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Question Number 210666 by universe last updated on 15/Aug/24

$$\mathrm{prove}\mathrm{that}\mathrm{p}\left(\mathrm{n}\right)\mathrm{is}\mathrm{integer}\mathrm{\forall }\mathrm{n}\in \mathbb{Z}$$$$\mathrm{p}\left(\mathrm{n}\right)=\frac{{3\mathrm{n}}^7+{7\mathrm{n}}^3+11\mathrm{n}}{21}$$

Answered by A5T last updated on 15/Aug/24

$$Letf\left(n\right)=3n^7+7n^3+11n=n(3n^6+7n^2+11)$$$$Notethatwhen7\mid n,then7\mid f\left(n\right)$$$$So,when(7,n)=1\Rightarrow n^6\equiv 1\left(mod7\right)$$$$\Rightarrow f\left(n\right)\equiv n(3+11)=14n\equiv 0\left(mod7\right)\Rightarrow 7\mid f\left(n\right)$$$$Similarlywhen3\mid n,then3\mid f\left(n\right)$$$$when(3,n)=1\Rightarrow n^2\equiv 1\left(mod7\right)\Rightarrow n^6\equiv 1\left(mod7\right)$$$$\Rightarrow f\left(n\right)\equiv n(3+7+11)=21n\equiv 0\left(mod3\right)\Rightarrow 3\mid f\left(n\right)$$$$\Rightarrow 21\mid f\left(n\right)\Rightarrow p\left(n\right)=\frac{f\left(n\right)}{21}\in \mathbb{Z}$$

Commented by universe last updated on 15/Aug/24

$$thankyousir$$

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