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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).

$${let}\:{p}_{{n}} \left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:−{x}^{\mathrm{6}{n}+\mathrm{1}} \:−\mathrm{1}\:{with}\:{n}\:{integr} \\ $$$${prove}\:{that}\:\forall{n}\:\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:{divide}\:{p}_{{n}} \left({x}\right). \\ $$

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

$${the}\:{roots}\:{of}\:{x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:{are}\:{j}\:{and}\:{j}^{\mathrm{2}} \:{with}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:{for}\:{that} \\ $$$${we}\:{must}\:{prove}\:{that}\:{p}_{{n}} \left({j}\right)={p}_{{n}} \left({j}^{\mathrm{2}} \right)=\mathrm{0}\:{andp}^{'} \left({j}\right)={p}^{'} \left({j}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${p}_{{n}} \left({j}\right)\:=\left({j}+\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:−{j}^{\mathrm{6}{n}+\mathrm{1}} \:−\mathrm{1} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:\left({j}^{\mathrm{2}} \right)^{\mathrm{6}{n}+\mathrm{1}} \:\:−{j}\:−\mathrm{1}\:=\:−{j}^{\mathrm{2}} \:−{j}−\mathrm{1}=\mathrm{0}\:{wit}\:{j}^{\mathrm{3}} =\mathrm{1} \\ $$$${p}_{{n}} \left({j}^{\mathrm{2}} \right)\:=\left({j}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:−\:\left({j}^{\mathrm{2}} \right)^{\mathrm{6}{n}+\mathrm{1}} \:−\mathrm{1} \\ $$$$=\left(−{j}\right)^{\mathrm{6}{n}+\mathrm{1}} \:\:−{j}^{\mathrm{2}} \:−\mathrm{1}\:\:=−{j}\:−{j}^{\mathrm{2}} \:−\mathrm{1}=\mathrm{0} \\ $$$${we}\:{have}\:{p}^{'} \left({x}\right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{\mathrm{6}{n}} \:−\left(\mathrm{6}{n}+\mathrm{1}\right){x}^{\mathrm{6}{n}} \Rightarrow \\ $$$${p}^{'} \left({j}\right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\:\left({j}+\mathrm{1}\right)^{\mathrm{6}{n}} \:−{j}^{\mathrm{6}{n}} \right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left({j}^{\mathrm{12}{n}} \:−{j}^{\mathrm{6}{n}} \right)=\mathrm{0} \\ $$$${p}^{'} \left({j}^{\mathrm{2}} \right)=\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\:\left(\mathrm{1}+{j}^{\mathrm{2}} \right)^{\mathrm{6}{n}} \:\:−\left({j}^{\mathrm{2}} \right)^{\mathrm{6}{n}} \right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\left(−{j}\right)^{\mathrm{6}{n}} \:−{j}^{\mathrm{12}{n}} \right)=\mathrm{0} \\ $$

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