cp's OEIS Frontend

A217470 The Diophantine equation x^2 - x*y - G*y^2 = -1, G a positive integer, D = 4*G + 1 not a perfect square, has no solution precisely for G = a(n).

%I A217470 #7 Oct 05 2012 12:34:29
%S A217470 5,8,11,14,17,19,23,26,29,32,33,35,38,40,41,44,47,50,51,52,53,54,55,
%T A217470 59,61,62,63,65,68,71,74,75,76,77,80,82,83,85,86,89,92,94,95,96,98,101
%N A217470 The Diophantine equation x^2 - x*y - G*y^2 = -1, G a positive integer, D = 4*G + 1 not a perfect square, has no solution precisely for G = a(n).
%C A217470 See the Perron reference for the theorem which by negation implies that this quadratic Diophantine equation has no solution if and only if A077427 is even.
%C A217470 See the pairs (x, y) = (A077057, A077058) which for these a(n) values are the smallest positive solutions of the Diophantine equation x^2 - x*y - a(n)*y^ = +1.
%C A217470 In the table on p. 108 of the Perron reference these a(n) values, called there also G, are the ones were in the third column numbers in brackets appear.
%C A217470 The case D = 4*G + 1 = m^2 > 1 has trivially no solutions: the equation is then X^2 - Y^2 = -4, with  X = |2*x-y|, Y = |m*y|. X and Y are either both even or both odd. In the first case one is led to the equation v^2 - w^2 = (v-w)*(v+w)= -1, with X = 2*v and Y = 2*w, and there is only the solution (v,w) = (0,1), hence 2*x = y, m*y = 2. But then m=2 and y=1 with non-integer x solution. In the other case X = 2*v+1 and Y = 2*w+1, v not w, leading to v + w + 1 = -1 with no positive integer solution. Thanks to T. D. Noe for pointing out that one has to mention that these values G = A002378(k), k>=1, with D a perfect (odd) square, are here not included.
%D A217470 O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).
%F A217470 a(n) gives the increasingly ordered values for G from A078358 which appear at position k where A077427(k) is even, for k>=1. The next even number in A077427 appears for k = 6 and
%F A217470   A078358(6) = 8, hence a(2) = 8.
%e A217470 a(1) = 5 because 5 = A078358(4) and A077427(4) = 2, which is even.
%Y A217470 Cf. A077057, A077058, A078358, A077427.
%K A217470 nonn
%O A217470 1,1
%A A217470 _Wolfdieter Lang_, Oct 04 2012