1 reply to this topic

[Turbo]

CMR:$n^{2}+n+1$ không chia hết cho 9 ($\forall n\epsilon N$)

[Hero Crab]

Giả sử $n^{2}+n+1 \vdots 9$

$\Rightarrow n^{2}+n+1 \vdots 3$

Ta có :$n^{2}+n+1=\left ( n+2 \right )^{2}-3\left ( n+1 \right )$

Vì $n^{2}+n+1 \vdots 3$

$\Rightarrow \left ( n+2 \right )^{2} \vdots 3$

$\Rightarrow \left ( n+2 \right ) \vdots 3$

$\Rightarrow \left ( n+2 \right )^{2} \vdots 9$

Mà $n^{2}+n+1 \vdots 9$

$\Rightarrow 3\left ( n+1 \right ) \vdots 9$ (Vô lý)

$\Rightarrow ĐPCM$

Edited by Hero Crab, 20-10-2018 - 13:07.