Để \(P\in Z\) thì :
\(2n-1⋮n-1\)
Mà \(n-1⋮n-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2n-1⋮n-1\\2n-2⋮n-1\end{matrix}\right.\)
\(\Leftrightarrow1⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}n-1=1\\n-1=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=2\\n=0\end{matrix}\right.\)
Vậy ..
Để \(P\in Z\), ta có
\(\Rightarrow\) \(2n-1⋮n-1\)
\(\Rightarrow\) \(2n-2+1⋮n-1\)
\(\Rightarrow\) \(2.\left(n-1\right)+1⋮n-1\)
Mà \(2.\left(n-1\right)⋮n-1\)
\(\Rightarrow\) \(1⋮n-1\)
\(\Rightarrow\) \(n-1=Ư\left(1\right)\left\{{}\begin{matrix}-1\\1\end{matrix}\right.\)
\(\Rightarrow\)\(n\left\{{}\begin{matrix}-2\\0\end{matrix}\right.\)
Vậy \(n\left\{{}\begin{matrix}-2\\0\end{matrix}\right.\)
