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

Question

Serenity Johnson · Accepted Answer

No

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#

Anna Allen · Answer

undefined No, there are no other pairs of consecutive Fibonacci numbers where both numbers are prime, apart from 2 and 3, and 3 and 5. This is because as the Fibonacci sequence progresses, the numbers become increasingly larger, making it less likely for consecutive numbers to both be prime. It is a well-known conjecture that there are infinitely many prime numbers in the Fibonacci sequence, but specific pairs of consecutive primes are rare and difficult to find as the sequence progresses.