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Question Number 197708 by karowan last updated on 26/Sep/23

$$$$$$x^a=x^{a+4}\mathrm{where}a\in \mathbb{Z}$$$$\mathrm{solve}\mathrm{for}x\mathrm{by}\mathrm{showing}\mathrm{steps}$$

Answered by Frix last updated on 26/Sep/23

$$x^{a+4}-x^a=0$$$$x^a(x-1)(x+1)(x^2+1)=0$$$$a>0\wedge x=0$$$$x=\pm 1$$$$x=\pm \mathrm{i}$$

Commented by karowan last updated on 26/Sep/23

correct

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