Funny system of equations in three variables

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Funny system of equations in three variables

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Source: Baltic Way 2020, Problem 5

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9027 posts

#1 h Nov 14, 2020, 2:59 PM • 1 Y

Y by Madaminmath

Find all real numbers $x,y,z$ so that
$\begin{align*} x^2 y + y^2 z + z^2 &= 0 \\ z^3 + z^2 y + z y^3 + x^2 y &= \frac{1}{4}(x^4 + y^4). \end{align*}$

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9027 posts

#2 h Nov 15, 2020, 6:55 PM • 3 Y

Y by Bassiskicking, Madaminmath, Arshia.esl

My solution
Motivation

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#3 h Nov 17, 2020, 1:29 PM • 1 Y

Y by Madaminmath

We should find discriminants from the first equation.
Let $D_{x}, D_{y}, D_{z}$ be discriminants for $x, y, z$ respectively. Now we will find them:
$D_{x}= b^2-4ac=-4y(y^2z+z^2)=-4y^3z-4yz^2$
$D_{y}= b^2-4ac=x^4-4z\cdot z^2=x^4-4z^3$
$D_{z}= b^2-4ac=y^4-4x^2y$
Now, $D_{x} \geq 0, D_{y} \geq 0, D_{z} \geq 0$ , so we have a system of inequalities:
$-4y^3z-4yz^2 \geq 0$
$x^4-4z^3 \geq 0$
$y^4-4x^2y \geq 0$
And when we add them we get $\frac{1}{4}(x^4+y^4) \geq z^3 + z^2 y + z y^3 + x^2 y$ which is same as our second condition, so we now have a system of equations from which we get that:
$z=-y^2, x^4=4z^3, y^3=4x^2$ . Now it is easy to get $\frac{y^6}{16}=-4y^6$ which holds only for $y=0$ . Finally $x=y=z=0$ is the only solution.

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#4 h Jan 19, 2021, 10:59 PM • 1 Y

Y by Yasamin.nbh

Notice we can derive two relations from the first equation, which are:

$(I): (z + \frac{y^2}{2})^2 = \frac{y^4}{4} - x^2y$
$(II): (\frac{x^2}{2} + yz)^2 = \frac{x^4}{4} - z^3$
Summing these two, we get:
$(z + \frac{y^2}{2})^2 + (\frac{x^2}{2} + yz)^2 = \frac{x^4 + y^4}{4} - z^3 -x^2y = z^2y + zy^3 = -(xy)^2$

Hence, $(z + \frac{y^2}{2})^2 + (\frac{x^2}{2} + yz)^2 + (xy)^2 = 0$ and now is just conclude that $x=y=z=0$

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#5 h Jan 7, 2023, 2:51 AM

Y by

Even though it's very contrived, I find this problem hilarious. Look at the first equation, where we have
$\begin{align*} \Delta_x \geq 0 &\implies y^3z+yz^2 \leq 0 \\ \Delta_y \geq 0 &\implies x^4-4z^3 \geq 0 \\ \Delta_z \geq 0 &\implies y^4-4x^2 y \geq 0. \end{align*}$ Summing $\frac 14$ times the second and third equations and subtracting the first, we yield the second equality as an equality. Thus, equality must hold in all three of these inequalities, so $x=y=z=0$ .

This post has been edited 1 time. Last edited by HamstPan38825, Jan 7, 2023, 2:52 AM

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#6 h Mar 22, 2023, 12:07 AM

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We can look at the discriminants when the first equation is considered as a quadratic in $x$ , $y$ , or $z$ , which gives us the inequalities
$\begin{align*}z^2y+zy^3&\le 0 \\ x^4-4z^3&\ge 0 \\ y^4-4x^2y&\ge 0. \end{align*}$ Now look at these, considering the second equation. Note that $\frac{1}{4}(x^4+y^4)\ge x^2y+z^3$ . So we can cancel out those terms, and the remaining inequality is
$z^2y+zy^3\ge 0.$ However, from our above discriminant-derived inequalites, $z^2y+zy^3\le 0$ , which means $yz(y^2+z)=0$ .
Assume now that none of $x$ , $y$ , or $z$ is $0$ . So $z=-y^2$ , so $x^4\ge -4y^6$ and $x^2\le \frac{y^3}{4}$ . Then $\frac{y^6}{16}\le -4y^6$ , clearly impossible unless $y=0$ . So the only solution is $(x, y, z)=(0, 0, 0)$ . $\blacksquare$

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pqr.

#7 h Aug 27, 2023, 3:50 PM

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We claim that the only solution is $(x, y, z) = (0, 0, 0)$ . This clearly works. Consider the discriminant of the first equation as a quadratic in $x, y, z$ respectively. We get $\Delta_x=-4y(zy^2+z^2) \ge 0 \Rightarrow 0 \ge zy^3+z^2y$ $\Delta_y=x^4-4z^3 \ge 0 \Rightarrow x^4 \ge 4z^3$ $\Delta_z=y^4-4x^2y \ge 0 \Rightarrow y^4 \ge 4x^2y.$ Adding the last two inequalities, $\frac14(x^4+y^4) \ge z^3+x^2y.$ But by the second equation, this means that the equality case of the first inequality must hold. From this we deduce that $z=-y^2$ , but plugging this into the first original equation yields $x^2y=0$ , so either $x=0$ or $y=0$ . We can check that both cases do indeed give $(x, y, z)=(0, 0, 0)$ , as desired.

This post has been edited 1 time. Last edited by pqr., Aug 27, 2023, 3:51 PM

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#8 h Oct 16, 2023, 2:05 AM

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Viewing the first equation as a quadratic in $x,y,z$ , we get

$\begin{align*} \Delta_x &= -4(y^3z+yz^2) \ge 0 \\ \Delta y &= x^4-4z^3 \ge 0 \\ \Delta z &= y^4-4x^2y \ge 0 \end{align*}$
Adding the equations and rearranging we get the second equality as an inequality, implying that $x=y=z=0$ . Thus, the only ordered pair is $\boxed{(0,0,0)}$ .

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eibc

#9 h Jan 1, 2024, 10:37 PM

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what in the world

The only solution is $(x, y, z) = (0, 0, 0)$ . This clearly works.

View the first equation as a quadratic in $x$ , and let $D_x$ be the discriminant; then $D_x = -4y^3z - 4yz^2$ . Defining $D_y$ and $D_z$ similarly, we have $D_y = x^4 - 4z^3, D_z = y^4 - 4x^2y$ . For $x$ , $y$ , and $z$ to be real, we must have $D_x, D_y, D_z \ge 0$ ; but the second equation rewrites as
$x^4 + y^4 - 4z^3 - 4z^2y - 4zy^3 - 4x^2y = 0 \implies D_x + D_y + D_z = 0,$ so in fact we have equality and $D_x = D_y = D_z = 0$ .

Looking at the equation for $D_x$ , we have $yz(y^2 - z) = 0$ . Then

If $y = 0$ , the original first equation gives $z = 0$ , and then the equation for $D_y$ gives $x = 0$
If $z = 0$ , the equation for $D_y$ gives $x = 0$ , and the original first equation gives $y = 0$
If $y^2 = z$ , then the equation for $D_y$ gives $x^2 = \pm 2y^3$ , so from the third equation we have $y = 0$ , which we already handled

Thus $(0, 0, 0)$ is indeed the only solution.

This post has been edited 1 time. Last edited by eibc, Jan 1, 2024, 10:37 PM

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#10 h Jan 22, 2024, 12:27 AM

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Consider the first equation as a quadratic wrt $x$ , $y$ , and $z$ .
Let $\Delta_{\chi}$ be the discriminant of a quadratic in variable $\chi$ .
Then
$\begin{align*} \\ \Delta_x = -4y^3z - 4yz^2 \geq 0 \implies 0 \geq zy(y^2 + z) \\ \Delta_y = x^4 - 4z^3 \geq 0 \implies x^4 \geq 4z^3 \\ \Delta_z = y^4 - 4x^2y \geq 0 \implies y^4 \geq 4x^2y \\ \end{align*}$
Summing the last two inequalities gives us
$x^4 + y^4 \geq 4(z^3 + x^2y) \implies \frac{1}{4}(x^4 + y^4) \geq z^3 + x^2y$ And adding the last inequality gives
$\frac{1}{4}(x^4 + y^4) \geq z^3 + x^2y + z^2y + zy^3$ And since this is the second original equality, we get that $x = y = z = 0$ must be the only solution.

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#11 h Mar 23, 2025, 5:23 PM

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Interesting problem

This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 23, 2025, 5:24 PM

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