Question

If $a$ and $b$ are positive integer such that $N =( a +i b )^3-107 i$ is a positive integer. Find $N$.

Answer 1

$ N =( a +i b )^3-107 i $
$= a ^3+3 a ^2 bi +3 ab ^2 i^2+ b ^3 i^3-107 i $
$=\left( a ^3-3 ab ^2\right)+i\left(3 a ^2 b - b ^3-107\right)$
since $N$ is + ve real hence its imaginary part $=$ zero
$b\left(3 a^2-b^2\right)=107$
since 107 is prime, hence it has exactly 2 divisiors which are 1 and 107 hence $b=1$ and $3 a^2-b^2=107$
if $3 a^2-b^2=1$ and $b=107$ then $a$ is not an integer.
$\therefore 3 a^2-b^2=107 \Rightarrow 3 a^2=108 \Rightarrow a^2=36 $
$\Rightarrow a=6(a \neq-6 \text { as it is given }) $
$\therefore N=216-3 \cdot 6 \cdot 1=198$