Giải thích các bước giải:
Ta có :
$\begin{cases}\overline{abc}=n^2-1\\ \overline{cba}=(n-2)^2=n^2-4n+4\end{cases}$
$\to\begin{cases}100a+10b+c=n^2-1\\100c+10b+a=(n-2)^2=n^2-4n+4\end{cases}$
$\to 100a+10b+c-(100c+10b+a)=n^2-1-(n^2-4n+4)$
$\to 99(a-c)=4n-5$
$\to 4n-5\quad\vdots\quad 99$
$\to n=99k+26$
$\to 99(a-c)=4(99k+26)-5$
$\to 99(a-c)=4.99k+99$
$\to a-c=4k+1$
Mà $-7\le a-c\le 7$ vì $1\le a,c\le 9$
$\to -7\le 4k+1\le 7$
$\to 4k+1\in\{-7, -3,1,5\}$
$\to k\in\{-2,-1,0,1\}$
$\to n\in\{-172,-73,26,125\}$
$\to \overline{abc}\in\{29583,5328,675,15624\}$
Vì $100\le \overline{abc}\le 999$
$\to \overline{abc}=675$
Thử lại $\overline{cba}=576=(26-2)^2$ đúng
Vậy $\overline{abc}=675$
