[Toán 9] chia hết

H

hoahuongduong50

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V

vy000

Ta cso:

$n-1 \equiv -1 (\mod n)$

$n+1 \equiv 1 (\mod n)$

$\Rightarrow \begin{cases}(n-1)^{n^{n+1}}\equiv -1 (\mod n)\\(n+1)^{n^{n+1}}\equiv 1 (\mod n)\end{cases}$

$\Rightarrow $ đpcm
ps: em nghĩ là $n^n$ mới đúng:|
 
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