Ta có: \(\overline{abc}=100a+10b+c=n^2-1\left(1\right)\)
\(\overline{cba}=100c+10b+a=\left(n-2\right)^2=n^2-4n+4\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\overline{abc}-\overline{cba}=\left(100a+10b+c\right)-\left(100c-10b-c\right)\)
\(\Rightarrow n^2-1-\left(n^2-4n+4\right)=99\left(a-c\right)\)
\(\Rightarrow n^2-1-n^2+4n-4=99\left(a-c\right)\)
\(\Rightarrow4n-5=99\left(a-c\right)\)
\(\Rightarrow4n-5⋮99\left(3\right)\)
Mặt khác: \(100\le n^2-1\le999\)
\(\Leftrightarrow101\le n^2\le1000\)
\(\Leftrightarrow11\le n\le31\)
\(\Leftrightarrow39\le4n-5\le119\left(4\right)\)
Từ (3) và (4) \(\Rightarrow4n-5=99\Rightarrow n=26\)
\(\Rightarrow\overline{abc}=26^2-1=675\)
Vậy \(\overline{abc}=675\)
