Question Number 200617 by sonukgindia last updated on 21/Nov/23
Answered by Frix last updated on 21/Nov/23
$${x}+{a}={y}^{\mathrm{2}} \\ $$$${y}+{a}={x}^{\mathrm{2}} \\ $$$${x}−{y}={y}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{x}−{y}=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}\right)+\left({x}−{y}\right)=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${y}={x}\vee{y}=−\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{Now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{do}\:\mathrm{the}\:\mathrm{rest}. \\ $$