Question

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

Answer 1

Answer: 198

$N=(a+ib{)}^3-107i$
$=a^3+3a^{2}bi+3a{b}^2{i}^2+b^3{i}^3-107i$
$=\left(a^3-3a{b}^2\right)+i\left(3a^{2}b-b^3-107\right)$
since $N$ is + ve real hence its imaginary part $=$ zero
$b\left(3a^2-b^2\right)=107$
since 107 is prime, hence it has exactly 2 divisiors 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.
$\therefore 3a^2-b^2=107\Rightarrow 3a^2=108\Rightarrow a^2=36$
$\Rightarrow a=6(a\ne -6\text{ as it is given })$
$\therefore N=216-3\cdot 6\cdot 1=198$