Để \(N\) nguyên thì \(n^2+3n-2⋮n^2-3\)
\(\Rightarrow n^2-3+3n+1⋮n^2-3\)
\(\Rightarrow3n+1⋮n^2-3\)
\(\Rightarrow\left(3n+1\right)\left(3n-1\right)⋮n^2-3\)
\(\Rightarrow9n^2-1⋮n^2-3\)
\(\Rightarrow9n^2-27+26⋮n^2-3\)
\(\Rightarrow9\left(n^2-3\right)+26⋮n^2-3\)
\(\Rightarrow26⋮n^2-3\)
\(\Rightarrow n^2-3\inƯ\left(26\right)=\left\{-26,-13,-2,-1,1,2,13,26\right\}\)
Vì \(n^2\ge0\Rightarrow n^2-3\ge-3\) nên \(n^2-3\in\left\{-2,-1,1,2,13,26\right\}\)
\(\Rightarrow n^2\in\left\{1,2,4,5,16,29\right\}\)
Vì \(n^2\) là số chính phương nên \(n^2\in\left\{1,4,16\right\}\)
\(\Rightarrow n\in\left\{-1,1,-2,2,-4,4\right\}\)
Thử lại thấy \(n\in\left\{-1,1,-2,2,4\right\}\) thỏa mãn
bao binh lam sai bét
