MQ47 --- Quartic equation

17x^4 + 34x^3y + 10x^2 + 25x^2y^2 + 10xy + 8xy^3 + y^4 + 25 = 0
==> (16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4) + ...
....... (x^4 + 2x^3y + x^2y^2 + 10x^2 + 10xy + 25) = 0
==> (2x + y)^4 + (x^2 + xy + 5)^2 = 0
As x, y are real numbers, so
2x + y = 0 . . . . . . . (i) and
x^2 + xy + 5 = 0 . . .(ii)
Solve (i), (ii), get
(x, y) = (根號5, -2根號5) or (-根號5, 2根號5)

2012-09-21 20:36:41 補充：
No odd degree, x^4 + 10x^2 + 25, x^4 + 10x^2y^2 + 25 are all perfect squares,
so it must be in the form A^2 + B^2 = 0
17x^4 is not a perfect square, so it must be break into two parts : 16x^4, x^4
16x^4 = (2x)^4, so I guess A should be (2x + y)^2

To 諸葛諭遜:
Can you tell me how do you think to figure out
17x⁴ + 34x³y + 10x² + 25x²y² + 10xy + 8xy³ + y⁴ + 25 = (2x + y)⁴ + (x² + xy + 5)²
?