n2+2n+4
=n2+n+n+1+3
=n(n+1) +(n+1)+3
=(n+1)2 +3
Để n2+2n+4 chia hết cho n+1 thì (n+1) thuộc ước của 3
Ta có bảng
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l n+1 l 1 l -1 l 3 l -3 l
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l n l 0 l -2 l 2 l -4 l
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Vậy ...
Ta có :
\(n^2+2n+4=n^2+n+n+1+3=n\left(n+1\right)+\left(n+1\right)+3=\left(n+1\right)\left(n+1\right)+3\)
Vì \(\left(n+1\right)\left(n+1\right)⋮\left(n+1\right)\\ \left(n+1\right)\left(n+1\right)+4⋮\left(n+1\right)\)
\(\Rightarrow4⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow n\in\left\{-5;-3;-2;0;1;3\right\}\)
n2 + n + n + 1 + 3 \(⋮\) n + 1
n (n + 1) + (n + 1) + 3 \(⋮\)n + 1
n (n + 1) \(⋮\)n + 1 ; n + 1 \(⋮\)n + 1
=> 3 \(⋮\)n + 1
n \(\in\) {- 4, 2}