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Question Number 32330 by abdo imad last updated on 23/Mar/18

$$let{p}_n\left(x\right)={(x+1)}^{6n+1}-x^{6n+1}-1withnintegr$$$$provethat\mathrm{\forall }n{(x^2+x+1)}^{2}divide{p}_n\left(x\right).$$

Commented by abdo imad last updated on 01/Apr/18

$$therootsof{x}^2+x+1arejand{j}^{2}withj=e^{i\frac{2\pi }{3}}forthat$$$$wemustprovethat{p}_n\left(j\right)=p_n\left(j^2\right)=0{andp}^{{}^{\prime }}\left(j\right)=p^{{}^{\prime }}\left(j^2\right)=0$$$$p_n\left(j\right)={(j+1)}^{6n+1}-j^{6n+1}-1$$$$={(-1)}^{6n+1}{\left(j^2\right)}^{6n+1}-j-1=-j^2-j-1=0wit{j}^3=1$$$$p_n\left(j^2\right)={(j^2+1)}^{6n+1}-{\left(j^2\right)}^{6n+1}-1$$$$={(-j)}^{6n+1}-j^2-1=-j-j^2-1=0$$$$wehave{p}^{{}^{\prime }}\left(x\right)=(6n+1){(x+1)}^{6n}-(6n+1)x^{6n}\Rightarrow$$$$p^{{}^{\prime }}\left(j\right)=(6n+1)({(j+1)}^{6n}-j^{6n})=(6n+1)(j^{12n}-j^{6n})=0$$$$p^{{}^{\prime }}\left(j^2\right)=(6n+1)({(1+j^2)}^{6n}-{\left(j^2\right)}^{6n})=(6n+1)({(-j)}^{6n}-j^{12n})=0$$

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