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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.

$${find}\:{n}\:{from}\:{N}\:\:{in}\:{ordre}\:{tohave}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:{divide} \\ $$$$\left({x}+\mathrm{1}\right)^{{n}} −{x}^{{n}} −\mathrm{1}. \\ $$

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.

$${let}\:{put}\:{p}\left({x}\right)=\:\left({x}+\mathrm{1}\right)^{{n}} −{x}^{{n}} −\mathrm{1}\:{and}\:{q}\left({x}\right)={x}^{\mathrm{2}} +{x}+\mathrm{1}\:{the}\:{roots} \\ $$$${of}\:{p}\left({x}\right){are}\:\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:{and}\:{j}^{−} ={e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\:{q}\:{divide}\:{p}\:\Leftrightarrow{p}\left({j}\right)={p}\left({j}^{−} \right)=\mathrm{0} \\ $$$${but}\:{p}\left({j}\right)={o}\:\Leftrightarrow\left({j}+\mathrm{1}\right)^{{n}} \:−{j}^{{n}} −\mathrm{1}=\mathrm{0}\:\Leftrightarrow\left(−\mathrm{1}\right)^{{n}} {j}^{\mathrm{2}{n}} \:−{j}^{{n}} −\mathrm{1}=\mathrm{0} \\ $$$${p}\left({j}^{−} \right)=\mathrm{0}\:\Leftrightarrow\:\left(\mathrm{1}+{j}^{−} \right)^{\mathrm{2}} \:−{j}^{−^{{n}} } \:−\mathrm{1}=\mathrm{0}\:\Leftrightarrow\left(−\mathrm{1}\right)^{{n}} {j}^{−^{\mathrm{2}{n}} } \:−{j}^{−^{{n}} } −\mathrm{1}=\mathrm{0}\Rightarrow\left(−\mathrm{1}\right) \\ $$$$\left(−\mathrm{1}\right)^{{n}} \left(\:\left({j}^{\mathrm{2}{n}} \:−{j}^{−^{\mathrm{2}{n}} } \right)−\left({j}^{{n}} \:−{j}^{−^{{n}} } \right)=\mathrm{0}\right. \\ $$$$\left(−\mathrm{1}\right)^{{n}} \mathrm{2}{i}\:{Im}\left({j}^{\mathrm{2}{n}} \right)−\:\mathrm{2}{i}\:{Im}\left({j}^{{n}} \right)=\mathrm{0} \\ $$$$\left(−\mathrm{1}\right)^{{n}} {sin}\left(\frac{\mathrm{4}{n}\pi}{\mathrm{3}}\right)−{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow{sin}\left(\frac{\mathrm{4}{n}\pi}{\mathrm{3}}\right)=\left(−\mathrm{1}\right)^{{n}} {sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)\:\:\left({e}\right) \\ $$$${case}\:\mathrm{1}\:\:\:\:{n}=\mathrm{2}{p} \\ $$$$\left({e}\right)\Leftrightarrow\:\frac{\mathrm{4}{n}\pi}{\mathrm{3}}\:=\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\:{or}\:\:\frac{\mathrm{4}{n}\pi}{\mathrm{3}}\:=\pi−\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\:\:\:\:\left({k}\in{Z}\right) \\ $$$$\Leftrightarrow\mathrm{4}{n}\:=\mathrm{2}{n}\:+\mathrm{6}{k}\:{or}\:\mathrm{4}{n}=\:\mathrm{3}−\mathrm{2}{n}\:+\mathrm{6}{k} \\ $$$${n}=\mathrm{3}{k}\:\:\:{or}\:\mathrm{6}{n}=\:\mathrm{6}{k}+\mathrm{3} \\ $$$${n}=\mathrm{3}{k}\:\:{or}\:{n}={k}\:+\frac{\mathrm{1}}{\mathrm{2}}\left({to}\:{eliminate}\right)\Leftrightarrow{n}=\mathrm{3}{k}\:\:=\mathrm{2}{p}\Rightarrow \\ $$$${k}=\mathrm{2}{s}\:\:\:{and}\:{n}=\mathrm{6}{s}\:\: \\ $$$${case}\:\mathrm{2}\:\:\:{n}=\mathrm{2}{p}+\mathrm{1} \\ $$$$\left({e}\right)\:\Leftrightarrow\:{sin}\left(\frac{\mathrm{4}{n}\pi}{\mathrm{3}}\right)={sin}\left(−\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)\:\Leftrightarrow\:\frac{\mathrm{4}{n}\pi}{\mathrm{3}}=\frac{−\mathrm{2}{n}\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:{or} \\ $$$$\frac{\mathrm{4}{n}\pi}{\mathrm{3}}=\:\pi\:+\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi \\ $$$$\Leftrightarrow\mathrm{4}{n}\:=−\mathrm{2}{n}\:+\mathrm{6}{k}\:\:{or}\:\mathrm{4}{n}=\:\mathrm{3}\:+\mathrm{2}{n}\:+\mathrm{6}{k} \\ $$$$\Leftrightarrow{n}={k}\:\:{or}\:\mathrm{2}{n}=\mathrm{3}+\mathrm{6}{k}\left({to}\:{eliminate}\right)\:\Leftrightarrow{n}={k}=\mathrm{2}{p}+\mathrm{1}\:{so} \\ $$$${q}\:{divide}\:{p}\:\:\Leftrightarrow\:{n}=\mathrm{6}{p}\:{or}\:{n}=\mathrm{2}{p}+\mathrm{1}. \\ $$$$ \\ $$

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