Question

If $a$ and $b$ are positive integers such that $N=(a+ib{)}^3-107i$ is a positive integer, then the value of $N$ is

Answer 1

$N=(a+ib{)}^3-107i=a^3+3a^{2}bi+3a{b}^2{i}^2+b^3{i}^3-107i=(a^3-3a{b}^2)+i(3a^{2}b-b^3-107)$
Since $N$ is positive integer, its imaginary part equals zero.
$b(3a^2-b^2)=107$
Since $107$ is prime, hence it has exactly two divisors which are $1$ and $107$
Hence, $b=1$ and $3a^2-b^2=107$
If $3a^2-b^2=1$ and $b=107,$ then $a$ is not an integer.
$\therefore 3a^2-b^2=107$
$\Rightarrow 3a^2=108\Rightarrow a^2=36$
$\Rightarrow a=6$ (Neglecting $a=-6$)
$\therefore N=216-3\cdot 6\cdot 1=198$