Find $c$ if $a,b, \; c$ satisfy $c = (a+bi)^3 - 107i$

Question

Find $c$ if $a,b, \; c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$

I can try expanding the cube, but that seems too direct. What other ways are there to go about this?

Answer 1

$$\Im((a+bi)^3-107i)=3a^2b-b^3-107=0,$$ $$(3a^2-b^2)b=107.$$ $107$ is a prime, so $b=1\lor b=107$. $$b=1\implies a=6,$$ $$b=107\implies 3a^2=11450\text{ (reject)}.$$ $$c=(a+bi)^3-107i=198.$$

Answer 2

Since $c \in \mathbb Z$, the right-hand side must have an imaginary component of $0$. Expanding, we get

$$c = (a+bi)^3 - 107i = (a^3-3ab^2) + (3a^2b-b^3-107)i$$

Therefore $3a^2b-b^3-107 = 0$, which implies that $(3a^2-b^2)b=107$. Since $107$ is prime, we conclude that $b$ is either $1$ or $107$.

Case 1: $b=1$. Then $3a^2 -1 = 107$, so $a=6$.

Case 2: $b=107$. Then $3a^2 - 107^2 = 1$. But there is no integer solution to this equation.

Therefore the unique solution is $(a,b,c) = (6,1,198)$.