\(A=n^{2018}+n^{2008}+1=n^2\left(n^{2016}+1\right)+n\left(n^{2007}+1\right)+\left(n^2+n+1\right)\)
Ta nhận xét: \(\left\{{}\begin{matrix}n^2\left(n^{2016}+1\right)⋮\left(n^2+n+1\right)\\n^2\left(n^{2007}+1\right)⋮\left(n^2+n+1\right)\\\left(n^2+n+1\right)⋮\left(n^2+n+1\right)\end{matrix}\right.\)
\(\Rightarrow A⋮\left(n^2+n+1\right)\)
\(\Rightarrow\left[{}\begin{matrix}n^2+n+1=1\\n^2+n+1=n^{2018}+n^{2008}+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}n\left(n+1\right)=0\\n\left(n-1\right)A\left(n\right)=0\end{matrix}\right.\)\(\left(A\left(n\right)>0\right)\)
\(\Rightarrow\left[{}\begin{matrix}n=0\\n=1\end{matrix}\right.\)
Thế lại A nhận n = 1, A = 3
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