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Question Number 156502 by MathSh last updated on 11/Oct/21

$$\mathrm{Solve}\mathrm{for}\mathrm{integers}:$$$$({\mathrm{x}}^2+{\mathrm{y}}^2)({\mathrm{x}}^4+{\mathrm{y}}^4)={(\mathrm{x}+\mathrm{y})}^6$$$$$$

Answered by Rasheed.Sindhi last updated on 12/Oct/21

$$$$$${(\mathrm{x}+\mathrm{y})}^6-({\mathrm{x}}^4+{\mathrm{y}}^4)({\mathrm{x}}^2+{\mathrm{y}}^2)=0$$$$6{\mathrm{x}}^5\mathrm{y}+14{\mathrm{x}}^4{\mathrm{y}}^2+20{\mathrm{x}}^3{\mathrm{y}}^3+14{\mathrm{x}}^2{\mathrm{y}}^4+6\mathrm{x}{\mathrm{y}}^5=0$$$$\mathrm{xy}({6\mathrm{x}}^4+{14\mathrm{x}}^3\mathrm{y}+{20\mathrm{x}}^2{\mathrm{y}}^2+{14\mathrm{xy}}^3+{6\mathrm{y}}^4)=0$$$$\mathrm{x}=0\mid \mathrm{y}=0\mid {6\mathrm{x}}^4+{14\mathrm{x}}^3\mathrm{y}+{20\mathrm{x}}^2{\mathrm{y}}^2+{14\mathrm{xy}}^3+{6\mathrm{y}}^4=0$$$${6\mathrm{x}}^4+{14\mathrm{x}}^3\mathrm{y}+{20\mathrm{x}}^2{\mathrm{y}}^2+{14\mathrm{xy}}^3+{6\mathrm{y}}^4=0$$$$2({\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2)({3\mathrm{x}}^2+4\mathrm{xy}+{3\mathrm{y}}^2)=0$$$$\underset{\underset{\mathrm{only}\mathrm{integer}\mathrm{solution}}{\mathrm{x}=0,\mathrm{y}=0}}{{\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2=0}\mid \underset{\underset{\mathrm{only}\mathrm{integer}\mathrm{solution}}{\mathrm{x}=0,\mathrm{y}=0}}{{3\mathrm{x}}^2+4\mathrm{xy}+{3\mathrm{y}}^2}=0$$$$\mathrm{Other}\mathrm{solutions}\mathrm{are}\mathrm{complex}:$$$${}^{\bullet }\mathrm{x}=\frac{-\mathrm{y}\pm \sqrt{{\mathrm{y}}^2-{4\mathrm{y}}^2}}{2}=\frac{-\mathrm{y}\pm \sqrt{-{3\mathrm{y}}^2}}{2}$$$$=\frac{-\mathrm{y}\pm \mathrm{iy}\sqrt{3}}{2}\notin \mathbb{Z}$$$${}^{\bullet }\mathrm{x}=\frac{-4\mathrm{y}\pm \sqrt{{16\mathrm{y}}^2-{36\mathrm{y}}^2}}{6}\notin \mathbb{Z}$$$$\mathrm{x}=0,\mathrm{y}=0$$

Commented by MathSh last updated on 12/Oct/21

$$\mathrm{Very}\mathrm{nice}\mathrm{dear}\mathbf{S}\mathrm{er},\mathrm{thank}\mathrm{you}$$

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