Question Number 139936 by Snail last updated on 02/May/21
$${Find}\:{all}\:{integers}\:{x}\:{and}\:{y}\:{i}.{e}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{9999}\left({x}−{y}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 04/May/21
$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{9999}\left({x}−{y}\right)+\mathrm{2}{xy} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{4}{xy}=\mathrm{9999}\left({x}−{y}\right)+\mathrm{2}{xy} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} −\mathrm{9999}\left({x}−{y}\right)+\mathrm{2}{xy}=\mathrm{0} \\ $$$$\:\:\:\:{D}=\mathrm{9999}^{\mathrm{2}} −\mathrm{8}{xy}={u}^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{9999}^{\mathrm{2}} −{u}^{\mathrm{2}} =\mathrm{8}{xy} \\ $$$${xy}=\mathrm{0},… \\ $$$$\:\:\:\left(\mathrm{9999}−{u}\right)\left(\mathrm{9999}+{u}\right)=\mathrm{8}{xy} \\ $$$$\:\mathrm{9999}−{u}={m}\wedge\mathrm{9999}+{u}={m}/\mathrm{8}{xy} \\ $$$$\:\:\mathrm{9999}−{u}=\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{8},{x},\mathrm{2}{x},\mathrm{4}{x},\mathrm{8}{x},{y},\mathrm{2}{y},… \\ $$$$\:\mathrm{9999}+{u}=\mathrm{8}{xy},\mathrm{4}{xy},\mathrm{2}{xy} \\ $$$$ \\ $$