Gọi số cần tìm là \(\overline{abcd}\).
Đặt \(\left\{{}\begin{matrix}\overline{abcd}=y^2\\\overline{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}=x^2\end{matrix}\right.\)(x,y > 0, x > y)
\(\Rightarrow x^2-y^2=1000\left(a+1\right)+100\left(b+1\right)+10\left(c+1\right)+\left(d+1\right)-1000a-100b-10c-d\)
\(\Rightarrow\left(x+y\right)\left(x-y\right)=1111=101\cdot11=1111\cdot1\)
Vì \(x,y>0\)
\(\Rightarrow x+y>x-y\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=101\\x-y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=1111\\x-y=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=56,y=45\\x=556,y=555\end{matrix}\right.\)
* x = 56, y = 45
\(\Rightarrow\overline{abcd}=45^2=2025\)
\(\Rightarrow\overline{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}=3136=56^2\) (thoả mãn)
* x = 556, y = 555
\(\Rightarrow x^2=309136>9999\left(loai\right)\)
Vậy số đó là 2025
