1. Offer an expression that identifies Fibonacci numbers whose terminal digits are only nines beginning with ...999.
0<n<2000, Fibonacci(n)+1=1000m, gives n=1498 in WolframAlpha. m has over 300 digits. Larger values exceed free computation time.
Another method:
A positive integer z is a Fibonacci number iff one of 5z^2+4 or 5z^24 is a perfect square [Gessel].
Using this we can obtain the Pellians (5(1000x1)^2+4=y^2, i.e. 5000000 x^210000 xy^2 = 9, and 5000000 x^210000 xy^2 = 1, both of which yield guaranteed solutions, as can be verified here (WolframAlpha times out before solving the recurrence): http://www.alpertron.com.ar/QUAD.HTM, however even the smallest solutions are too big for me to work with.
Edited on November 12, 2012, 3:21 am

Posted by broll
on 20121112 00:16:21 