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

Solve : x^2 +2x−1≡0(mod 3)

Answered by MJS_new last updated on 19/Sep/20

x^2 +2x−1=3k with k∈Z  ⇒ x=−1±(√(3k+2))  if x∈R ⇒ k∈N

Commented byAziztisffola last updated on 19/Sep/20

 x also in Z. wich k?

Commented byAziztisffola last updated on 19/Sep/20

2+3k   is not a perfect square.

Commented byMJS_new last updated on 19/Sep/20

there′s no x∈Z

Commented byAziztisffola 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 +2x≡1(mod3)  x≡0(mod3)⇒x^2 +2x≢1(mod3)  x≡1(mod3)⇒x^2 +2x≢1(mod3)  x≡2(mod3)⇒x^2 +2x≢1(mod3)  ⇒∄ x∈Z such that: x^2 +2x−1≡0(mod3)

Commented byAziztisffola last updated on 19/Sep/20

yes sir that′s it.   x^2 +2x−1≡0(mod3)  ⇔(x+1)^2 ≡2[3]  ⇔(x+1)^2 =2+3k  /k∈Z   and 2+3k≠ n^2   for n in Z.  then no solution.

Answered by 1549442205PVT last updated on 20/Sep/20

 x^2 +2x−1≡0(mod 3)(∗)  ∀x∈Z we always have x=3k±1or x=3k  i)If x=3k then x^2 +2x−1=−1(mod3)  ⇒(∗)has no roots(1)  ii)If x=3k+1 then (x+1)^2 −2=  (3k+2)^2 −2=9k^2 +12k+2)=2(mod3)(2)  iii)If x=3k−1 then (x+1)^2 −2=9k^2 −2  =−2(mod3)(3)  From (1)(2)(3) infer the equation(∗)  has no roots in Z