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Source: IMC 2000 day 1 problem 2

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Peter

#1 h Oct 29, 2005, 1:03 PM • 2 Y

Y by Adventure10, brain_idea

Let $p(x)=x^5+x$ and $q(x)=x^5+x^2$ , Find al pairs $(w,z)\in \mathbb{C}\times\mathbb{C}$ , $w\not=z$ for which $p(w)=p(z),q(w)=q(z)$ .

This post has been edited 1 time. Last edited by Peter, Oct 29, 2005, 1:50 PM

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#2 h Oct 29, 2005, 7:06 PM • 3 Y

Y by Adventure10, brain_idea, Mango247

see http://www.mathlinks.ro/Forum/viewtopic.php?t=56172

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#3 h Jun 20, 2023, 10:28 AM

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From $p(w)-q(w) = p(z)-q(z)$ we have
$w-w^2 = z-z^2 \quad \Longrightarrow \quad (w-z)(w+z-1) = 0 \quad \Longrightarrow \quad w = 1-z$ $p(w) = p(z) \quad \Longrightarrow \quad w^5 + w = z^5 + z \quad \Longrightarrow \quad (w-z)(w^4+w^3z+w^2z^2+wz^3+z^4+1) = 0$ $w^4+w^3z+w^2z^2+wz^3+z^4+1 = 0 \; ........(1)$ Substitute $w = 1-z$ to $(1)$ , we get
$(z^2-z+1)(z^2-z+2) = 0$ $z_{1,2} = \frac{1}{2}\pm\frac{i\sqrt{3}}{2} \quad \textrm{or} \quad z_{3,4} = \frac{1}{2}\pm\frac{i\sqrt{7}}{2}$ The answers are
$(w,z) = \left(\frac{1}{2}+\frac{i\sqrt{3}}{2}, \frac{1}{2}-\frac{i\sqrt{3}}{2}\right), \left(\frac{1}{2}-\frac{i\sqrt{3}}{2}, \frac{1}{2}+\frac{i\sqrt{3}}{2}\right), \left(\frac{1}{2}+\frac{i\sqrt{7}}{2}, \frac{1}{2}-\frac{i\sqrt{7}}{2}\right), \left(\frac{1}{2}-\frac{i\sqrt{7}}{2}, \frac{1}{2}+\frac{i\sqrt{7}}{2}\right).$ Note: It is clear to verify that these solutions satisfy for the condition of the problem.

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lifeismathematics

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lifeismathematics

#4 h Jul 16, 2023, 1:37 PM

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first we have $w^{5}+w=z^5+z$ and $w^5+w^2=z^5+z^2 \implies z-w=z^2-w^2 \implies z+w=1$

we have $w^2+z^2=1-2wz$ and $w^4+z^4=1+2w^2z^2-4wz$

So,alo we have $w^4+z^4+wz(w^2+z^2+wz)=-1\implies wz=1,2$

so we get: $w=(\varepsilon ,\varepsilon^2)$ and $z=(1-\varepsilon , 1-\varepsilon^2)$ , where $\varepsilon^3=1$

and also $w=\frac{1 \pm \sqrt{7}i}{2}$ , so $z=\frac{1 \mp \sqrt{7}i}{2}$

$\square$

This post has been edited 2 times. Last edited by lifeismathematics, Jul 16, 2023, 1:38 PM

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