THE NUMBER OF SOLUTIONS OF PELL EQUATIONS x(2) - ky(2) = N AND x(2) + xy - ky(2) = N OVER F-p

Tekcan A.

ARS COMBINATORIA, vol.102, pp.225-236, 2011 (SCI-Expanded) identifier

  • Publication Type: Article / Article
  • Volume: 102
  • Publication Date: 2011
  • Journal Name: ARS COMBINATORIA
  • Journal Indexes: Science Citation Index Expanded (SCI-EXPANDED), Scopus
  • Page Numbers: pp.225-236
  • Bursa Uludag University Affiliated: Yes


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).