## Task A-2.7. (20 points)

Determine the natural numbers $a, b$ and $c$ such that the equality $(a+b i)^{3}-107 i=c$ holds. ( $i$ is the imaginary unit.)
## Solution.

After cubing the binomial and simplifying the expression, we get

$$
a^{3}-3 a b^{2}+\left(3 a^{2} b-b^{3}-107\right) i=c
$$

Since $a$, $b$, and $c$ are real, by equating the real and imaginary parts, we obtain

$$
a^{3}-3 a b^{2}=c, \quad 3 a^{2} b-b^{3}-107=0
$$

From the second equation, we get $\left(3 a^{2}-b^{2}\right) b=107$. Since 107 is a prime number and $a$ and $b$ are natural numbers, we conclude

$$
3 a^{2}-b^{2}=107, \quad b=1
$$

or

$$
3 a^{2}-b^{2}=1, \quad b=107
$$

In the first case, $3 a^{2}=108$, so $a=6$, and in the second case, $3 a^{2}=11450$, so $a \notin \mathbb{N}$. Now, from the equality of the real parts of the original equation, we get $c=a^{3}-3 a b^{2}=6^{3}-3 \cdot 6 \cdot 1^{2}=198$.

(5 points)