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

Solve for integers: (x^2 + y^2 )(x^4 + y^4 ) = (x + y)^6

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

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

(x+y)^6 −(x^4 +y^4 )(x^2 +y^2 )=0 6 x^5 y + 14 x^4 y^2 + 20 x^3 y^3 + 14 x^2 y^4 + 6 x y^5 =0 xy(6x^4 +14x^3 y+20x^2 y^2 +14xy^3 +6y^4 )=0 x=0∣y=0∣6x^4 +14x^3 y+20x^2 y^2 +14xy^3 +6y^4 =0 6x^4 +14x^3 y+20x^2 y^2 +14xy^3 +6y^4 =0 2(x^2 +xy+y^2 )(3x^2 +4xy+3y^2 )=0 x^2 +xy+y^2 =0_(x=0,y=0_(only integer solution) ) ∣ 3x^2 +4xy+3y^2 _(x=0,y=0_(only integer solution) ) =0 Other solutions are complex: ^• x=((−y±(√(y^2 −4y^2 )))/2)=((−y±(√(−3y^2 )))/2) =((−y±iy(√3))/2)∉Z ^• x=((−4y±(√(16y^2 −36y^2 )))/6)∉Z x=0,y=0

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

Commented by MathSh last updated on 12/Oct/21

Very nice dear Ser, thank you

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

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