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let-p-n-x-x-1-6n-1-x-6n-1-1-with-n-integr-prove-that-n-x-2-x-1-2-divide-p-n-x-

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Question Number 32330 by abdo imad last updated on 23/Mar/18
let p_n (x)=(x+1)^(6n+1) −x^(6n+1) −1 with n integr prove that ∀n (x^2 +x+1)^2 divide p_n (x).
letpn(x)=(x+1)6n+1−x6n+1−1withnintegrprovethat∀n(x2+x+1)2dividepn(x).
Commented by abdo imad last updated on 01/Apr/18
the roots of x^2 +x+1 are j and j^2 with j=e^(i((2π)/3)) for that we must prove that p_n (j)=p_n (j^2 )=0 andp^′ (j)=p^′ (j^2 )=0 p_n (j) =(j+1)^(6n+1) −j^(6n+1) −1 =(−1)^(6n+1) (j^2 )^(6n+1) −j −1 = −j^2 −j−1=0 wit j^3 =1 p_n (j^2 ) =(j^2 +1)^(6n+1) − (j^2 )^(6n+1) −1 =(−j)^(6n+1) −j^2 −1 =−j −j^2 −1=0 we have p^′ (x) =(6n+1)(x+1)^(6n) −(6n+1)x^(6n) ⇒ p^′ (j) =(6n+1)( (j+1)^(6n) −j^(6n) ) =(6n+1)(j^(12n) −j^(6n) )=0 p^′ (j^2 )=(6n+1)( (1+j^2 )^(6n) −(j^2 )^(6n) ) =(6n+1)((−j)^(6n) −j^(12n) )=0
therootsofx2+x+1arejandj2withj=ei2π3forthatwemustprovethatpn(j)=pn(j2)=0andp′(j)=p′(j2)=0pn(j)=(j+1)6n+1−j6n+1−1=(−1)6n+1(j2)6n+1−j−1=−j2−j−1=0witj3=1pn(j2)=(j2+1)6n+1−(j2)6n+1−1=(−j)6n+1−j2−1=−j−j2−1=0wehavep′(x)=(6n+1)(x+1)6n−(6n+1)x6n⇒p′(j)=(6n+1)((j+1)6n−j6n)=(6n+1)(j12n−j6n)=0p′(j2)=(6n+1)((1+j2)6n−(j2)6n)=(6n+1)((−j)6n−j12n)=0

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