prove that if n∈N, n>1 and n is odd then 1^n +...+(n−1)^n is divisible by n (dont use ≡(modn))


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Question Number 192220 by gatocomcirrose last updated on 12/May/23

$prove that if n \in N, n > 1 and n is odd then$ $1^{n} + . . . + {(n - 1)}^{n} is divisible by n$ $(dont use \equiv (modn))$
Commented byAST last updated on 12/May/23
$It follows from the fact that a^n +b^n is divisible by a+b when n is odd. {1^n +(n−1)^n }+{2^n +(n−2)^n }+...+{(((n−1)/2))^n +(((n+1)/2))^n }$
$I t f o l l o w s f r o m t h e f a c t t h a t a^{n} + b^{n} i s d i v i s i b l e$ $b y a + b w h e n n i s o d d .$ ${1^{n} + {(n - 1)}^{n}} + {2^{n} + {(n - 2)}^{n}} + . . . + {{(\frac{n - 1}{2})}^{n} + {(\frac{n + 1}{2})}^{n}}$
Commented bygatocomcirrose last updated on 12/May/23

$ohhhh yeahh thanks bro$
	Answered by Frix last updated on 12/May/23

	$1^{n} + {(n - 1)}^{n} = 1^{n} + n \times (. . .) - 1^{n} = n \times (. . .)$ $2^{n} + {(n - 2)}^{n} = 2^{n} + n \times (. . .) - 2^{n} = n \times (. . .)$ $. . .$ $You get tbe idea ?$
	Commented bygatocomcirrose last updated on 12/May/23

	$yeah thank you!$
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