⎩ ⎪ ⎨ ⎪ ⎧ x 5 − x 4 + 2 x 2 y y 5 − y 4 + 2 y 2 z z 5 − z 4 + 2 z 2 x = 2 = 2 = 2
How many ordered triples of real numbers ( x , y , z ) satisfy the system of equations above?
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+) If x = 1 , we have y = 1 (from the first equation) and hence z = 1 (from the second equation).
+) If x > 1 , from the third equation we have:
2 = z 5 − z 4 + 2 z 2 x > z 5 − z 4 + 2 z 2 ⟹ ( z − 1 ) ( z 4 + 2 z + 2 ) < 0
Since z 4 + 2 z + 2 = ( z 2 − 2 1 ) 2 + ( z + 1 ) 2 + 4 3 > 0 ⟹ z < 1 , that leads to y > 1 (if we consider the same way in the second equation) and hence gets that x < 1 (if we consider the same way in the first equation) ⟹ contradiction.
+) If x < 1 , we can similarly get the final contradiction.
So ( x , y , z ) = ( 1 , 1 , 1 ) is the only solution of the systems of equations above.