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Question Number 27816 by Rasheed.Sindhi last updated on 15/Jan/18

If N is perfect nth power, prove that

n ∣ (d (N) - 1)

[W h e r e d (N) d e n o t e s n u m b e r

o f d i v i s o r s o f N]

Also show by an example that its

vice versa is not necessarily correct .

Commented by Rasheed.Sindhi last updated on 16/Jan/18

You can't use 'macro parameter character #' in math mode

Answered by mrW2 last updated on 15/Jan/18

N = a^{n}

a = \underset{i = 1}{\prod^{k}} p_{i}^{e_{i}}

N = a^{n} = \underset{i = 1}{\prod^{k}} p^{{n e}_{i}}

d (N) = \underset{i = 1}{\prod^{k}} ({n e}_{i} + 1) = ({n e}_{1} + 1) \underset{i = 2}{\prod^{k}} ({n e}_{i} + 1) = {n e}_{1} \underset{i = 2}{\prod^{k}} ({n e}_{i} + 1) + \underset{i = 2}{\prod^{k}} ({n e}_{i} + 1)

d (N) m o d n = [\underset{i = 1}{\prod^{k}} ({n e}_{i} + 1)] m o d n = [\underset{i = 2}{\prod^{k}} ({n e}_{i} + 1)] m o d n = [\underset{i = 3}{\prod^{k}} ({n e}_{i} + 1)] m o d n = \dots = ({n e}_{k} + 1) m o d n = 1

\Rightarrow n ∣ [d (N) - 1]

{l e t}^{'} s l o o k a t N = 48 = 2^{4} \times 3^{1}

d (N) = 5 \times 2 = 10

d (N) - 1 = 9 m o d 3 = 0

b u t N i s n o t a p e r f e c t 3 r d p o w e r, s i n c e

t h e r e i s n o i n t e g e r a w i t h a^{3} = 48 .

Commented by Rasheed.Sindhi last updated on 16/Jan/18

e^{x} cellent Sir!