Question Number 210666 by universe last updated on 15/Aug/24
$$\:\:\mathrm{prove}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{integer}\:\forall\:\mathrm{n}\in\mathbb{Z} \\ $$$$\:\:\:\mathrm{p}\left(\mathrm{n}\right)\:=\:\frac{\mathrm{3n}^{\mathrm{7}} +\mathrm{7n}^{\mathrm{3}} +\mathrm{11n}}{\mathrm{21}} \\ $$
Answered by A5T last updated on 15/Aug/24
$${Let}\:{f}\left({n}\right)=\mathrm{3}{n}^{\mathrm{7}} +\mathrm{7}{n}^{\mathrm{3}} +\mathrm{11}{n}={n}\left(\mathrm{3}{n}^{\mathrm{6}} +\mathrm{7}{n}^{\mathrm{2}} +\mathrm{11}\right) \\ $$$${Note}\:{that}\:{when}\:\mathrm{7}\mid{n},\:{then}\:\mathrm{7}\mid{f}\left({n}\right) \\ $$$${So},\:{when}\:\left(\mathrm{7},{n}\right)=\mathrm{1}\Rightarrow{n}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow{f}\left({n}\right)\equiv{n}\left(\mathrm{3}+\mathrm{11}\right)=\mathrm{14}{n}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{7}\mid{f}\left({n}\right) \\ $$$${Similarly}\:{when}\:\mathrm{3}\mid{n},{then}\:\mathrm{3}\mid{f}\left({n}\right) \\ $$$${when}\:\left(\mathrm{3},{n}\right)=\mathrm{1}\Rightarrow{n}^{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow{n}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow{f}\left({n}\right)\equiv{n}\left(\mathrm{3}+\mathrm{7}+\mathrm{11}\right)=\mathrm{21}{n}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow\mathrm{3}\mid{f}\left({n}\right) \\ $$$$\Rightarrow\mathrm{21}\mid{f}\left({n}\right)\Rightarrow{p}\left({n}\right)=\frac{{f}\left({n}\right)}{\mathrm{21}}\in\mathbb{Z} \\ $$
Commented by universe last updated on 15/Aug/24
$${thank}\:{you}\:{sir} \\ $$