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∈Z thì :
2n−1⋮n−12n−1⋮n−1
Mà n−1⋮n−1n−1⋮n−1
⇔⎧⎨⎩2n−1⋮n−12n−2⋮n−1⇔{2n−1⋮n−12n−2⋮n−1
⇔1⋮n−1⇔1⋮n−1
⇔n−1∈Ư(1)⇔n−1∈Ư(1)
⇔[n−1=1n−1=−1⇔[n−1=1n−1=−1
⇔[n=2n=0⇔[n=2n=0
