \mathbb N$?

The edit - to include prime powers, rather than just primes - makes it very unlikely that there is an easy answer. What motivates this question, and do you have any reason to think the answer might be "yes"? — Mark Bennet, Jul 27 '13 at 09:22
Every Fibonacci number bigger than 1 [except F(6)=8 and F(12)=144] has at least one prime factor that is not a factor of any earlier Fibonacci number. Working on this — ARi, Jul 27 '13 at 11:03
You will find that every integer is a factor of infinitely many Fibonacci numbers, in particular $p^i$ will be factor of infinitely many - so you will find that there are pairs of Fibonacci numbers both divisible by $p^i$ whose difference is divisible by $p^i$. What I don't see is how you hope to get anywhere close to equality. — Mark Bennet, Jul 27 '13 at 11:07

score 3 · Accepted Answer · answered Jul 27 '13 at 08:36

3

No. $p=17$ is not a difference of any pair of Fibonacci numbers up to 34 (by inspection). If either $F_k$ or $F_n$ is $>34$, then their difference is either zero or $\ge 55-34=21$. Therefore $17$ is a counterexample.

answered Jul 27 '13 at 08:36

Jyrki Lahtonen

133,153

@JyrkiLahtonen Got it. Erasing comment... – coffeemath Jul 27 '13 at 09:06

score 3 · Answer 2 · answered Jul 27 '13 at 08:37

3

What have you tried? $17$?

The terms of the Fibonacci sequence show exponential growth, and the gaps grow faster than the primes.

answered Jul 27 '13 at 08:37

Mark Bennet

100,194

score 0 · Answer 3 · answered Jul 27 '13 at 11:27

0

There are lots of instances where $F_m - F_n$ give rise to primes, but there are many more primes than there are $F_n$ or differences of thus.

All the same, if $m-n$ is a multiple of $3$ or $4$, then the outcome is a multiple of $2$ or $3$.

answered Jul 27 '13 at 11:27

wendy.krieger

6,856

Fibonacci terms with primal distances

3 Answers3