3 equations with 3 (supposedly) real unknowns

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3 equations with 3 (supposedly) real unknowns

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#1 h Jul 15, 2023, 5:58 PM • 4 Y

Y by Snark_Graphique, David_Kim_0202, KST2003, Amir Hossein

Find all real numbers $a, b, c$ that satisfy
$2a - b =a^2b, \qquad 2b-c = b^2 c, \qquad 2c-a= c^2 a.$

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#2 h Jul 15, 2023, 6:33 PM • 2 Y

Y by OptimalFEian, Amir Hossein

Note that $a=f(f(f(a)))$ where $f(x)=\frac{2x}{1+x^2}$ .

Fixed points: $\frac{2x}{1+x^2}=x\implies x^3-x=0\implies x(x+1)(x-1)=0\implies x\in\{-1,0,1\}$ .

Three-cycle points:
Since $\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$ , we have $f(i\tan\theta)=i\tan 2\theta$ and so we need $\tan 8\theta=\tan\theta\implies\theta=\frac{\pi k}{7}$ for $k\in\mathbb{Z}$ .

This gives the following results:

Real:
$(-1,-1,-1)$ , $(0,0,0)$ , $(1,1,1)$

Complex:
$\left(i\tan\frac{\pi}{7},i\tan\frac{2\pi}{7},i\tan\frac{4\pi}{7}\right)$ , $\left(i\tan\frac{2\pi}{7},i\tan\frac{4\pi}{7},i\tan\frac{\pi}{7}\right)$ , $\left(i\tan\frac{4\pi}{7},i\tan\frac{\pi}{7},i\tan\frac{2\pi}{7}\right)$

Only three real solutions!

This post has been edited 1 time. Last edited by rchokler, Jul 15, 2023, 6:34 PM

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#3 h Jul 5, 2024, 5:39 AM

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So, here is the solution by the author.
It is obvious that the triplets $(0, 0, 0), (1, 1, 1), (-1, -1, -1)$ satisfy the given system. Now, we will prove that these are the only solutions.
Firstly, one can easily verify that if one of $a, b, c$ is zero, then all are zero. Thus, we look for non-zero solutions. We can rewrite the first equation as
$\frac{2a}{b} = a^2 + 1$ implying that $a$ and $b$ have the same sign. Similarly, we can deduce that $b$ and $c$ , $c$ and $a$ must have the same sign. Thus, $a, b, c$ are either all positive or all negative. W.L.O.G we assume the first case; otherwise, we can multiply each of the equations by $-1$ .
Adding the three original equations gives
$a + b + c = a^2 b + b^2 c + c^2 a.$ Moreover, we can rewrite the given equations as
$\frac{2}{b}-\frac{1}{a} = a, \qquad \frac{2}{c} - \frac{1}{b} = b, \qquad \frac{2}{a} - \frac{1}{c} = c.$ Adding these equations gives
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = a+b+c$ To recall, we have
$a^2 b +b^2 c +c^2a = a + b+ c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ and let us denote this value by $S$ . Now, we can safely apply Cauchy's Inequality so that
$S^2 = (a^2 b+b^2c+c^2a)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right) \geq (a+b+c)^2 = S^2 .$ Thus, we must have
$a^2 b^2 = b^2 c^2 = c^2 a^2 \implies a=b=c$ and substituting back in the given system gives $a=b=c=1$ .

This post has been edited 1 time. Last edited by OptimalFEian, Jul 5, 2024, 5:40 AM

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