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Difference between revisions of "2007 Indonesia MO Problems/Problem 6"

(Solution to Problem 6 -- systems with monotonics)
 
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Assume that <math>x > 2</math>.  Thus, <math>x^3 > 8</math>, so <math>x^3 - 8 > 0</math> and <math>x^3 + x - 8 > x</math>, so <math>y > x > 2</math>.  By doing the same steps, we can show that <math>z > y</math> and <math>x > z</math>.  However, that would mean that <math>x > x</math>, which does not work, so there are no solutions where <math>x > 2</math>.
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Assume that <math>x > 2</math>.  Because <math>x^3</math> is a monotonically increasing function, <math>x^3 > 8</math>, so <math>x^3 - 8 > 0</math> and <math>x^3 + x - 8 > x</math>, so <math>y > x > 2</math>.  By doing the same steps, we can show that <math>z > y</math> and <math>x > z</math>.  However, that would mean that <math>x > x</math>, which does not work, so there are no solutions where <math>x > 2</math>.
  
 
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Similarly, assume that <math>x < 2</math>.  Thus, <math>x^3 < 8</math>, so <math>x^3 - 8 < 0</math> and <math>x^3 + x - 8 < x</math>, so <math>y < x < 2</math>.  By doing the same steps, we can show that <math>z < y</math> and <math>x < z</math>.  However, that would mean that <math>x < x</math>, which does not work, so there are no solutions where <math>x < 2</math>.
+
Similarly, assume that <math>x < 2</math>.  Because <math>x^3</math> is a monotonically increasing function, <math>x^3 < 8</math>, so <math>x^3 - 8 < 0</math> and <math>x^3 + x - 8 < x</math>, so <math>y < x < 2</math>.  By doing the same steps, we can show that <math>z < y</math> and <math>x < z</math>.  However, that would mean that <math>x < x</math>, which does not work, so there are no solutions where <math>x < 2</math>.
  
 
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Latest revision as of 00:16, 19 March 2020

Problem

Find all triples $(x,y,z)$ of real numbers which satisfy the simultaneous equations

\[x = y^3 + y - 8\]

\[y = z^3 + z - 8\]

\[z = x^3 + x - 8.\]

Solution

To start, since all three equations have a similar form, we can let $x = y = z$ to see if there are any solutions. Doing so results in \begin{align*} x &= x^3 + x - 8 \\ 0 &= x^3 - 8 \\ 0 &= (x-2)(x^2 + 2x + 4). \end{align*} Note that $x^2 + 2x + 4 = 0$ has complex solutions, so the solution where $x = y = z$ is $(2,2,2)$.


Additionally, note that $x^3$ and $x$ are monotonically increasing functions, so $x^3 + x - 8$ is a monotonically increasing function. Thus, we can suspect that $(2,2,2)$ is the only solution. To prove this, we can use proof by contradiction.


Assume that $x > 2$. Because $x^3$ is a monotonically increasing function, $x^3 > 8$, so $x^3 - 8 > 0$ and $x^3 + x - 8 > x$, so $y > x > 2$. By doing the same steps, we can show that $z > y$ and $x > z$. However, that would mean that $x > x$, which does not work, so there are no solutions where $x > 2$.


Similarly, assume that $x < 2$. Because $x^3$ is a monotonically increasing function, $x^3 < 8$, so $x^3 - 8 < 0$ and $x^3 + x - 8 < x$, so $y < x < 2$. By doing the same steps, we can show that $z < y$ and $x < z$. However, that would mean that $x < x$, which does not work, so there are no solutions where $x < 2$.


Thus, we proved that $x = 2$ is the only solution, and by substituting the value into the original equations, we get the only solution of $\boxed{(2,2,2)}$.

See Also

2007 Indonesia MO (Problems)
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 7
All Indonesia MO Problems and Solutions
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