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A111200 Solution of Diophantine equation (1/x^2 + 1/y^2 = 1/z^2); x values in increasing order.

%I A111200 #10 Jul 07 2016 23:48:47
%S A111200 15,20,30,40,45,60,65,75,80,90,100,105,120,130,135,136,140,150,156,
%T A111200 160,165,175,180,195,200,210,220,225,240,255,260,270,272,280,285,300,
%U A111200 312,315,320,325,330,340,345,350,360,369,375,380,390,400,405,408,420,435
%N A111200 Solution of Diophantine equation (1/x^2 + 1/y^2 = 1/z^2); x values in increasing order.
%C A111200 Similar to integral Pythagorean triangles, but with reciprocal of integers: 1/x and 1/y are the legs and 1/z is the hypotenuse. Properties:
%C A111200 1) x*y is congruent to 0 mod z;
%C A111200 2) x = u*v *(u^2+v^2) * m /2 or x = (u^2-v^2)*(u^2+v^2)*m/4; y = (u^2-v^2)*(u^2+v^2)*m/4 or y = u*v *(u^2+v^2)*m /2; z = (u^2-v^2)*u*v*m /2; u and v are odd and relatively prime, u > v, m is an integer.
%C A111200 To use the equations: set n, evaluate (2*n^2+2*n+1), derive m = (2*n^2+2*n+1)*integer, then evaluate x, y, z.
%C A111200 3) Primitive integral solution, i.e., those solutions in which there is no factor common to x, y and z, are (15, 20, 12), (65, 156, 60), (136,255,120), (175, 600, 168), (369,1640, 360), (671, 3660, 660), ...
%C A111200 Primitive integral solution are those where m = 1.
%C A111200 4) If the triad (a, b, c) is a solution of a Pythagorean triangle, i.e., a^2 + b^2 = c^2, then (x = a*c, y = b*c, z= a*b) is a solution of (1/x^2 + 1/y^2 = 1/z^2). For Pythagorean triads see for example http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html [broken link]
%C A111200 5) First multiple solutions (more values for y and z given x ) are x = 60,120,180,195. Multiple solution can be derived by equations at point 2.
%C A111200 6) No prime is in the sequence.
%C A111200 Related topics: the Diophantine equation (1/x^i + 1/y^i = 1/z^i) has no integral solutions for i>2, as it is easy to demonstrate by means of the Fermat-Wiles theorem.
%e A111200 a(1) = 15 because 1/15^2 + 1/20^2 = 1/12^2 (smallest solution); a(2)= 20 as 1/20^2 + 1/15^2 = 1/12^2.
%p A111200 recPy:=proc(n) local x,y,z,Rx2,Ry2; for x from 1 by 1 to n do Rx2:=1/x^2; for y from 1 by 1 to x^2 do Ry2:=1/y^2; for z from 1 by 1 to x do if (Rx2 + Ry2 =1/z^2) then print(x);fi; od; od; od; end: # convert into set
%Y A111200 Cf. A094807.
%K A111200 nonn
%O A111200 1,1
%A A111200 _Giorgio Balzarotti_, _Paolo P. Lava_, Oct 24 2005
%E A111200 More terms from _Vladeta Jovovic_, Oct 25 2005