Apart from 2, 3 and 3, 5 is there any pair of consecutive Fibonacci numbers which are both prime?

Oct 15, 2016

No

Explanation:

The Fibonacci sequence is defined by:

${F}_{0} = 0$

${F}_{1} = 1$

${F}_{n} = {F}_{n - 2} + {F}_{n - 1} \text{ }$ for $n > 1$

Starting with ${F}_{0}$, it begins:

$0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987 , \ldots$

Prove by induction that:

${F}_{m + n} = {F}_{m - 1} {F}_{n} + {F}_{m} {F}_{n + 1}$

for any $m , n \ge 1$

Base cases

${F}_{m + \textcolor{b l u e}{1}} = {F}_{m - 1} + {F}_{m} = {F}_{m - 1} {F}_{\textcolor{b l u e}{1}} + {F}_{m} {F}_{\textcolor{b l u e}{1} + 1}$

${F}_{m + \textcolor{b l u e}{2}} = {F}_{m + 1} + {F}_{m} = {F}_{m - 1} + 2 {F}_{m} = {F}_{m - 1} {F}_{\textcolor{b l u e}{2}} + {F}_{m} {F}_{\textcolor{b l u e}{2} + 1}$

Induction step

${F}_{m + k + 1} = {F}_{m + k - 1} + {F}_{m + k}$

$\textcolor{w h i t e}{{F}_{m + k + 1}} = {F}_{m - 1} {F}_{k - 1} + {F}_{m} {F}_{k} + {F}_{m - 1} {F}_{k} + {F}_{m} {F}_{k + 1}$

$\textcolor{w h i t e}{{F}_{m + k + 1}} = {F}_{m - 1} \left({F}_{k - 1} + {F}_{k}\right) + {F}_{m} \left({F}_{k} + {F}_{k + 1}\right)$

$\textcolor{w h i t e}{{F}_{m + k + 1}} = {F}_{m - 1} {F}_{k + 1} + {F}_{m} {F}_{k + 2}$

$\square$

Hence:

${F}_{2 n} = {F}_{n + n} = {F}_{n - 1} {F}_{n} + {F}_{n} {F}_{n + 1} = {F}_{n} \left({F}_{n - 1} + {F}_{n + 1}\right)$

So: ${F}_{2 n}$ is divisible by ${F}_{n}$ for any $n \ge 1$.

If $n = 2$ then ${F}_{2} = 1$ so that does not imply that ${F}_{4}$ is composite - it is actually prime. But for any larger values of $n$, ${F}_{2 n}$ is composite.

So for any two consecutive Fibonacci numbers after $\left({F}_{4} , {F}_{5}\right) = \left(3 , 5\right)$, since one of them will have even index greater than $4$, it will be divisible by an earlier non-unit Fibonacci number greater than ${F}_{2}$.

e.g. ${F}_{22} = 17711$ is divisible by ${F}_{11} = 89$