Question 3. If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are positive integers, satisfying $\mathrm{c}=$ $(a+b i)^{3}-107 i$, find $c$. (where $i^{2}=-1$)
$$
\text { Solve } \begin{array}{l} 
c=(a+b i)^{3}-107 i \text { can be written as } \\
c=a\left(a^{2}-3 b^{2}\right)+i\left[b \left(3 a^{2}\right.\right. \\
\left.\left.-b^{2}\right)-107\right] .
\end{array}
$$

Since $c$ is a real number, the imaginary part must be 0, i.e.,
$$
\text { b }\left(3 a^{2}-b^{2}\right)-107 \text {. }
$$

Since $a$ and $b$ are positive integers, and 107 is a prime number, from (2) we can get one of the following two cases: $b=107$ and $3 a^{2}-b^{2}$ $=1$ or $b=1$ and $3 a^{2}-b^{2}=107$. In the first case, $3 a^{2}=107^{2}+1$. This is impossible, because $107^{2}+1$ is not a multiple of three. Therefore, the first case should be excluded. In the second case, we can find $a=6$. From (1), we get $c=a\left(a^{2}-3 b^{2}\right)=6\left(6^{2}-3.1^{2}\right)=198$.