Question

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

Answer 1

No

Explanation:

The Fibonacci sequence is defined by:

`F_0 = 0`

`F_1 = 1`

`F_n = F_(n-2) + F_(n-1)" "` 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,...`

Prove by induction that:

`F_(m+n) = F_(m-1)F_n+F_mF_(n+1)`

for any `m, n >= 1`

Base cases

`F_(m+color(blue)(1)) = F_(m-1) + F_m = F_(m-1)F_(color(blue)(1)) + F_m F_(color(blue)(1)+1)`

`F_(m+color(blue)(2)) = F_(m+1) + F_m = F_(m-1) + 2F_m = F_(m-1)F_color(blue)(2) + F_mF_(color(blue)(2)+1)`

Induction step

`F_(m+k+1) = F_(m+k-1) + F_(m+k)`

`color(white)(F_(m+k+1)) = F_(m-1)F_(k-1) + F_mF_k + F_(m-1)F_k + F_mF_(k+1)`

`color(white)(F_(m+k+1)) = F_(m-1)(F_(k-1) + F_k) + F_m(F_k +F_(k+1))`

`color(white)(F_(m+k+1)) = F_(m-1)F_(k+1) + F_m F_(k+2)`

`square`

Hence:

`F_(2n) = F_(n+n) = F_(n-1)F_n + F_nF_(n+1) = F_n(F_(n-1) + F_(n+1))`

So: `F_(2n)` is divisible by `F_n` for any `n >= 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_(2n)` is composite.

So for any two consecutive Fibonacci numbers after `(F_4, F_5) = (3, 5)`, 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`