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Question Number 137848 by Bekzod Jumayev last updated on 07/Apr/21

Solve the equation  (x^2 −x+1)^2 +(y^2 +y+1)^2 =2021  x,y∈Z

$$\boldsymbol{{Solve}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}} \\ $$$$\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2021} \\ $$$$\boldsymbol{{x}},\boldsymbol{{y}}\in\boldsymbol{{Z}} \\ $$

Answered by mr W last updated on 07/Apr/21

x^2 −x=(x−1)x=even  x^2 −x+1=odd  (x^2 −x+1)^2 =odd^2 =odd  similarly  (y^2 +y+1)^2 =odd  (x^2 −x+1)^2 +(y^2 +y+1)^2 =odd+odd=even  but 2021 is odd, therefore  (x^2 −x+1)^2 +(y^2 +y+1)^2 =2021  has no solution.

$${x}^{\mathrm{2}} −{x}=\left({x}−\mathrm{1}\right){x}={even} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}={odd} \\ $$$$\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} ={odd}^{\mathrm{2}} ={odd} \\ $$$${similarly} \\ $$$$\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)^{\mathrm{2}} ={odd} \\ $$$$\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)^{\mathrm{2}} ={odd}+{odd}={even} \\ $$$${but}\:\mathrm{2021}\:{is}\:{odd},\:{therefore} \\ $$$$\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2021} \\ $$$${has}\:{no}\:{solution}. \\ $$

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