Đặt \(n^2-n+2=k^2\left(k\ge n\right)\)
\(\Rightarrow n^2-n-2=k^2-4\)
\(\Rightarrow\left(n+1\right)\left(n-2\right)=\left(k+2\right)\left(k-2\right)\)
\(\circledast k=-2\Leftrightarrow n=-1\left(tm\right)\)
\(\circledast k=2\Rightarrow n=2\left(tm\right)\)
\(\circledast k\ne\pm2\)
Do \(n-2\le k-2\Leftrightarrow n+1\ge k+2\)
Mà: \(n+1\le k+1\)
\(\Rightarrow k+2\le n+1\le k+1\) (vô lí)
Vậy n = -1; 2
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