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Question Number 111650 by mohammad17 last updated on 04/Sep/20

Commented by mohammad17 last updated on 04/Sep/20

$p l e a s s i r h e l p m e (b y t h e d i v i s i b i l i t y)$
Commented by mohammad17 last updated on 04/Sep/20

$a r e y o u c a n h e l p$
	Answered by Her_Majesty last updated on 04/Sep/20

	$n^{2} ∣ (n^{n - 1} - 1) \Rightarrow n^{2} k = n^{n - 1} - 1; k \in N$ $k = \frac{n^{n - 1} - 1}{n^{2}} = n^{n - 3} - \frac{1}{n^{2}}$ $b u t n^{n - 3} \in N f o r n ⩾ 3 \land \frac{1}{n^{2}} \in N o n l y f o r n = 1$ $\Rightarrow$ $s t r i c t l y n^{2} ∤ (n^{n - 1} - 1) \land n \in N i s w r o n g b e c a u s e$ $1 \in N a n d o f c o u r s e 1 ∣ 0$ $b u t {i t}^{'} s p r o v e n f o r n \in N \land n > 1$
	Commented by mohammad17 last updated on 04/Sep/20

	$t h a n k y o u s i r$
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