\(\left(mn-2\right)⋮3\Rightarrow mn\) chia cho 3 dư 2
Đặt \(m=3k+r;n=3p+q\left(p;q;r;k\in N;r\ne q;1\le r;q\le2\right)\)
Vì m;n bình đẳng nên giả sử \(m\ge n\) \(\Rightarrow r\ge q\Rightarrow r=1;q=2\)
Ta có : \(x^m+x^n+1=x^{3k+1}+x^{3p+2}+1\)
\(=\left(x^{3k+1}-x\right)+\left(x^{3p+2}-x^2\right)+\left(x^2+x+1\right)\)
\(=x\left(x^{3k}-1\right)+x^2\left(x^{3p}-1\right)+\left(x^2+x+1\right)\)
Ta thấy \(x\left(x^{3k}-1\right)+x^2\left(x^{3p}-1\right)⋮x^3-1⋮x^2+x+1\)
\(\Rightarrow\)\(x\left(x^{3k}-1\right)+x^2\left(x^{3p}-1\right)+\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\)
Hay \(x^m+x^n+1⋮x^2+x+1\)