Question Number 27816 by Rasheed.Sindhi last updated on 15/Jan/18
![If N is perfect nth power, prove that n ∣ (d(N)−1) [Where d(N) denotes number of divisors of N] Also show by an example that its vice versa is not necessarily correct.](https://www.tinkutara.com/question/Q27816.png)
$$\mathrm{If}\:\mathrm{N}\:\mathrm{is}\:\boldsymbol{\mathrm{perfect}}\:\boldsymbol{\mathrm{nth}}\:\boldsymbol{\mathrm{power}},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\mathrm{n}\:\mid\:\left(\mathrm{d}\left(\mathrm{N}\right)−\mathrm{1}\right)\: \\ $$$$\left[{Where}\:\mathrm{d}\left(\mathrm{N}\right)\:{denotes}\:\boldsymbol{{number}}\right. \\ $$$$\left.\boldsymbol{{of}}\:\boldsymbol{{divisors}}\:\boldsymbol{{of}}\:\mathrm{N}\right] \\ $$$$\mathrm{Also}\:\mathrm{show}\:\mathrm{by}\:\mathrm{an}\:\mathrm{example}\:\mathrm{that}\:\mathrm{its} \\ $$$$\mathrm{vice}\:\mathrm{versa}\:\mathrm{is}\:\mathrm{not}\:\mathrm{necessarily}\:\mathrm{correct}. \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jan/18
$$\mathrm{Generalization}\:\mathrm{ofQ}#\mathrm{27767} \\ $$
Answered by mrW2 last updated on 15/Jan/18
![N=a^n a=Π_(i=1) ^k p_i ^e_i N=a^n =Π_(i=1) ^k p^(ne_i ) d(N)=Π_(i=1) ^k (ne_i +1)=(ne_1 +1)Π_(i=2) ^k (ne_i +1)=ne_1 Π_(i=2) ^k (ne_i +1)+Π_(i=2) ^k (ne_i +1) d(N) mod n=[Π_(i=1) ^k (ne_i +1)] mod n=[Π_(i=2) ^k (ne_i +1)] mod n=[Π_(i=3) ^k (ne_i +1)] mod n=...=(ne_k +1) mod n=1 ⇒n ∣ [d(N)−1] let′s look at N=48=2^4 ×3^1 d(N)=5×2=10 d(N)−1=9 mod 3=0 but N is not a perfect 3rd power, since there is no integer a with a^3 =48.](https://www.tinkutara.com/question/Q27861.png)
$$\mathrm{N}={a}^{{n}} \\ $$$${a}=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}_{{i}} ^{{e}_{{i}} } \\ $$$$\mathrm{N}={a}^{{n}} =\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}^{{ne}_{{i}} } \\ $$$${d}\left({N}\right)=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\left({ne}_{{i}} +\mathrm{1}\right)=\left({ne}_{\mathrm{1}} +\mathrm{1}\right)\underset{{i}=\mathrm{2}} {\overset{{k}} {\prod}}\left({ne}_{{i}} +\mathrm{1}\right)={ne}_{\mathrm{1}} \underset{{i}=\mathrm{2}} {\overset{{k}} {\prod}}\left({ne}_{{i}} +\mathrm{1}\right)+\underset{{i}=\mathrm{2}} {\overset{{k}} {\prod}}\left({ne}_{{i}} +\mathrm{1}\right) \\ $$$${d}\left({N}\right)\:{mod}\:{n}=\left[\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\left({ne}_{{i}} +\mathrm{1}\right)\right]\:{mod}\:{n}=\left[\underset{{i}=\mathrm{2}} {\overset{{k}} {\prod}}\left({ne}_{{i}} +\mathrm{1}\right)\right]\:{mod}\:{n}=\left[\underset{{i}=\mathrm{3}} {\overset{{k}} {\prod}}\left({ne}_{{i}} +\mathrm{1}\right)\right]\:{mod}\:{n}=…=\left({ne}_{{k}} +\mathrm{1}\right)\:{mod}\:{n}=\mathrm{1} \\ $$$$\Rightarrow{n}\:\mid\:\left[{d}\left({N}\right)−\mathrm{1}\right] \\ $$$$ \\ $$$${let}'{s}\:{look}\:{at}\:{N}=\mathrm{48}=\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{1}} \\ $$$${d}\left({N}\right)=\mathrm{5}×\mathrm{2}=\mathrm{10} \\ $$$${d}\left({N}\right)−\mathrm{1}=\mathrm{9}\:{mod}\:\mathrm{3}=\mathrm{0} \\ $$$${but}\:{N}\:{is}\:{not}\:{a}\:{perfect}\:\mathrm{3}{rd}\:{power},\:{since} \\ $$$${there}\:{is}\:{no}\:{integer}\:{a}\:{with}\:{a}^{\mathrm{3}} =\mathrm{48}. \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jan/18
$$\mathrm{e}^{\mathrm{x}} \mathrm{cellent}\:\mathrm{Sir}! \\ $$