ON kTH-POWER NUMERICAL CENTRES

نویسنده

  • PATRICK INGRAM
چکیده

We call the integer N a kth-power numerical centre for n if 1 + 2 + · · · + N = N + (N + 1) + · · · + n. We prove, using the explicit lower bounds on linear forms in elliptic logarithms, that there are no nontrivial fifth-power numerical centres for any n, and demonstrate that there are only finitely many pairs (N, n) satisfying the above for any given k > 1. The problem of finding kth-power centres for k = 1, 2, 3 has been treated in [9]. We will call an integer N a kth power numerical centre for n if (1) 1 + 2 + · · ·+N = N + (N + 1) + · · ·+ n. This equation is trivial in the case k = 0, while the solutions to the problem in the case k = 1 correspond to the solutions of the Pell equation X − 2Y 2 = 1, with X = 2n + 1, Y = 2N . In [5] and [9] the cases with k = 2, 3 were treated, and it was shown that the only solutions to (1) were the trivial ones, i.e., those with (N,n) ∈ {(0, 0), (1, 1)}. We will prove the following: Proposition. For fixed k > 1, equation (1) has only finitely many solutions. In particular, for k = 5 there are only the trivial solutions. Equation (1) is, of course, equivalent to Sk(N) + Sk(N − 1) = Sk(n), where Sk(x) = 1 + 2 + · · ·+ x. For k even the above curves are smooth and so have genus 1 2k(k− 1) by a straightforward application of a theorem of Hurwitz (see [7], p. 41). When k is odd, the above admits the change of variables x = (2n + 1), y = (2N) and the resulting curves are smooth in x and y of degree k+1 2 , and so have genus 1 8 (k−1)(k−3). The general claim, then, follows from the celebrated result of Faltings [4]. For a more direct proof we apply results of Bilu and Tichy [1] on the number of solutions to the Diophantine equation f(x) = g(y). More specifically, we apply a refinement of this by Rakaczki [6] which applies just in case f(x) = Sk(x). For the result in the case k = 5, we apply the results of David [3], as presented in [11] (see also [10] and [12]). The problem of finding all integral points on the curve (1) in the case k = 5 reduces to that of locating all integral points on a certain non-Weierstrass model of an elliptic curve. As it turns out, integer points 1991 Mathematics Subject Classification. Primary: 11D25, 11J89.

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تاریخ انتشار 2005