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.


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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

	$N = a^{n}$ $a = \underset{i = 1}{\prod^{k}} p_{i}^{e_{i}}$ $N = a^{n} = \underset{i = 1}{\prod^{k}} p^{{n e}_{i}}$ $d (N) = \underset{i = 1}{\prod^{k}} ({n e}_{i} + 1) = ({n e}_{1} + 1) \underset{i = 2}{\prod^{k}} ({n e}_{i} + 1) = {n e}_{1} \underset{i = 2}{\prod^{k}} ({n e}_{i} + 1) + \underset{i = 2}{\prod^{k}} ({n e}_{i} + 1)$ $d (N) m o d n = [\underset{i = 1}{\prod^{k}} ({n e}_{i} + 1)] m o d n = [\underset{i = 2}{\prod^{k}} ({n e}_{i} + 1)] m o d n = [\underset{i = 3}{\prod^{k}} ({n e}_{i} + 1)] m o d n = . . . = ({n e}_{k} + 1) m o d n = 1$ $\Rightarrow n ∣ [d (N) - 1]$ ${l e t}^{'} s l o o k a t N = 48 = 2^{4} \times 3^{1}$ $d (N) = 5 \times 2 = 10$ $d (N) - 1 = 9 m o d 3 = 0$ $b u t N i s n o t a p e r f e c t 3 r d p o w e r, s i n c e$ $t h e r e i s n o i n t e g e r a w i t h a^{3} = 48 .$
	Commented by Rasheed.Sindhi last updated on 16/Jan/18

	$e^{x} cellent Sir!$
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