Câu hỏi của Phương Lê

Question

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

Trần Tuấn Hoàng · Accepted Answer

- Ta có: $x^3-1=\left(x-1\right)\left(x^2+x+1\right)$$\Rightarrow\left(x^3-1\right)⋮\left(x^2+x+1\right)$- Áp dụng hệ quả nhị thức Newton ta có: $x^n-1⋮x-1$ với $n\in N$.- Vì $n$ không chia hết cho $3$ $\Rightarrow n$ có dạng $3k+1$ hoặc $3k+2$ $\left(k\in N\right)$- Với $n=3k+1$ thì:$x^{2n}+x^n+1=x^{2\left(3k+1\right)}+x^{3k+1}+1=x^{6k+2}+x^{3k+1}+1=x^{3k+2}\left(x^{3k}-1\right)+x^{3k}\left(x^2+x+1\right)-\left(x^{3k}-1\right)$- Do $\left\{{}\begin{matrix}x^{3k+2}\left(x^{3k}-1\right)⋮\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\x^{3k}\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\end{matrix}\right.$ $\Rightarrow x^{3k+2}\left(x^{3k}-1\right)+x^{3k}\left(x^2+x+1\right)-\left(x^{3k}-1\right)⋮\left(x^2+x+1\right)$hay $x^{2n}+x^n+1⋮x^2+x+1$ khi $n=3k+1\left(k\in N\right)$ (1).- Với $n=3k+2$ thì:$x^{2n}+x^n+1=x^{2\left(3k+2\right)}+x^{3k+2}+1=x^{6k+4}+x^{3k+2}+1=x^{3k+4}\left(x^{3k}-1\right)+x^{3k+2}\left(x^2+x+1\right)-\left(x^{3k+3}-1\right)$- Do $\left\{{}\begin{matrix}x^{3k+4}\left(x^{3k}-1\right)⋮\left(x^{3k}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\x^{3k+2}\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\\left(x^{3\left(k+1\right)}-1\right)⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\end{matrix}\right.$ $\Rightarrow x^{3k+4}\left(x^{3k}-1\right)+x^{3k+2}\left(x^2+x+1\right)-\left(x^{3k+3}-1\right)⋮\left(x^2+x+1\right)$ hay  $x^{2n}+x^n+1⋮x^2+x+1$ khi $n=3k+2\left(k\in N\right)$ (2).- Từ (1), (2) ta suy ra đpcmCâu trả lời: - Ta có: $x^3-1=\left(x-1\right)\left(x^2+x+1\right)$$\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

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