Ajutor!

[Simona_Lorena]

Fie σϵ S6, σ=(1 2 3 4 5 6)
2 4 5 3 6 1
Sa se rezolve in S6 ecuatia: x^2014=σ.
Va multumesc!

[DD]

x^2014=(x^1007)^2=y^2,unde y=x^1007 sau y^2=sigma=
(1 2 3 4 5 6 )
(2 4 5 3 6 1 ) dar nu exista permutare in S6 al carei patrat sa dea permutarea ''sigma''
Fie;
( 1 2 3 4 5 6 ) ( 1 2 3 4 5 6 ) ( 1 2 3 4 5 6 )
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( 2 2............). ( 2 2............)=( 2 2............)-fals
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( 3 5 2 6 4 3 ). ( 3 5 2 6 4 3 )=( 2 4 5 3 6 2 )-fals
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( 4 3 4 2......). ( 4 3 4 2.......)=( 2 4...3.......)-fals
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( 5 6...1 2 4 ). ( 5 6....1 2 4 )=( 2 4......6 1 )-fals
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( 6 1........2 ) .( 6 1..........2 )=(2...............)-fals
Se incearca gasirea permutarii y testand pe rand, sub 1 ,celelalte cifre

[gigelmarga]

y^2=sigma=
(1 2 3 4 5 6 )
(2 4 5 3 6 1 ) dar nu exista permutare in S6 al carei patrat sa dea permutarea ''sigma''

Asa e, dar justificarea nu se face, barbar, prin incercari.
E suficient sa observam ca "sigma" e permutare impara, iar patratul oricarei permutari e permutare para.