Question

Chứng minh rằng với mọi số tự nhiên m, n thì :

\(x^{6m+4}+x^{6n+2}+1\) chia hết cho \(x^4+x^2+1\)

Answer 1

\(x^{6m+4}-x^4+x^{6n+2}-x^2+x^4+x^2+1\)

\(=x^4\left(x^{6m}-1\right)+x^2\left(x^{6n}-1\right)+x^4+x^2+1\)(1)

Ta có \(x^{6n}-1=\left(x^6-1\right)\left(x^{6\left(n-1\right)}+x^{6\left(n-2\right)}+...+x^6+1\right)⋮\left(x^6-1\right)\)

Tương tự \(\left(x^{6n}-1\right)⋮\left(x^6-1\right)\)

Mà \(x^6-1=\left(x^2\right)^3-1=\left(x^2-1\right)\left(x^4+x^2+1\right)⋮\left(x^4+x^2+1\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x^{6m}-1\right)⋮\left(x^4+x^2+1\right)\\\left(x^{6n}-1\right)⋮\left(x^4+x^2+1\right)\end{matrix}\right.\) (2)

Từ (1);(2) \(\Rightarrow\left(x^{6m+4}+x^{6n+4}+1\right)⋮\left(x^4+x^2+1\right)\)