(x^2 +x+1)^2 +(y^2 −y+1)^2 =2004^2 x;y∈Z x;y=?


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Question Number 185087 by SEKRET last updated on 16/Jan/23

${(x^{2} + x + 1)}^{2} + {(y^{2} - y + 1)}^{2} = 2004^{2}$ $x; y \in Z$ $x; y = ?$
	Answered by floor(10²Eta[1]) last updated on 16/Jan/23

	$a \equiv 0 (\mod 8) \Rightarrow a^{2} \equiv 0 (\mod 8)$ $a \equiv 1 (\mod 8) \Rightarrow a^{2} \equiv 1 (\mod 8)$ $a \equiv 2 (\mod 8) \Rightarrow a^{2} \equiv 4 (\mod 8)$ $a \equiv 3 (\mod 8) \Rightarrow a^{2} \equiv 1 (\mod 8)$ $a \equiv 4 (\mod 8) \Rightarrow a^{2} \equiv 0 (\mod 8)$ $a \equiv 5 (\mod 8) \Rightarrow a^{2} \equiv 1 (\mod 8)$ $a \equiv 6 (\mod 8) \Rightarrow a^{2} \equiv 4 (\mod 8)$ $a \equiv 7 (\mod 8) \Rightarrow a^{2} \equiv 1 (\mod 8)$ ${(x^{2} + x + 1)}^{2} + {(y^{2} - y + 1)}^{2} \equiv 0 (\mod 8)$ $1) x^{2} + x + 1 \equiv 0 (\mod 8) \land y^{2} - y + 1 \equiv 4 (\mod 8)$ $2) x^{2} + x + 1 \equiv 4 (\mod 8) \land y^{2} - y + 1 \equiv 0 (\mod 8)$ $3) x^{2} + x + 1 \equiv 2 (\mod 8) \land y^{2} - y + 1 \equiv 6 (\mod 8)$ $4) x^{2} + x + 1 \equiv 6 (\mod 8) \land y^{2} - y + 1 \equiv 2 (\mod 8)$ $1) no solution$ $2) no sol .$ $3) no sol .$ $4) no sol .$
	Commented by Rasheed.Sindhi last updated on 17/Jan/23

	$W h y n o t$ $5) x^{2} + x + 1 \equiv 4 (\mod 8) \land y^{2} - y + 1 \equiv 4 (\mod 8) ?$ $H o w e v e r i n t h i s c a s e a l s o n o s o l u t i o n$
	Answered by Frix last updated on 17/Jan/23

	$(x^{2} + x + 1) > 0$ $(y^{2} - y + 1) > 0$ $a^{2} + b^{2} = 2004^{2} with a, b \neq 0$ ${there}^{'} s no such Pythagorean triple \Rightarrow$ $no solution$
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