Question Number 192839 by Mingma last updated on 29/May/23
Answered by witcher3 last updated on 02/Jun/23
![we see that 2,7 can′t bee factor of n and 11∣n if n≢0[11] ((n^2 +4^n +7^n )/n)∈N⇔((4^n +7^n )/n)∈N 4^n +7^n ≡4^n +(−4)^n ≡0[11],n=2k+1... but ((n^2 +4^n +7^n )/n)≡n≠0[11] let n=11^b m b>1,m∧11=1;m∧2=1 we have to show that 4^n +7^n ≡0[11^(b+1) ] first 4^(11) ≡(4^4 )^2 .4^3 =(256)^2 .64[121] ≡(14)^2 .64=75.4.16[121] =58.2.8[121]=81[121] 7^(11) ≡40[121]=(−81)[121] (4^(11) )^(11) =(11^2 k+81)^(11) =Σ_(j=0) ^(11) C_j ^(11) (11^2 k)^j 81^(11−j) ∀j≥1 11^3 ∣C_j ^(11) (11^2 k)^j ⇒4^(11^2 ) ≡81^(11) [11^3 ] By recursion 4^(11^b ) ≡81^(11^(b−1) ) [11^(b+1) ] ⇒4^n =(4^(11^b ) )^m =81^(m.11^(b−1) ) [11^(b+1) ] 7^n =7^(m.11^b ) =(−81)^(m11^(b−1) ) [11^(b+1) ] m11^(b−1) ≡1[2] 4^n +7^n ≡0[11^(b+1) ] ((n^2 +4^n +7^n )/n)=n+((11^(b+1) t)/(11^b .m))=n+11(t/m)∈N m∣t ;(t/m)=h n+11h=11^b m+11h=11(h+11^(b−1) m)≡0[11]](https://www.tinkutara.com/question/Q193022.png)
$$\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{2},\mathrm{7}\:\mathrm{can}'\mathrm{t}\:\mathrm{bee}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{and}\:\mathrm{11}\mid\mathrm{n}\: \\ $$$$\mathrm{if}\:\mathrm{n}≢\mathrm{0}\left[\mathrm{11}\right] \\ $$$$\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}} }{\mathrm{n}}\in\mathbb{N}\Leftrightarrow\frac{\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}} }{\mathrm{n}}\in\mathbb{N} \\ $$$$\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}} \equiv\mathrm{4}^{\mathrm{n}} +\left(−\mathrm{4}\right)^{\mathrm{n}} \equiv\mathrm{0}\left[\mathrm{11}\right],\mathrm{n}=\mathrm{2k}+\mathrm{1}… \\ $$$$\mathrm{but}\:\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}} }{\mathrm{n}}\equiv\mathrm{n}\neq\mathrm{0}\left[\mathrm{11}\right] \\ $$$$\mathrm{let}\:\mathrm{n}=\mathrm{11}^{\mathrm{b}} \mathrm{m}\:\:\:\mathrm{b}>\mathrm{1},\mathrm{m}\wedge\mathrm{11}=\mathrm{1};\mathrm{m}\wedge\mathrm{2}=\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}} \equiv\mathrm{0}\left[\mathrm{11}^{\mathrm{b}+\mathrm{1}} \right] \\ $$$$\mathrm{first} \\ $$$$\mathrm{4}^{\mathrm{11}} \equiv\left(\mathrm{4}^{\mathrm{4}} \right)^{\mathrm{2}} .\mathrm{4}^{\mathrm{3}} =\left(\mathrm{256}\right)^{\mathrm{2}} .\mathrm{64}\left[\mathrm{121}\right] \\ $$$$\equiv\left(\mathrm{14}\right)^{\mathrm{2}} .\mathrm{64}=\mathrm{75}.\mathrm{4}.\mathrm{16}\left[\mathrm{121}\right] \\ $$$$=\mathrm{58}.\mathrm{2}.\mathrm{8}\left[\mathrm{121}\right]=\mathrm{81}\left[\mathrm{121}\right] \\ $$$$\mathrm{7}^{\mathrm{11}} \equiv\mathrm{40}\left[\mathrm{121}\right]=\left(−\mathrm{81}\right)\left[\mathrm{121}\right] \\ $$$$\left(\mathrm{4}^{\mathrm{11}} \right)^{\mathrm{11}} =\left(\mathrm{11}^{\mathrm{2}} \mathrm{k}+\mathrm{81}\right)^{\mathrm{11}} \\ $$$$=\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{11}} {\sum}}\mathrm{C}_{\mathrm{j}} ^{\mathrm{11}} \left(\mathrm{11}^{\mathrm{2}} \mathrm{k}\right)^{\mathrm{j}} \mathrm{81}^{\mathrm{11}−{j}} \\ $$$$\forall\mathrm{j}\geqslant\mathrm{1}\:\:\:\:\mathrm{11}^{\mathrm{3}} \mid\mathrm{C}_{\mathrm{j}} ^{\mathrm{11}} \left(\mathrm{11}^{\mathrm{2}} \mathrm{k}\right)^{\mathrm{j}} \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{11}^{\mathrm{2}} } \equiv\mathrm{81}^{\mathrm{11}} \left[\mathrm{11}^{\mathrm{3}} \right] \\ $$$$\mathrm{By}\:\mathrm{recursion} \\ $$$$\mathrm{4}^{\mathrm{11}^{\mathrm{b}} } \equiv\mathrm{81}^{\mathrm{11}^{\mathrm{b}−\mathrm{1}} } \left[\mathrm{11}^{\mathrm{b}+\mathrm{1}} \right] \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{n}} =\left(\mathrm{4}^{\mathrm{11}^{\mathrm{b}} } \right)^{\mathrm{m}} =\mathrm{81}^{\mathrm{m}.\mathrm{11}^{\mathrm{b}−\mathrm{1}} } \left[\mathrm{11}^{\mathrm{b}+\mathrm{1}} \right] \\ $$$$\mathrm{7}^{\mathrm{n}} =\mathrm{7}^{\mathrm{m}.\mathrm{11}^{\mathrm{b}} } =\left(−\mathrm{81}\right)^{\mathrm{m11}^{\mathrm{b}−\mathrm{1}} } \left[\mathrm{11}^{\mathrm{b}+\mathrm{1}} \right] \\ $$$$\mathrm{m11}^{\mathrm{b}−\mathrm{1}} \equiv\mathrm{1}\left[\mathrm{2}\right]\: \\ $$$$\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}} \equiv\mathrm{0}\left[\mathrm{11}^{\mathrm{b}+\mathrm{1}} \right] \\ $$$$\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}} }{\mathrm{n}}=\mathrm{n}+\frac{\mathrm{11}^{\mathrm{b}+\mathrm{1}} \mathrm{t}}{\mathrm{11}^{\mathrm{b}} .\mathrm{m}}=\mathrm{n}+\mathrm{11}\frac{\mathrm{t}}{\mathrm{m}}\in\mathbb{N} \\ $$$$\mathrm{m}\mid\mathrm{t}\:;\frac{\mathrm{t}}{\mathrm{m}}=\mathrm{h} \\ $$$$\mathrm{n}+\mathrm{11h}=\mathrm{11}^{\mathrm{b}} \mathrm{m}+\mathrm{11h}=\mathrm{11}\left(\mathrm{h}+\mathrm{11}^{\mathrm{b}−\mathrm{1}} \mathrm{m}\right)\equiv\mathrm{0}\left[\mathrm{11}\right] \\ $$
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 AST last updated on 05/Jun/23
![⇒n∣4^n +7^n ⇒n is odd Let p be the minimal prime that divides n Then 4^n +7^n ≡0(mod p)⇒4^n (1+((7/4))^n )≡0(mod p) ⇒((7/4))^(2n) ≡1(mod p) [Since (n,4)=1] ⇒ord_p ((7/4))∣2n and ord_p ((7/4))∣p−1 ⇒ord_p ((7/4))∣(2n,p−1)=2(n,((p−1)/2))=2 [Since if n and ((p−1)/2) have any similar factors other than 1, it contradicts the minimality of p] ⇒ord_p ((7/4))=1 or 2 ⇒7≡4(mod p) or 49≡16(mod p) ⇒3≡0(mod p) or 33≡0(mod p)⇒p=3 or 11 Checking⇒p=11 v_(11) (4^n +7^n )=v_(11) (11)+v_(11) (n)=1+v_(11) (n)>v_(11) (n) ⇒n=11^k x where (x,11)=1 Similarly, checking the minimal prime of x gives 11. This contradicts (x,11)=1 ⇒n=11^k So n∣4^n +7^n only if n=11^k ⇒((n^2 +4^n +7^n )/n)=11^k +((11^(k+1) q)/(11^k ))=11^k +11q =11(11^(k−1) +q) which is divisible by 11.](https://www.tinkutara.com/question/Q193165.png)
$$\Rightarrow{n}\mid\mathrm{4}^{{n}} +\mathrm{7}^{{n}} \:\Rightarrow{n}\:{is}\:{odd} \\ $$$${Let}\:{p}\:{be}\:{the}\:{minimal}\:{prime}\:{that}\:{divides}\:{n} \\ $$$${Then}\:\mathrm{4}^{{n}} +\mathrm{7}^{{n}} \equiv\mathrm{0}\left({mod}\:{p}\right)\Rightarrow\mathrm{4}^{{n}} \left(\mathrm{1}+\left(\frac{\mathrm{7}}{\mathrm{4}}\right)^{{n}} \right)\equiv\mathrm{0}\left({mod}\:{p}\right) \\ $$$$\Rightarrow\left(\frac{\mathrm{7}}{\mathrm{4}}\right)^{\mathrm{2}{n}} \equiv\mathrm{1}\left({mod}\:{p}\right)\:\left[{Since}\:\left({n},\mathrm{4}\right)=\mathrm{1}\right] \\ $$$$\Rightarrow{ord}_{{p}} \left(\frac{\mathrm{7}}{\mathrm{4}}\right)\mid\mathrm{2}{n}\:{and}\:{ord}_{{p}} \left(\frac{\mathrm{7}}{\mathrm{4}}\right)\mid{p}−\mathrm{1} \\ $$$$\Rightarrow{ord}_{{p}} \left(\frac{\mathrm{7}}{\mathrm{4}}\right)\mid\left(\mathrm{2}{n},{p}−\mathrm{1}\right)=\mathrm{2}\left({n},\frac{{p}−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$$\left[{Since}\:{if}\:{n}\:{and}\:\frac{{p}−\mathrm{1}}{\mathrm{2}}\:{have}\:{any}\:{similar}\:{factors}\right. \\ $$$$\left.{other}\:{than}\:\mathrm{1},\:{it}\:{contradicts}\:{the}\:{minimality}\:{of}\:{p}\right] \\ $$$$\Rightarrow{ord}_{{p}} \left(\frac{\mathrm{7}}{\mathrm{4}}\right)=\mathrm{1}\:{or}\:\mathrm{2} \\ $$$$\Rightarrow\mathrm{7}\equiv\mathrm{4}\left({mod}\:{p}\right)\:{or}\:\mathrm{49}\equiv\mathrm{16}\left({mod}\:{p}\right) \\ $$$$\Rightarrow\mathrm{3}\equiv\mathrm{0}\left({mod}\:{p}\right)\:{or}\:\mathrm{33}\equiv\mathrm{0}\left({mod}\:{p}\right)\Rightarrow{p}=\mathrm{3}\:{or}\:\mathrm{11} \\ $$$${Checking}\Rightarrow{p}=\mathrm{11} \\ $$$${v}_{\mathrm{11}} \left(\mathrm{4}^{{n}} +\mathrm{7}^{{n}} \right)={v}_{\mathrm{11}} \left(\mathrm{11}\right)+{v}_{\mathrm{11}} \left({n}\right)=\mathrm{1}+{v}_{\mathrm{11}} \left({n}\right)>{v}_{\mathrm{11}} \left({n}\right) \\ $$$$\Rightarrow{n}=\mathrm{11}^{{k}} {x}\:{where}\:\left({x},\mathrm{11}\right)=\mathrm{1} \\ $$$${Similarly},\:{checking}\:{the}\:{minimal}\:{prime}\:{of}\:{x} \\ $$$${gives}\:\mathrm{11}.\:{This}\:{contradicts}\:\left({x},\mathrm{11}\right)=\mathrm{1} \\ $$$$\Rightarrow{n}=\mathrm{11}^{{k}} \\ $$$${So}\:{n}\mid\mathrm{4}^{{n}} +\mathrm{7}^{{n}} \:\:\:\:\:{only}\:{if}\:{n}=\mathrm{11}^{{k}} \\ $$$$\Rightarrow\frac{{n}^{\mathrm{2}} +\mathrm{4}^{{n}} +\mathrm{7}^{{n}} }{{n}}=\mathrm{11}^{{k}} +\frac{\mathrm{11}^{{k}+\mathrm{1}} {q}}{\mathrm{11}^{{k}} }=\mathrm{11}^{{k}} +\mathrm{11}{q} \\ $$$$=\mathrm{11}\left(\mathrm{11}^{{k}−\mathrm{1}} +{q}\right)\:{which}\:{is}\:{divisible}\:{by}\:\mathrm{11}. \\ $$