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Question Number 134615 by mathocean1 last updated on 05/Mar/21

show that for x; y ∈ Z: (x+6y)^4 −x^4 is divisible by 24.

$${show}\:{that}\:{for}\:{x};\:{y}\:\in\:\mathbb{Z}: \\ $$$$\left({x}+\mathrm{6}{y}\right)^{\mathrm{4}} −{x}^{\mathrm{4}} \:{is}\:{divisible}\:{by}\:\mathrm{24}. \\ $$

Answered by Olaf last updated on 05/Mar/21

Let X = (x+6y)^4 −x^4 X = (x+6y−x)[(x+6y)^3 +x(x+6y)^2 +x^2 (x+6y)+x^3 ] X = 6y[(x+6y)^3 +x(x+6y)^2 +x^2 (x+6y)+x^3 ] ⇒ X is divisible by 6, X = 6Y Y = (x+6y)^3 +x(x+6y)^2 +x^2 (x+6y)+x^3 Y = (x^3 +18x^2 y+108xy^2 +196y^3 ) +(x^3 +12x^2 y+36xy^2 )+(x^3 +6x^2 y)+x^3 Y = 4(x^3 +9x^2 y+36xy^2 +49y^3 ) ⇒ Y is divisible by 4, Y = 4p Finally, X = 6Y = 24p ⇒ X is divisible by 24

$$\mathrm{Let}\:\mathrm{X}\:=\:\left({x}+\mathrm{6}{y}\right)^{\mathrm{4}} −{x}^{\mathrm{4}} \\ $$$$\mathrm{X}\:=\:\left({x}+\mathrm{6}{y}−{x}\right)\left[\left({x}+\mathrm{6}{y}\right)^{\mathrm{3}} +{x}\left({x}+\mathrm{6}{y}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left({x}+\mathrm{6}{y}\right)+{x}^{\mathrm{3}} \right] \\ $$$$\mathrm{X}\:=\:\mathrm{6}{y}\left[\left({x}+\mathrm{6}{y}\right)^{\mathrm{3}} +{x}\left({x}+\mathrm{6}{y}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left({x}+\mathrm{6}{y}\right)+{x}^{\mathrm{3}} \right] \\ $$$$\Rightarrow\:\mathrm{X}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{6},\:\mathrm{X}\:=\:\mathrm{6Y} \\ $$$$ \\ $$$$\mathrm{Y}\:=\:\left({x}+\mathrm{6}{y}\right)^{\mathrm{3}} +{x}\left({x}+\mathrm{6}{y}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left({x}+\mathrm{6}{y}\right)+{x}^{\mathrm{3}} \\ $$$$\mathrm{Y}\:=\:\left({x}^{\mathrm{3}} +\mathrm{18}{x}^{\mathrm{2}} {y}+\mathrm{108}{xy}^{\mathrm{2}} +\mathrm{196}{y}^{\mathrm{3}} \right) \\ $$$$+\left({x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} {y}+\mathrm{36}{xy}^{\mathrm{2}} \right)+\left({x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} {y}\right)+{x}^{\mathrm{3}} \\ $$$$\mathrm{Y}\:=\:\mathrm{4}\left({x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{2}} {y}+\mathrm{36}{xy}^{\mathrm{2}} +\mathrm{49}{y}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\:\mathrm{Y}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{4},\:\mathrm{Y}\:=\:\mathrm{4}{p} \\ $$$$ \\ $$$$\mathrm{Finally},\:\mathrm{X}\:=\:\mathrm{6Y}\:=\:\mathrm{24}{p} \\ $$$$\Rightarrow\:\mathrm{X}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{24} \\ $$

Answered by Rasheed.Sindhi last updated on 06/Mar/21

(x+6y)^4 −x^4 ={(x+6y)^2 −x^2 }{(x+6y)^2 +x^2 } =(x+6y−x)(x+6y+x)(2x^2 +12xy+36y^2 ) =6y.2(x+3y).2(x^2 +6xy+18y^2 ) =24y(x+3y)(x^2 +6xy+18y^2 ) ∴ The given expression is divisibile by 24.

$$\left({x}+\mathrm{6}{y}\right)^{\mathrm{4}} −{x}^{\mathrm{4}} \\ $$$$=\left\{\left({x}+\mathrm{6}{y}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right\}\left\{\left({x}+\mathrm{6}{y}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} \right\} \\ $$$$=\left({x}+\mathrm{6}{y}−{x}\right)\left({x}+\mathrm{6}{y}+{x}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{12}{xy}+\mathrm{36}{y}^{\mathrm{2}} \right) \\ $$$$=\mathrm{6}{y}.\mathrm{2}\left({x}+\mathrm{3}{y}\right).\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{6}{xy}+\mathrm{18}{y}^{\mathrm{2}} \right) \\ $$$$=\mathrm{24}{y}\left({x}+\mathrm{3}{y}\right)\left({x}^{\mathrm{2}} +\mathrm{6}{xy}+\mathrm{18}{y}^{\mathrm{2}} \right) \\ $$$$\therefore\:{The}\:{given}\:{expression}\:{is}\:{divisibile} \\ $$$${by}\:\mathrm{24}. \\ $$

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