An easy 3 variable equation

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An easy 3 variable equation

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Source: Turkey National Mathematical Olympiad 2022 P4

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#1 h Dec 23, 2022, 6:39 PM

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For which real numbers $a$ , there exist pairwise different real numbers $x, y, z$ satisfying
$\frac{x^3+a}{y+z}=\frac{y^3+a}{x+z}=\frac{z^3+a}{x+y}= -3.$

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#2 h Dec 24, 2022, 9:47 AM • 2 Y

Y by farhad.fritl, Japnoor2411

Let $p=\sum x$ and $q=\sum xy$ . We have that $x^3-3x=y^3-3y=z^3-3z=a-3x-3y-3z$ . So we get that $x,y,z$ are roots of polynomial $P(t)=t^3-3t+c$ , where $c=3p-a$ . From Vieta we get $p=0$ and $q=-3$ which means $c=-a$ and we should find possible values of $a=t^3-3t=x^3-3x$ , where $x,y$ are roots of $x^2+xy+y^2=3$ . Since $x\ne y\ne z=-x-y\ne x$ , we get $x\ne y, x\ne -2y, y\ne -2x$ . So, $x,y\ne \pm 1,\pm2$ . Also since $y^2-y\cdot x+ (x^2-3)$ has roots, we get $|x|\le 2 \implies -2\le (x-1)^2(x+2)-2= x^3-3x=a=(x+1)^2(x-2)+2\le 2$ . Since $x\ne \pm 1, \pm 2$ , we get $a\in(-2,2)$ .
Since we should also have $xy(x+y)\ne 0$ because of denominator can't be $0$ . So, $a=0$ doesn't work. So, the answer is $a\in(-2,2)\setminus \{0\}$ .
@below thanks.

This post has been edited 2 times. Last edited by Jalil_Huseynov, Dec 26, 2022, 7:04 PM
Reason: Below instead of above

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#3 h Dec 24, 2022, 12:14 PM

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Note that $a \neq 0$ since sum of the two variables has to be $0$ which makes the denominator $0$

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#4 h Dec 25, 2022, 10:15 AM • 2 Y

Y by Mango247, Japnoor2411

From the ratio identities we can easily obtain that $x^2+xy+y^2 = 3$ , $y^2+yz+z^2 = 3$ and $z^2+zx+x^2 = 3$ . Use any two of these three new equations to get $x + y + z = 0$ . Plugging this back to the given equations, we get that $x$ , $y$ and $z$ are the three distinct real roots of $m^3-3m+a = 0$ . Also, note that since the denominators cannot be zero, $0$ cannot be a root of this equation and therefore $a \neq 0$ . Now observe that for $a = 2$ , $m^3-3m+2=(m-1)^2(m+2)$ and for $a = -2$ , $m^3-3m-2=(m+1)^2(m-2)$ . One may easily observe that using graph shifting about the $y$ -axis, there cannot be three distinct real roots if $a > 2$ or $a < -2$ . And finally, if $a \neq 2$ and $a \neq -2$ , the graph of $m^3-3m+a$ will cut the $x$ -axis at three points, so our final answer is $a \in (-2, 2) - \{0\}$ .

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lifeismathematics

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lifeismathematics

#5 h Jan 21, 2023, 3:21 PM • 1 Y

Y by Japnoor2411

nice problem

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#8 h Nov 3, 2023, 9:48 AM • 1 Y

Y by Japnoor2411

With some algebratic calculations and equating the ratios, we get $x + y + z = 0$ . If we write $y + z = -x$ , then we have $x^3 +a = 3x$ . At the end, we have a equation of the third degree $x^3 -3x +a =0$ .
The situations are symmetric. So :
$y^3 -3y +a =0$ .
$z^3 -3z + a=0$ .
Notice that $a\neq{0}$ .
Due to $x, y, z$ are different real numbers, the equation $t^3 -3t +a=0$ must have three distinct real roots. And these roots are $x, y$ and $z$ .
If $a \in (-2,2)-$ $\{ 0 \}$ , then, the equation $t^3 -3t +a$ has three distinct real roots.
If $a= \{2,2\}$ , then the equation has two distinct real roots.
If $a<-2$ , then the equation has just one real root.
The case of $a>2$ is same with $a<-2$ .
So, the answer is : $a \in (-2,2)- \{0\}$ . $\square$

This post has been edited 24 times. Last edited by Turker31, Aug 9, 2024, 2:45 PM

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#10 h Apr 4, 2025, 2:45 PM • 1 Y

Y by Japnoor2411

So $b=x+y+z$ $x^3+a=-3y-3z$ $P(t)=t^3-3t+a-3b\implies x+y+z=0$ from vieta $x^3+a=-3y-3z,y^3+a=-3x-3z\implies x^2+xy+y^2=3\implies y^2+xy+(x^2-3)=0\implies \Delta\ge 0 \implies 12\ge 3x^2 \implies x\in [-2,2]$ . Notice in similar fashion we get $y\in [-2,2] , z\in [-2,2]$ , also its easy to observe that none of $\{x,y,z\} \in \{-2,2,0\}$ . So we have $\boxed{a \in (-2,2)-\{0\}}$

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