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Question Number 139936 by Snail last updated on 02/May/21

Find all integers x and y i.e x^2 +y^2 =9999(x−y)

$${Find}\:{all}\:{integers}\:{x}\:{and}\:{y}\:{i}.{e}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{9999}\left({x}−{y}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 04/May/21

(x+y)^2 =9999(x−y)+2xy (x−y)^2 +4xy=9999(x−y)+2xy (x−y)^2 −9999(x−y)+2xy=0 D=9999^2 −8xy=u^2 9999^2 −u^2 =8xy xy=0,... (9999−u)(9999+u)=8xy 9999−u=m∧9999+u=m/8xy 9999−u=1,2,4,8,x,2x,4x,8x,y,2y,... 9999+u=8xy,4xy,2xy

$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{9999}\left({x}−{y}\right)+\mathrm{2}{xy} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{4}{xy}=\mathrm{9999}\left({x}−{y}\right)+\mathrm{2}{xy} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} −\mathrm{9999}\left({x}−{y}\right)+\mathrm{2}{xy}=\mathrm{0} \\ $$$$\:\:\:\:{D}=\mathrm{9999}^{\mathrm{2}} −\mathrm{8}{xy}={u}^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{9999}^{\mathrm{2}} −{u}^{\mathrm{2}} =\mathrm{8}{xy} \\ $$$${xy}=\mathrm{0},... \\ $$$$\:\:\:\left(\mathrm{9999}−{u}\right)\left(\mathrm{9999}+{u}\right)=\mathrm{8}{xy} \\ $$$$\:\mathrm{9999}−{u}={m}\wedge\mathrm{9999}+{u}={m}/\mathrm{8}{xy} \\ $$$$\:\:\mathrm{9999}−{u}=\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{8},{x},\mathrm{2}{x},\mathrm{4}{x},\mathrm{8}{x},{y},\mathrm{2}{y},... \\ $$$$\:\mathrm{9999}+{u}=\mathrm{8}{xy},\mathrm{4}{xy},\mathrm{2}{xy} \\ $$$$ \\ $$

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