Resolve the system of equations

Algebra Level 3

{ x 5 x 4 + 2 x 2 y = 2 y 5 y 4 + 2 y 2 z = 2 z 5 z 4 + 2 z 2 x = 2 \begin{cases} \begin{aligned} { x }^{ 5 }-{ x }^{ 4 }+2{ x }^{ 2 }y&=2 \\ { y }^{ 5 }-{ y }^{ 4 }+2{ y }^{ 2 }z&=2 \\ { z }^{ 5 }-{ z }^{ 4 }+2{ z }^{ 2 }x&=2\end{aligned} \end{cases}

How many ordered triples of real numbers ( x , y , z ) (x,y,z) satisfy the system of equations above?


The answer is 1.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Linkin Duck
Mar 24, 2017

+) If x = 1 x=1 , we have y = 1 y=1 (from the first equation) and hence z = 1 z=1 (from the second equation).

+) If x > 1 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 2={ z }^{ 5 }-{ z }^{ 4 }+2{ z }^{ 2 }x > { z }^{ 5 }-{ z }^{ 4 }+2{ z }^{ 2 }\\ \Longrightarrow \left( z-1 \right) \left( { z }^{ 4 }+2z+2 \right) < 0

Since z 4 + 2 z + 2 = ( z 2 1 2 ) 2 + ( z + 1 ) 2 + 3 4 > 0 z < 1 , { z }^{ 4 }+2z+2={ \left( { z }^{ 2 }-\frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 }+\frac { 3 }{ 4 } >0 \Longrightarrow z<1, that leads to y > 1 y>1 (if we consider the same way in the second equation) and hence gets that x < 1 x<1 (if we consider the same way in the first equation) \Longrightarrow contradiction.

+) If x < 1 x<1 , we can similarly get the final contradiction.

So ( x , y , z ) = ( 1 , 1 , 1 ) (x,y,z)=(1,1,1) is the only solution of the systems of equations above.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...