Abstract
Let rge 1 be an integer and mathbf{U}:=(U_{n})_{nge 0} be the Lucas sequence given by U_0=0, U_1=1, and U_{n+2}=rU_{n+1}+U_n, for all nge 0 . In this paper, we show that there are no positive integers rge 3,~xne 2,~nge 1 such that U_n^x+U_{n+1}^x is a member of mathbf{U}.
Highlights
Let r ≥ 1 be an integer and U := (Un)n≥0 be the Lucas sequence given by U0 = 0, U1 = 1, and Un+2 = rUn+1 + Un, for all n ≥ 0
We show that there are no positive integers r ≥ 3, x = 2, n ≥ 1 such that Unx + Unx+1 is a member of U
U(n−1)/2V(n+1)/2 if n ≡ 1 U(n+1)/2V(n−1)/2 if n ≡ 3,. Both sequences U and V can be extended to negative indices either by allowing n to be negative in the Binet formulas (4) and (7) or by using the recurrence relation to extend U to negative indices via U−n = −rU−(n−1) + U−(n−2) for all n ≥ 1
Summary
For r = 1 Luca and Oyono [8] proved that Eq (3) has no positive integer solutions (n, m, x) with n ≥ 2 and x ≥ 3. Rihane et al [11] studied Eq (3) when r = 2 and proved that there is no positive integer solution (n, m, x) of it with x = 2. When k = 2, this sequence coincides with the sequence of Fibonacci numbers They proved that Eq (3) has no positive integer solution (k, n, m, x) with k ≥ 3, n ≥ 2, and x ≥ 1. Another related result involving the balancing numbers was studied by Rihane et al in [10].
Sign-in/Register to view highlights & summary Sign-in/Register to access full text options 
Free) 
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Schedule a call
