Solve : x^2 +2x−1≡0(mod 3)


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Question Number 114592 by Aziztisffola last updated on 19/Sep/20

$Solve : x^{2} + 2 x - 1 \equiv 0 (\mod 3)$
	Answered by MJS_new last updated on 19/Sep/20

	$x^{2} + 2 x - 1 = 3 k with k \in Z$ $\Rightarrow x = - 1 \pm \sqrt{3 k + 2}$ $if x \in R \Rightarrow k \in N$
	Commented by Aziztisffola last updated on 19/Sep/20

	$x a l s o i n Z . wich k ?$
	Commented by Aziztisffola last updated on 19/Sep/20

	$2 + 3 k is not a perfect square .$
	Commented by MJS_new last updated on 19/Sep/20

	${there}^{'} s no x \in Z$
	Commented by Aziztisffola last updated on 19/Sep/20

	$yes sir no solution in Z .$
	Answered by floor(10²Eta[1]) last updated on 19/Sep/20

	$x^{2} + 2 x \equiv 1 (\mod 3)$ $x \equiv 0 (\mod 3) \Rightarrow x^{2} + 2 x ≢ 1 (\mod 3)$ $x \equiv 1 (\mod 3) \Rightarrow x^{2} + 2 x ≢ 1 (\mod 3)$ $x \equiv 2 (\mod 3) \Rightarrow x^{2} + 2 x ≢ 1 (\mod 3)$ $\Rightarrow ∄ x \in Z such that : x^{2} + 2 x - 1 \equiv 0 (\mod 3)$
	Commented by Aziztisffola last updated on 19/Sep/20

	$yes sir {that}^{'} s it .$ $x^{2} + 2 x - 1 \equiv 0 (\mod 3)$ $\Leftrightarrow {(x + 1)}^{2} \equiv 2 [3]$ $\Leftrightarrow {(x + 1)}^{2} = 2 + 3 k / k \in Z$ $and 2 + 3 k \neq n^{2} for n in Z .$ $then no solution .$
	Answered by 1549442205PVT last updated on 20/Sep/20

	$x^{2} + 2 x - 1 \equiv 0 (\mod 3) ()$ $\forall x \in Z we always have x = 3 k \pm 1 or x = 3 k$ $i) If x = 3 k then x^{2} + 2 x - 1 = - 1 (\mod 3)$ $\Rightarrow () has no roots (1)$ $ii) If x = 3 k + 1 then {(x + 1)}^{2} - 2 =$ ${(3 k + 2)}^{2} - 2 = {9 k}^{2} + 12 k + 2) = 2 (\mod 3) (2)$ $iii) If x = 3 k - 1 then {(x + 1)}^{2} - 2 = {9 k}^{2} - 2$ $= - 2 (\mod 3) (3)$ $From (1) (2) (3) infer the equation (*)$ $has no roots in Z$
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