Question

[Phương Lê]

Chứng minh rằng nếu số tự nhiên n không chia hết cho 3 thì x^2n +x^n +1 chia hết cho x^2 +x+1

Answer 1

[Trần Tuấn Hoàng]

- Ta có: \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

\(\Rightarrow\left(x^3-1\right)⋮\left(x^2+x+1\right)\)

- Áp dụng hệ quả nhị thức Newton ta có: \(x^n-1⋮x-1\) với \(n\in N\).

- Vì \(n\) không chia hết cho \(3\) \(\Rightarrow n\) có dạng \(3k+1\) hoặc \(3k+2\) \(\left(k\in N\right)\)

- Với \(n=3k+1\) thì:

\(x^{2n}+x^n+1=x^{2\left(3k+1\right)}+x^{3k+1}+1=x^{6k+2}+x^{3k+1}+1=x^{3k+2}\left(x^{3k}-1\right)+x^{3k}\left(x^2+x+1\right)-\left(x^{3k}-1\right)\)

- Do \(\left\{{}\begin{matrix}x^{3k+2}\left(x^{3k}-1\right)⋮\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\\x^{3k}\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\\\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\end{matrix}\right.\) 

\(\Rightarrow x^{3k+2}\left(x^{3k}-1\right)+x^{3k}\left(x^2+x+1\right)-\left(x^{3k}-1\right)⋮\left(x^2+x+1\right)\)

hay \(x^{2n}+x^n+1⋮x^2+x+1\) khi \(n=3k+1\left(k\in N\right)\) (1).

- Với \(n=3k+2\) thì:

\(x^{2n}+x^n+1=x^{2\left(3k+2\right)}+x^{3k+2}+1=x^{6k+4}+x^{3k+2}+1=x^{3k+4}\left(x^{3k}-1\right)+x^{3k+2}\left(x^2+x+1\right)-\left(x^{3k+3}-1\right)\)- Do \(\left\{{}\begin{matrix}x^{3k+4}\left(x^{3k}-1\right)⋮\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\\x^{3k+2}\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\\\left(x^{3\left(k+1\right)}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\end{matrix}\right.\) 

\(\Rightarrow x^{3k+4}\left(x^{3k}-1\right)+x^{3k+2}\left(x^2+x+1\right)-\left(x^{3k+3}-1\right)⋮\left(x^2+x+1\right)\)

 hay  \(x^{2n}+x^n+1⋮x^2+x+1\) khi \(n=3k+2\left(k\in N\right)\) (2).

- Từ (1), (2) ta suy ra đpcm