System of equations

hmm well

i guess the source is http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=33505&p=2414098#p2414098

so it will ve y² and not y⁴.

We will use geometric interpretations. Let AB denote |x| , BC denote |y|. Since x^2 + y^2 = 16, AC = 4 , and \triangle ABC is right angled at B. Let BD = |z| . \angle ABD = \pi/3 and |AD| = 2 (By cosine rule). \angle DBC = \pi/6. |CD| = 2. In \triangle ADC , AD=CD = 2 but AC= 4 . That contradicts triangle inequality and hence \triangle ADC must be degenerate. Now its obvious that AD= CD= BD = 2= |z|. Further \triangle ABD must be equilateral. Hence AB = 2 = |x|. By cosine rule on \triangle BDC, BC = 2\sqrt{3}= |y|. Now its trivial to conclude that the solutions are (x,y,z) = (2, 2\sqrt{3}, 2 ) and (-2, -2\sqrt{3},-2). (And that no other combination works)

thnx. though i have already seen that post.