A Special 4-digit number

Let $\overline{cd}=x$ , so that $\overline{ab}=x+1$ and $\overline{abcd}=100(x+1)+x=101x+100$ .

This has to be a square number; so say $y^2=101x+100$ . Note that $y$ must also be a two-digit number (since $\overline{abcd}<10000=100^2$ ).

Factoring using the difference of two squares, we have $y^2-100=(y+10)(y-10)=101x$ . Since $101$ is prime, the only way we can satisfy this with a two-digit number $y$ is if $y+10=101$ , ie $y=91$ , and $y^2=\boxed{8281}$ .

We note that 4-digit perfect squares are $32^2, 33^2, 34^3, \cdots 99^2$ . Therefore we can assume the perfect square we are looking for be $\overline{mn}^2 = (10m+n)^2 = 100m^2 + 20mn + n^2$ , For $n=1$ , we have:

$\begin{cases} \overline{ab} = \left \lfloor \dfrac {100m^2 + 20m + 1}{100} \right \rfloor = m^2 + \left \lfloor \dfrac m5 \right \rfloor \\ \overline{cd} = 20m + 1 - 100 \left \lfloor \dfrac m5 \right \rfloor \end{cases}$

From $\overline{ab} - \overline{cd} = 1$ ,

$\begin{aligned} m^2 + \left \lfloor \dfrac m5 \right \rfloor - 20m - 1 + 100 \left \lfloor \dfrac m5 \right \rfloor & = 1 \\ m^2 - 20m + 101 \left \lfloor \dfrac m5 \right \rfloor - 2 & = 0 \end{aligned}$

$\begin{cases} m^2 - 20m - 2 = 0 & \text{for }m < 5 & \small \color{#D61F06} \text{No integer solution} \\ m^2 - 20m + 99 = 0 & \text{for }m \ge 5 \\ (m-9)(m-11) = 0 \\ \implies m = 9 & & \small \color{#3D99F6} \text{Acceptable solution} \end{cases}$

Therefore the perfect square is $\overline{mn}^2 = 91^2 = \boxed{8281}$ .

$\text{Table}\left[\text{If}\left[\sqrt{100 (n+1)+n}\in \mathbb{Z},100 (n+1)+n,\text{Nothing}\right],\{n,10,99\}\right] \Rightarrow 8281$

English translation: make a table of when the value of $\100 (n+1)+n$ when the square root of the value is an integer while varying n from 10 to 99 inclusive. That range was selected because of the two digit number used to construct the four digit number. The resultant listconsisted of just one element, 8281. The expression used is just the composition of the four digit number from the two digit number.