Question

If $a$ and $b$ are two positive integers such that $N=(a+ib{)}^3-107i$ is a positive integer, then $\frac{N}{6}$ is

Answer 1

$N=(a+ib{)}^3-107i=a^3-i{b}^3+3a^{2}ib-3a{b}^2-107i=(a^3-3a{b}^2)+i(-b^3+3a^{2}b-107)$
Since, $N$ is a positive integer so,
$Im\left(z\right)=0\Rightarrow -b^3+3a^{2}b-107=0\Rightarrow b(3a^2-b^2)=107\times 1=1\times 107[\because 107\text{is a prime number}]$

$b=1,3a^2-b^2=107\Rightarrow a=6N=a^3-3a{b}^2=198\therefore \frac{N}{6}=33$
OR,
$b=107,3a^2-b^2=1\Rightarrow a^2=\frac{1+{107}^2}{3}\notin I\text{(reject)}$

Hence, $\frac{N}{6}=33$