Solution:

$8281$

Step by step solution:

Let $\overline{abcd}$ be the giving square number.

So it can be rewrite as $\overline{abcd}= k^2\quad (32 \leqslant k < 100)$

Or: $1000a + 100b + 10c + d = k^2$

In the other hand, we have:

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

Or: $10a + b - (10c + d) = 1$

Then we get:

$1000a + 100b + 10c + d + 10a + b - (10c + d)= k^2 +1$

It's equal to: $\quad 1010a + 101b = k^2 + 1$

Or: $\quad 101(10a + b) = k^2 + 1$

It means: $\quad k^2 + 1\ \vdots\ 101$

So that: $\quad k^2 + 101 - 100\ \vdots\ 101$

Thus: $\quad k^2 - 100\ \vdots\ 101$

Hence: $\quad (k-10)(k+10)\ \vdots\ 101$

Because $\quad k - 10 < 101$

Then: $\quad k + 10\ \vdots\ 101$

Therefore: $\quad k = 91$

We get: $\quad \overline{abcd}= 91^2 = 8281$

Answer: $8281$

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