Task B-3.1. Determine all four-digit natural numbers divisible by 45 for which the difference of the squares of the hundreds digit and the tens digit is 24.
Solution. Let the desired number be $\overline{a b c d}$. We have

\[
\begin{aligned}
& \overline{a b c d}=45 \cdot x \\
& b^{2}-c^{2}=24
\end{aligned}
\]

Since the product $45 \cdot x$ ends with the digit 0 or 5, the number $\overline{a b c d}$ is of the form $\overline{a b c 0}$ or $\overline{a b c 5}$.

From $b^{2}-c^{2}=24$ it follows that $(b-c)(b+c)=24$ and $b, c \in\{0,1,2, \ldots, 9\}, b<c$.

Since $b+c<18$, we can write the number 24 as the product of the numbers:

\[
2 \cdot 12, \quad 3 \cdot 8 \quad \text { and } \quad 4 \cdot 6
\]

Then we have

(a)

\[
(b-c)(b+c)=2 \cdot 12 \quad \Rightarrow \quad\left\{\begin{array}{l}
b-c=2 \\
b+c=12
\end{array}\right.
\]

which gives $b=7, c=5$,

(b)

\[
(b-c)(b+c)=3 \cdot 8 \quad \Rightarrow \quad\left\{\begin{array}{l}
b-c=3 \\
b+c=8
\end{array}\right.
\]

so $b, c \notin\{0,1,2, \ldots, 9\}$,

(c)

\[
(b-c)(b+c)=4 \cdot 6 \quad \Rightarrow \quad\left\{\begin{array}{l}
b-c=4 \\
b+c=6
\end{array}\right.
\]

which gives $b=5, c=1$.

The possible forms of the number $\overline{a b c d}$ are now $\overline{a 750}, \overline{a 510}, \overline{a 755}$ or $\overline{a 515}$.

Each of these numbers is divisible by 5, and to be divisible by 45, it must also be divisible by 9. The sum of the digits of each number must be divisible by 9.

For $\overline{a 750}$, the sum of the digits is $a+7+5+0=a+12$, which with $a \in\{1,2,3, \ldots, 9\}$ gives $a=6$.

For $\overline{a 510}$, the sum of the digits is $a+5+1+0=a+6$ so $a=3$.

For $\overline{a 755}$, the sum of the digits is $a+7+5+5=a+17$ so $a=1$.

For $\overline{a 515}$, the sum of the digits is $a+5+1+5=a+11$ so $a=7$.

Therefore, the desired numbers are 6750, 3510, 1755, and 7515.