The two numbers such that each one is the squ | vedclass

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(D) Let the two numbers be $x$ and $y$. According to the problem,$y = x^2$ and $x = y^2$. Substituting the first into the second,we get $x = (x^2)^2 =

Vedclass · Accepted Answer

$\omega, \omega^2$

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(D) Let the two numbers be $x$ and $y$. According to the problem,$y = x^2$ and $x = y^2$. Substituting the first into the second,we get $x = (x^2)^2 = x^4$. This implies $x^4 - x = 0$,or $x(x^3 - 1) = 0$. The roots are $x = 0$ or $x^3 = 1$. If $x = 0$,then $y = 0^2 = 0$. If $x^3 = 1$,the roots are $1, \omega, \omega^2$. For the pair $(\omega, \omega^2)$,we have $(\omega)^2 = \omega^2$ and $(\omega^2)^2 = \omega^4 = \omega \cdot \omega^3 = \omega \cdot 1 = \omega$. Thus,the numbers are $\omega$ and $\omega^2$.

The two numbers such that each one is the square of the other are:

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