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Question Number 192839 by Mingma last updated on 29/May/23

Answered by witcher3 last updated on 02/Jun/23

$$\mathrm{we}\mathrm{see}\mathrm{that}2,7{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{bee}\mathrm{factor}\mathrm{of}\mathrm{n}$$$$\mathrm{and}11\mid \mathrm{n}$$$$\mathrm{if}\mathrm{n}\not\equiv 0\left[11\right]$$$$\frac{{\mathrm{n}}^2+4^{\mathrm{n}}+7^{\mathrm{n}}}{\mathrm{n}}\in \mathbb{N}\iff \frac{4^{\mathrm{n}}+7^{\mathrm{n}}}{\mathrm{n}}\in \mathbb{N}$$$$4^{\mathrm{n}}+7^{\mathrm{n}}\equiv 4^{\mathrm{n}}+{(-4)}^{\mathrm{n}}\equiv 0\left[11\right],\mathrm{n}=2\mathrm{k}+1...$$$$\mathrm{but}\frac{{\mathrm{n}}^2+4^{\mathrm{n}}+7^{\mathrm{n}}}{\mathrm{n}}\equiv \mathrm{n}\ne 0\left[11\right]$$$$\mathrm{let}\mathrm{n}={11}^{\mathrm{b}}\mathrm{m}\mathrm{b}>1,\mathrm{m}\wedge 11=1;\mathrm{m}\wedge 2=1$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}$$$$4^{\mathrm{n}}+7^{\mathrm{n}}\equiv 0\left[{11}^{\mathrm{b}+1}\right]$$$$\mathrm{first}$$$$4^{11}\equiv {\left(4^4\right)}^2.4^3={\left(256\right)}^2.64\left[121\right]$$$$\equiv {\left(14\right)}^2.64=75.4.16\left[121\right]$$$$=58.2.8\left[121\right]=81\left[121\right]$$$$7^{11}\equiv 40\left[121\right]=(-81)\left[121\right]$$$${\left(4^{11}\right)}^{11}={({11}^2\mathrm{k}+81)}^{11}$$$$=\underset{\mathrm{j}=0}{\sum ^{11}}{\mathrm{C}}_{\mathrm{j}}^{11}{\left({11}^2\mathrm{k}\right)}^{\mathrm{j}}{81}^{11-j}$$$$\mathrm{\forall }\mathrm{j}⩾1{11}^3\mid {\mathrm{C}}_{\mathrm{j}}^{11}{\left({11}^2\mathrm{k}\right)}^{\mathrm{j}}$$$$\Rightarrow 4^{{11}^2}\equiv {81}^{11}\left[{11}^3\right]$$$$\mathrm{By}\mathrm{recursion}$$$$4^{{11}^{\mathrm{b}}}\equiv {81}^{{11}^{\mathrm{b}-1}}\left[{11}^{\mathrm{b}+1}\right]$$$$\Rightarrow 4^{\mathrm{n}}={\left(4^{{11}^{\mathrm{b}}}\right)}^{\mathrm{m}}={81}^{\mathrm{m}.{11}^{\mathrm{b}-1}}\left[{11}^{\mathrm{b}+1}\right]$$$$7^{\mathrm{n}}=7^{\mathrm{m}.{11}^{\mathrm{b}}}={(-81)}^{{\mathrm{m}11}^{\mathrm{b}-1}}\left[{11}^{\mathrm{b}+1}\right]$$$${\mathrm{m}11}^{\mathrm{b}-1}\equiv 1\left[2\right]$$$$4^{\mathrm{n}}+7^{\mathrm{n}}\equiv 0\left[{11}^{\mathrm{b}+1}\right]$$$$\frac{{\mathrm{n}}^2+4^{\mathrm{n}}+7^{\mathrm{n}}}{\mathrm{n}}=\mathrm{n}+\frac{{11}^{\mathrm{b}+1}\mathrm{t}}{{11}^{\mathrm{b}}.\mathrm{m}}=\mathrm{n}+11\frac{\mathrm{t}}{\mathrm{m}}\in \mathbb{N}$$$$\mathrm{m}\mid \mathrm{t};\frac{\mathrm{t}}{\mathrm{m}}=\mathrm{h}$$$$\mathrm{n}+11\mathrm{h}={11}^{\mathrm{b}}\mathrm{m}+11\mathrm{h}=11(\mathrm{h}+{11}^{\mathrm{b}-1}\mathrm{m})\equiv 0\left[11\right]$$

Commented by Mingma last updated on 04/Jun/23

Perfect ��

Commented by witcher3 last updated on 04/Jun/23

$$\mathrm{thanx}\mathrm{sir}$$

Answered by deleteduser1 last updated on 05/Jun/23

$$\Rightarrow n\mid 4^n+7^n\Rightarrow nisodd$$$$Letpbetheminimalprimethatdividesn$$$$Then{4}^n+7^n\equiv 0\left(modp\right)\Rightarrow 4^n(1+{\left(\frac{7}{4}\right)}^n)\equiv 0\left(modp\right)$$$$\Rightarrow {\left(\frac{7}{4}\right)}^{2n}\equiv 1\left(modp\right)[Since(n,4)=1]$$$$\Rightarrow {ord}_p\left(\frac{7}{4}\right)\mid 2nand{ord}_p\left(\frac{7}{4}\right)\mid p-1$$$$\Rightarrow {ord}_p\left(\frac{7}{4}\right)\mid (2n,p-1)=2(n,\frac{p-1}{2})=2$$$$[Sinceifnand\frac{p-1}{2}haveanysimilarfactors$$$$otherthan1,itcontradictstheminimalityofp]$$$$\Rightarrow {ord}_p\left(\frac{7}{4}\right)=1or2$$$$\Rightarrow 7\equiv 4\left(modp\right)or49\equiv 16\left(modp\right)$$$$\Rightarrow 3\equiv 0\left(modp\right)or33\equiv 0\left(modp\right)\Rightarrow p=3or11$$$$Checking\Rightarrow p=11$$$$v_{11}(4^n+7^n)=v_{11}\left(11\right)+v_{11}\left(n\right)=1+v_{11}\left(n\right)>v_{11}\left(n\right)$$$$\Rightarrow n={11}^{k}xwhere(x,11)=1$$$$Similarly,checkingtheminimalprimeofx$$$$gives11.Thiscontradicts(x,11)=1$$$$\Rightarrow n={11}^k$$$$Son\mid 4^n+7^{n}onlyifn={11}^k$$$$\Rightarrow \frac{n^2+4^n+7^n}{n}={11}^k+\frac{{11}^{k+1}q}{{11}^k}={11}^k+11q$$$$=11({11}^{k-1}+q)whichisdivisibleby11.$$

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