Please use this identifier to cite or link to this item: http://hdl.handle.net/11452/24509
Title: The number of solutions of pell equations x2 -ky2 = N and x2+xy- Ky2 = N over Fp
Authors: Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Anabilim Dalı.
Tekcan, Ahmet
AAH-8518-2021
55883777900
Keywords: Mathematics
Pell equation
Solutions of the Pell equation
Finite field
Issue Date: Oct-2011
Publisher: Charles Babbage Res CTR
Citation: Tekcan, A. (2011). "The number of solutions of pell equations x2 -ky2 = N and x2+xy- Ky2 = N over Fp". Ars Combinatoria, 102, 225-236.
Abstract: Let p be a prime number such that p equivalent to 1, 3(mod 4), let F-p, be a finite field, let N is an element of F-p* = F-p - {0} be a fixed. Let P-p(k) (N) : x(2) - ky(2) = N and (P) over tilde (k)(p)(N) : x(2) + xy - ky(2) = N be two Pell equations over F-p, where k = p-1/4 or k = p-3/4, respectively. Let P-p(k)(N)(F-p) and (P) over tilde (k)(p)(N)(F-p) denote the set of integer solutions of the Pell equations P-p(k)(N) and (P) over tilde (k)(p)(N), respectively. In the first section we give some preliminaries from general Pell equation x(2) - ky(2) = +/- N. In the second section, we determine the number of integer solutions of P-p(k)(N). We proved that P-p(k)(N)(F-p) = p+ 1 if p equivalent to 1(mod 4) or p equivalent to 7(mod 12) and P-p(k)(N)(F-p) = p - 1 if p equivalent to 11(mod 12). In the third section we consider the Pell equation (P) over tilde (k)(p)(N). We proved that (P) over tilde (k)(p)(N)(F-p) = 2p if p equivalent to 1(mod 4) and N is an element of Q(p); (P) over tilde (k)(p)(N)(F-p) = 0 if p equivalent to 1(mod 4) and N is not an element of Q(p) ; (P) over tilde (k)(p)(N)(F-p) = p + 1 if p equivalent to 3(mod 4).
URI: http://hdl.handle.net/11452/24509
ISSN: 0381-7032
Appears in Collections:Scopus
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