Solution. Arbitrary m from the divisible m -mel Fibonacci numbers indices form an arithmetic progression, in which 0 is included; denote this difference of d m -mel. Easy to control in that d 2 = 3, d = 4 3, d 4 = 6, d = 5 5, d 6 and d = 12 12 = 12th If a, b, is divisible by 5 c between, say 5 | the then 5 \ big_b \ rightarrow ~ 5 = d_5 \ big | b \ big | u_c ~ \ Rightarrow ~ 5 = d_5 \ big | c, i.e., b and c can each be divided by 5. If, between b and c is divisible by 3, say 3 | the then 3 \ big | a \ big | u_b ~ \ Rightarrow ~ 4 = d_3 \ big | b \ big | u_c ~ \ Rightarrow ~ 6 = d_4 \ big | c \ big | u_a ~ \ Rightarrow ~ 12 = d_6 \ big | a \ Big | u_c} \ rightarrow ~ 12 = d_ {12} \ big | c \ big | u_c} \ rightarrow ~ 12 = d_ { i.e., a, b, c are all divisible by 12. If there is an even among b, c, say 2 | the then 2 | the | Because u 3 = d b 2 | b, that is between the tracks also divisible by 3, so, according to the preceding paragraph, b, c is divisible by 12. Hereinafter, it is assumed that alphabet smallest prime factors p \G 7 a, b, c can assume the role due to the symmetry of p | a. The divisor number p is called d \ pi (P) Pisano interval does. It is also known that when p is prime and p \not 5, then \ pi (P) | p 2 -1, so d p | \ pi (P) | p 2 -1 and thus p | u d p | u p 2 -1. (More generally, if p \ equiv \ pm1 \ pmod5 Then d p | p -1, p \ equiv \ pm2 \ pmod5 case is d p | p +1). On the other hand p | the | u b, which p | gcd (p 2 -1 u, b u) = u gcd (p 2 -1, b). The number b of each prime divisor of p or p + 2 at least, none of these divisors? 2 -1 = (p -1) (p + 1) = s. Therefore u gcd (p 2 -1, b) = u 1 = 1, e p | 1, which is a contradiction. This is not possible.