7.4. Determine all four-digit natural numbers $\overline{a b c d}$ such that $\overline{a b c d}=\overline{a b}+(\overline{c d})^{2}$. (Here, $\overline{p q r s}$ and $\overline{p q}$ denote the numbers written with the digits $p, q, r, s$ and $p, q$ respectively, where $p \neq 0$.)
## Solution:

From $\overline{a b c d}=\overline{a b}+(\overline{c d})^{2}$, we get $100 \cdot \overline{a b}+\overline{c d}=\overline{a b}+(\overline{c d})^{2}, 99 \cdot \overline{a b}=(\overline{c d})^{2}-\overline{c d}, 99 \cdot \overline{a b}=\overline{c d} \cdot (\overline{c d}-1)$. From the last equality, it follows that $\overline{c d} \cdot (\overline{c d}-1) \vdots 11$ and $\overline{c d} \cdot (\overline{c d}-1) \vdots 9$. Consider $\overline{c d} \cdot (\overline{c d}-1) \vdots 11$. Since the numbers $\overline{c d}$ and $\overline{c d}-1$ are coprime, we get $\overline{c d} \vdots 11$ or $(\overline{c d}-1) \vdots 11$.

If $\overline{c d} \vdots 11$ then $\overline{c d} \in\{11,22,33,44,55,66,77,88,99\}$.

For $\overline{c d}=11$ we get $99 \cdot \overline{a b}=11 \cdot 10, \overline{a b}=\frac{10}{9}$ - impossible. For $\overline{c d}=22$ we get $99 \cdot \overline{a b}=22 \cdot 21$, $\overline{a b}=\frac{14}{3}$ - impossible. For $\overline{c d}=33$ we get $99 \cdot \overline{a b}=33 \cdot 32, \overline{a b}=\frac{32}{3}$ - impossible. For $\overline{c d}=44$ we get $99 \cdot \overline{a b}=44 \cdot 43, \overline{a b}=\frac{172}{9}$ - impossible. For $\overline{c d}=55$ we get $99 \cdot \overline{a b}=55 \cdot 54, \overline{a b}=30$. For $\overline{c d}=66$ we get $99 \cdot \overline{a b}=66 \cdot 65, \overline{a b}=\frac{130}{3}$ - impossible. For $\overline{c d}=77$ we get $99 \cdot \overline{a b}=77 \cdot 76, \overline{a b}=\frac{532}{9}$ - impossible. For $\overline{c d}=88$ we get $99 \cdot \overline{a b}=88 \cdot 87, \overline{a b}=\frac{232}{3}$ - impossible. For $\overline{c d}=99$ we get $99 \cdot \overline{a b}=99 \cdot 98, \overline{a b}=98$.

If $(\overline{c d}-1) \vdots 11$ then $\overline{c d} \in\{12,23,34,45,56,67,78,89\}$.

For $\overline{c d}=12$ we get $99 \cdot \overline{a b}=12 \cdot 11, \overline{a b}=\frac{4}{3}$ - impossible. For $\overline{c d}=23$ we get $99 \cdot \overline{a b}=23 \cdot 22$, $\overline{a b}=\frac{46}{3}$ - impossible. For $\overline{c d}=34$ we get $99 \cdot \overline{a b}=34 \cdot 33, \overline{a b}=\frac{34}{3}$ - impossible. For $\overline{c d}=45$ we get $99 \cdot \overline{a b}=45 \cdot 44, \overline{a b}=20$. For $\overline{c d}=56$ we get $99 \cdot \overline{a b}=56 \cdot 55, \overline{a b}=\frac{280}{9}$ - impossible. For $\overline{c d}=67$ we get $99 \cdot \overline{a b}=67 \cdot 66, \overline{a b}=\frac{134}{3}$ - impossible. For $\overline{c d}=78$ we get $99 \cdot \overline{a b}=78 \cdot 77, \overline{a b}=\frac{182}{3}$ - impossible. For $\overline{c d}=89$ we get $99 \cdot \overline{a b}=89 \cdot 88, \overline{a b}=\frac{712}{9}$ - impossible.

Thus, we get $\overline{a b c d} \in\{2045,3055,9899\}$.

Answer: $\overline{a b c d} \in\{2045,3055,9899\}$