On Reciprocal Sums Formed by Solutions of Pell’s Equation

Let (Xn, Yn)n 1 denote the positive integer solutions of Pell’s equation X 2 DY 2 = 1 or X2 DY 2 = 1. We introduce the Dirichlet series ⇣X(s) = P1 n=1 1/X s n and ⇣Y (s) = P1 n=1 1/Y s n for <(s) > 0 and prove that both functions do not satisfy any nontrivial algebraic di↵erential equation. For any positive integers s1 and s2 the two numbers ⇣X(2s1) and ⇣Y (2s2) are algebraically independent over a transcendental field extension of Q, whereas the three numbers ⇣X(2), ⇣Y (2), and P1 n=1 1/(XnYn) 2 are linearly dependent over Q. From the transcendence of ⇣Y (2) and the corresponding alternating series we obtain an application to the Archimedean cattle problem. Irrationality results for series of the form P1 n=1 ( 1) /Xn, P1 n=1 ( 1) /Yn, and P1 n=1 ( 1) /XnYn are obtained by a theorem of R.André-Jeannin. 1. A Summary on the Solutions of Pell’s Equation Let D denote a positive integer which is not a perfect square. In this paper we investigate positive integer solutions (X,Y ) of Pell’s equations X2 DY 2 = 1 and X2 DY 2 = 1, provided that the Pell-Minus-equation X2 DY 2 = 1 is solvable in integers. Let (Xn, Yn) 2 N2 : X2 n DY 2 n = ±1 ^ n = 1, 2, . . . be the set of all integer solutions, which are given recursively by Xn + Yn p D = X1 + Y1 p D n (n = 1, 2, . . . ) if X DY 2 = 1 , (1) and Xn + Yn p D = X1 + Y1 p D 2n 1 (n = 1, 2, . . . ) if X DY 2 = 1 , (2) where (X1, Y1) is the primitive solution of the corresponding equation with smallest coordinates X and Y . Let k be the length of the period of the continued fraction INTEGERS: 16 (2016) 2 expansion of p D, and let pm/qm denote the m-th convergent of p D. First we treat the integer solutions of X2 DY 2 = 1. It is well-known [8, § 27] that (Xn, Yn) = (pmk 1, qmk 1) (3) holds for n = 1, 2, 3 . . . with m = n, when k is even; in particular we have (X1, Y1) = (pk 1, qk 1). When k is odd, all solutions Xn and Yn are given by (3) for m = 2n. Then, (X1, Y1) = (p2k 1, q2k 1). Pell’s equation X2 DY 2 = 1 is solvable by integers if and only if k is odd. Then we obtain all solutions Xn and Yn from (3) for m = 2n 1, where (X1, Y1) = (pk 1, qk 1). Expanding the right-hand sides of (1) and (2) using the Binomial theorem and rearranging the terms, we find the representations