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Question Number 194942 by horsebrand11 last updated on 20/Jul/23

Commented by MJS_new last updated on 20/Jul/23

$$-2$$

Commented by horsebrand11 last updated on 20/Jul/23

$$\mathrm{how}?$$

Answered by MJS_new last updated on 20/Jul/23

$$2x=2y^3+(1+\sqrt{5})y$$$$2y=2x^3+(1+\sqrt{5})x$$$$y=px$$$$2x=2p^3{x}^3+(1+\sqrt{5})px$$$$2px=2x^3+(1+\sqrt{5})x$$$$x=0\Rightarrow y=0\mathrm{but}x\ne y\Rightarrow x\ne 0$$$$2=p(2p^2{x}^2+1+\sqrt{5})$$$$2p=2x^2+1+\sqrt{5}$$$$==========$$$$x^2=\frac{2-(1+\sqrt{5})p}{2p^3}=\frac{2p-(1+\sqrt{5})}{2}$$$$p^4-\frac{1+\sqrt{5}}{2}{p}^3+\frac{1+\sqrt{5}}{2}p-1=0$$$$\mathrm{we}\mathrm{know}\mathrm{that}x=\pm y\Rightarrow p=\pm 1\mathrm{are}\mathrm{solutions},$$$$\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{factorize}$$$$(p-1)(p+1)(p^2-\frac{1+\sqrt{5}}{2}+1)=0$$$$\mathrm{but}x\ne y\wedge x+y\ne 0$$$$p^2-\frac{1+\sqrt{5}}{2}+1=0$$$$p={\mathrm{e}}^{\pm \mathrm{i}\pi /5}$$$$y=px\Rightarrow \frac{x}{y}=\frac{1}{p}\wedge \frac{y}{x}=p$$$${\left(\frac{x}{y}\right)}^{2025}+{\left(\frac{y}{x}\right)}^{2025}={\mathrm{e}}^{405\pi \mathrm{i}}+{\mathrm{e}}^{-405\pi \mathrm{i}}={\mathrm{e}}^{\mathrm{i}\pi }+{\mathrm{e}}^{-\mathrm{i}\pi }=$$$$=-2$$

Commented by horsebrand11 last updated on 20/Jul/23

$$\mathrm{it}\mathrm{does}\mathrm{meant}\mathrm{x},\mathrm{y}\in \mathrm{C}?$$

Commented by MJS_new last updated on 20/Jul/23

$$\mathrm{yes}$$

Commented by horsebrand11 last updated on 20/Jul/23

$$\mathrm{thanks}\mathrm{sir}$$

Commented by MJS_new last updated on 20/Jul/23

$$x,y\in \mathbb{R}\Rightarrow$$$$x=y=0\vee x=y=\frac{1-\sqrt{5}}{2}\vee (x=\pm \frac{3+\sqrt{5}}{2}\wedge y=-x)$$

Commented by MM42 last updated on 20/Jul/23

$$Accordingtothegivenequtions$$$$x\ne y\mathrm{\&}x\ne -y$$

Commented by Frix last updated on 20/Jul/23

$$\mathrm{He}\mathrm{only}\mathrm{stated}\mathrm{that}\mathrm{if}x,y\mathrm{being}\mathrm{real}\mathrm{these}$$$$\mathrm{solutions}\mathrm{follow}.\mathrm{But}\mathrm{you}\mathrm{changed}\mathrm{your}$$$$\mathrm{answer}\mathrm{below}...$$

Answered by MM42 last updated on 20/Jul/23

$$2xy=2y^4+(\sqrt{5}+1)y^2$$$$2xy=2x^4+(\sqrt{5}+1)x^2$$$$2(y^4-x^4)+(\sqrt{5}+1)(y^2-x^2)=0$$$$(y^2-x^2)(2y^2+2x^2+\sqrt{5}+1)=0$$$$ifx,y\in \mathbb{R}\Rightarrow y=\pm x\Rightarrow Accordingtotheissueis$$$$unanswerd.$$$$ifx,y\in \mathbb{C}\Rightarrow y^2+x^2=-\frac{\sqrt{5}+1}{2}$$$$$$

Commented by MJS_new last updated on 20/Jul/23

$$\mathrm{but}\mathrm{it}\mathrm{says}x\ne y\mathrm{and}x+y\ne 0$$

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