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^2-4 is a perfect square [Gessel].

Using this we can obtain the Pellians (5(1000x-1)^2+4=y^2, i.e. 5000000 x^2-10000 x-y^2 = -9, and 5000000 x^2-10000 x-y^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 2012-11-12 00:16:21