Question Number 32330 by abdo imad last updated on 23/Mar/18
$${let}\:{p}_{{n}} \left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:−{x}^{\mathrm{6}{n}+\mathrm{1}} \:−\mathrm{1}\:{with}\:{n}\:{integr} \\ $$$${prove}\:{that}\:\forall{n}\:\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:{divide}\:{p}_{{n}} \left({x}\right). \\ $$
Commented by abdo imad last updated on 01/Apr/18
$${the}\:{roots}\:{of}\:{x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:{are}\:{j}\:{and}\:{j}^{\mathrm{2}} \:{with}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:{for}\:{that} \\ $$$${we}\:{must}\:{prove}\:{that}\:{p}_{{n}} \left({j}\right)={p}_{{n}} \left({j}^{\mathrm{2}} \right)=\mathrm{0}\:{andp}^{'} \left({j}\right)={p}^{'} \left({j}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${p}_{{n}} \left({j}\right)\:=\left({j}+\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:−{j}^{\mathrm{6}{n}+\mathrm{1}} \:−\mathrm{1} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:\left({j}^{\mathrm{2}} \right)^{\mathrm{6}{n}+\mathrm{1}} \:\:−{j}\:−\mathrm{1}\:=\:−{j}^{\mathrm{2}} \:−{j}−\mathrm{1}=\mathrm{0}\:{wit}\:{j}^{\mathrm{3}} =\mathrm{1} \\ $$$${p}_{{n}} \left({j}^{\mathrm{2}} \right)\:=\left({j}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:−\:\left({j}^{\mathrm{2}} \right)^{\mathrm{6}{n}+\mathrm{1}} \:−\mathrm{1} \\ $$$$=\left(−{j}\right)^{\mathrm{6}{n}+\mathrm{1}} \:\:−{j}^{\mathrm{2}} \:−\mathrm{1}\:\:=−{j}\:−{j}^{\mathrm{2}} \:−\mathrm{1}=\mathrm{0} \\ $$$${we}\:{have}\:{p}^{'} \left({x}\right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{\mathrm{6}{n}} \:−\left(\mathrm{6}{n}+\mathrm{1}\right){x}^{\mathrm{6}{n}} \Rightarrow \\ $$$${p}^{'} \left({j}\right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\:\left({j}+\mathrm{1}\right)^{\mathrm{6}{n}} \:−{j}^{\mathrm{6}{n}} \right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left({j}^{\mathrm{12}{n}} \:−{j}^{\mathrm{6}{n}} \right)=\mathrm{0} \\ $$$${p}^{'} \left({j}^{\mathrm{2}} \right)=\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\:\left(\mathrm{1}+{j}^{\mathrm{2}} \right)^{\mathrm{6}{n}} \:\:−\left({j}^{\mathrm{2}} \right)^{\mathrm{6}{n}} \right)\:=\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\left(−{j}\right)^{\mathrm{6}{n}} \:−{j}^{\mathrm{12}{n}} \right)=\mathrm{0} \\ $$