vì \(n^2+n+1589\) là số chính phương .
\(\Rightarrow n^2+n+1589=n^2+n+\left(2a+1\right)n+\left(a+1\right)^2\) (với \(a\in Z\))
\(\Rightarrow\left(2a+1\right)n+\left(a+1\right)^2=1589\) \(\Leftrightarrow n=\dfrac{1589-\left(a+1\right)^2}{2a+1}\)
mà vì \(n\in N\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}1589-\left(a+1\right)^2\ge0\\2a+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}1589-\left(a+1\right)^2\le0\\2a+1< 0\end{matrix}\right.\end{matrix}\right.\\1589-\left(a+1\right)^2⋮2a+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}-40,9\le a\le38,9\\a>\dfrac{-1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}a\ge38,9\\a\le-40,9\end{matrix}\right.\\a< \dfrac{-1}{2}\end{matrix}\right.\end{matrix}\right.\\1588-a^2-2a⋮2a+1\end{matrix}\right.\)
tới đây bn giải bình thường nha :)
Giúp mk nha !
