Do \(n^2+4⋮n+2\)
\(\Rightarrow\left(n^2+2n\right)-\left(2n+4\right)+8⋮n+2\)
\(\Rightarrow n\left(n+2\right)-2\left(n+2\right)+8⋮n+2\)
\(\Rightarrow\left(n-2\right)\left(n+2\right)+8⋮n+2\)
Lại có \(\left(n-2\right)\left(n+2\right)⋮n+2\)
\(\Rightarrow8⋮n+2\)
Mà \(n\in N\)
\(\Rightarrow n+2\inƯ\left(8\right)=\left\{1;2;4;8\right\}\)
Mặt khác \(n+2\ge2\)
| n+2 | 2 | 4 | 8 |
| n | 0 | 2 | 6 |
Vậy n=0;2;6
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