Solve-x-2-2x-1-0-mod-3-

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