Task B-4.7. Determine all integers $a, b$ and $c$ such that $c=(a+b i)^{3}-7 i$.
Solution. Let's write the number c in a suitable form

$$
\begin{aligned}
& c=(a+b i)^{3}-7 i \\
& c=a^{3}+3 a^{2} b i-3 a b^{2}-b^{3} i-7 i
\end{aligned}
$$

Equating the real and imaginary parts:

$$
\begin{aligned}
& \left\{\begin{array}{l}
c=a^{3}-3 a b^{2} \\
0=3 a^{2} b-b^{3}-7
\end{array}\right. \\
& \left\{\begin{array}{l}
a\left(a^{2}-3 b^{2}\right)=c \\
b\left(3 a^{2}-b^{2}\right)=7
\end{array}\right.
\end{aligned}
$$

Now we have four possibilities:

$$
\begin{aligned}
& 1^{\circ} b=1,3 a^{2}-b^{2}=7 \\
& 3 a^{2}=8, \text { this equation has no solution in the set of integers } \\
& 2^{\circ} b=-1,3 a^{2}-b^{2}=-7 \\
& 3 a^{2}=-6, \text { this equation has no solution in the set of integers } \\
& 3^{\circ} \quad b=7,3 a^{2}-b^{2}=1 \\
& 3 a^{2}=50, \text { this equation has no solution in the set of integers } \\
& 4^{\circ} \quad b=-7,3 a^{2}-b^{2}=-1 \\
& 3 a^{2}=48, a^{2}=16, a= \pm 4 \\
& c= \pm 4(16-3 \cdot 49) \\
& c_{1}=-524, c_{2}=524
\end{aligned}
$$

The solutions are $a=4, b=-7, c=-524$ or $a=-4, b=-7, c=524$.