Đáp án:
n∈{0;1}
Giải thích các bước giải:
\(n^5+1⋮n^3+1\)
\(\Rightarrow n^6+n⋮n^3+1\)
Viết về dạng phân số:
\(\dfrac{n^6+n}{n^3+1}=\dfrac{n^6-1+n+1}{n^3+1}=\dfrac{n^6-1}{n^3+1}+\dfrac{n+1}{n^3+1}\)
\(=\dfrac{\left(n^3-1\right)\left(n^3+1\right)}{n^3+1}+\dfrac{n+1}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(=n^3-1+\dfrac{1}{n^2-n+1}\)
Để \(n^5+1⋮n^3+1\) thì: \(\dfrac{1}{n^2-n+1}\in Z\Leftrightarrow n^2-n+1\in\left\{1;-1\right\}\)
\(\circledast\)Với: \(n^2-n+1=1\Leftrightarrow n^2-n=0\Leftrightarrow n\left(n-1\right)=0\Leftrightarrow\left[{}\begin{matrix}n=0\\n=1\end{matrix}\right.\)
\(\circledast\) Với \(n^2-n+1=-1\Leftrightarrow n^2-n+2=0\)
\(\Rightarrow n^2-n+\dfrac{1}{4}+\dfrac{7}{4}=0\)
\(\Rightarrow\left(n-\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\Leftrightarrow\) pt vô nghiệm
Vậy \(n\in\left\{0;1\right\}\)
