| Popis: |
Partial solution by the editors. We show that the conclusion holds if fn is not a kth power, where k > 2 and n > 4k 3, or if fn is a kth power and k is divisible by 2 or 3. (The assertion is false for k = 1.) First suppose that fn is a kth power for some k divisible by 2 or by 3. This makes fn a square or a cube. It is known that the only squares among the Fibonacci numbers are fi, f2, and f12 = 144 (J. H. E. Cohn, On square Fibonacci numbers, J. London Math. Soc. 39 (1964) 537-540), and that the only cubes among the Fibonacci numbers are fi, f2, and f6 = 8 (H. London and R. Finkelstein, On Fibonacci and Lucas numbers which are perfect powers, Fibonacci Quart. 7 (1969) 476-481, 487; errata, ibid. 8 (1970) 248). The assertion is easily checked in these cases. Now suppose that fn is not a perfect kth power, where k > 2 and n > 4k 3. Our plan is to show that !f/T-7 + kfn-2k is closer to VT than VT is to an integer. This is easy to check when n 12. We use the Binet formula for Fibonacci numbers |
