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The pairs (x,y) such that (x^2 + y^2)/(x*y + 1) is a perfect square, i.e., 4.
1

%I #22 Jul 21 2017 12:35:12

%S 0,2,2,8,8,30,30,112,112,418,418,1560,1560,5822,5822,21728,21728,

%T 81090,81090,302632,302632,1129438,1129438,4215120,4215120,15731042,

%U 15731042,58709048,58709048,219105150,219105150,817711552,817711552,3051741058,3051741058,11389252680,11389252680

%N The pairs (x,y) such that (x^2 + y^2)/(x*y + 1) is a perfect square, i.e., 4.

%C Essentially A052530, each term besides the first repeated. - _R. J. Mathar_, Jul 21 2009

%H Vincenzo Librandi, <a href="/A162959/b162959.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,4,0,-1).

%F From _Colin Barker_, Feb 21 2013: (Start)

%F a(n) = 4*a(n-2) - a(n-4).

%F G.f.: 2*x^2*(x+1) / (x^4-4*x^2+1). (End)

%e Pairs are (8,30) with (8^2 + 30^2)/(8*30 + 1) = 4, or (30,112) with (30^2 + 112^2)/(30*112 + 1) = 4.

%t CoefficientList[Series[2 x (x + 1) / (x^4 - 4 x^2 + 1), {x, 0, 40}], x] (* _Vincenzo Librandi_, May 14 2013 *)

%o (PARI) x='x+O('x^66); concat([0],Vec(2*x^2*(x+1)/(x^4-4*x^2+1))) \\ _Joerg Arndt_, May 15 2013

%K nonn,less,easy

%O 1,2

%A _Vincenzo Librandi_, Jul 19 2009