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Question Number 28534 by abdo imad last updated on 26/Jan/18

find n from N in ordre tohave x^2 +x+1 divide (x+1)^n −x^n −1.

findnfromNinordretohavex2+x+1divide(x+1)n−xn−1.

Commented by abdo imad last updated on 28/Jan/18

let put p(x)= (x+1)^n −x^n −1 and q(x)=x^2 +x+1 the roots of p(x)are j=e^(i((2π)/3)) and j^− =e^(−i((2π)/3)) q divide p ⇔p(j)=p(j^− )=0 but p(j)=o ⇔(j+1)^n −j^n −1=0 ⇔(−1)^n j^(2n) −j^n −1=0 p(j^− )=0 ⇔ (1+j^− )^2 −j^−^n −1=0 ⇔(−1)^n j^−^(2n) −j^−^n −1=0⇒(−1) (−1)^n ( (j^(2n) −j^−^(2n) )−(j^n −j^−^n )=0 (−1)^n 2i Im(j^(2n) )− 2i Im(j^n )=0 (−1)^n sin(((4nπ)/3))−sin(((2nπ)/3))=0 ⇔sin(((4nπ)/3))=(−1)^n sin(((2nπ)/3)) (e) case 1 n=2p (e)⇔ ((4nπ)/3) =((2nπ)/3) +2kπ or ((4nπ)/3) =π−((2nπ)/3) +2kπ (k∈Z) ⇔4n =2n +6k or 4n= 3−2n +6k n=3k or 6n= 6k+3 n=3k or n=k +(1/2)(to eliminate)⇔n=3k =2p⇒ k=2s and n=6s case 2 n=2p+1 (e) ⇔ sin(((4nπ)/3))=sin(−((2nπ)/3)) ⇔ ((4nπ)/3)=((−2nπ)/3) +2kπ or ((4nπ)/3)= π +((2nπ)/3) +2kπ ⇔4n =−2n +6k or 4n= 3 +2n +6k ⇔n=k or 2n=3+6k(to eliminate) ⇔n=k=2p+1 so q divide p ⇔ n=6p or n=2p+1.

letputp(x)=(x+1)n−xn−1andq(x)=x2+x+1therootsofp(x)arej=ei2π3andj−=e−i2π3qdividep⇔p(j)=p(j−)=0butp(j)=o⇔(j+1)n−jn−1=0⇔(−1)nj2n−jn−1=0p(j−)=0⇔(1+j−)2−j−n−1=0⇔(−1)nj−2n−j−n−1=0⇒(−1)(−1)n((j2n−j−2n)−(jn−j−n)=0(−1)n2iIm(j2n)−2iIm(jn)=0(−1)nsin(4nπ3)−sin(2nπ3)=0⇔sin(4nπ3)=(−1)nsin(2nπ3)(e)case1n=2p(e)⇔4nπ3=2nπ3+2kπor4nπ3=π−2nπ3+2kπ(k∈Z)⇔4n=2n+6kor4n=3−2n+6kn=3kor6n=6k+3n=3korn=k+12(toeliminate)⇔n=3k=2p⇒k=2sandn=6scase2n=2p+1(e)⇔sin(4nπ3)=sin(−2nπ3)⇔4nπ3=−2nπ3+2kπor4nπ3=π+2nπ3+2kπ⇔4n=−2n+6kor4n=3+2n+6k⇔n=kor2n=3+6k(toeliminate)⇔n=k=2p+1soqdividep⇔n=6porn=2p+1.

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