Perfect powers among Fibonomial coefficients

作者：Marques Diego^*; Togbe Alain

来源：Comptes Rendus Mathematique, 2010, 348(13-14): 717-720.

DOI：10.1016/j.crma.2010.06.006

Let F(n) be the nth Fibonacci number. For 1 <= k <= m, let
[mk](f) = F(m)F(m-1) ... F(m-k+1)/F(1) ... F(k)
be the corresponding Fibonomial coefficient. In 2003, the problem of determining the perfect powers in the Fibonacci sequence was completely solved. In fact, the only solutions of F(m)=y(t), with m > 2, are (m, y, t) = (6, 2, 3), (12, 12, 2). In this paper, we prove that the only solutions of the Diophantine equation
[mk](f) = y(t),
with m > k + 1 and t > 1, are those related to k = 1, that is (m, k, y, t) = (6, 1, 2, 3) and (12, 1, 12, 2).

• 出版日期2010-7