Equating factor form $ u^2 + 1 = 2v^3$

Question

From the equation , $ u^2 + 1 = 2v^3$, we can write, $(u + i)(u − i) = 2v^3$.

I read in a book [1] that

Because $gcd(u +i, u − i) = 1$ and $2 = (1 + i)(1 − i)$, using again the uniqueness of prime factorization in $Z[i]$, we obtain $u + i = (1+ i)(a + bi)^3$

My question is that why $u + i = (1+ i)(a + bi)^3$ (this is what I call equating factor) specifically? Why not -

$u + i = (1- i)(a + bi)^3$

or

$u + i = (1- i)(a + bi)^2$

or

$u + i = (1- i)(a + bi)$

? is there a specific reason for $u + i = (1+ i)(a + bi)^3$ or there are other possibilities ?

Reference:

1. Page 158 of An Introduction to Diophantine Equations by Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

Answer 1

Part 1:

Since $gcd(u +i, u − i) = 1$ and $2 = (1 + i)(1 − i)$, due to the uniqueness of prime factorization in $Z[i]$, you can not write $u + i = (1+ i)(a + bi)^2$ or $u + i = (1+ i)(a + bi)^1$ because then $$gcd(u +i, u − i) = (a + bi)$$ which contradicts $gcd(u +i, u − i) = 1$

Part 2:

So, there are 2 possible cases, those are-

1. $u + i = (1+ i)(a + bi)^3$
2. $u + i = (1- i)(a + bi)^3$

Now, we show that both cases require similar solution.

Case 1: The imaginary part of $(1+ i)(a + bi)^3$ is $a^3+3a^2b-3ab^2-b^3$. Since, $u + i = (1+ i)(a + bi)^3$, so, $a^3+3a^2b-3ab^2-b^3=1\implies (a-b)(a^2+4ab+b^2)=1$, from this we get two systems, they are-

$$(a-b)=1; (a^2+4ab+b^2)=1 \cdots \text{(System 1)}$$

and

$$(a-b)=-1; (a^2+4ab+b^2)=-1 \cdots \text{(System 2)}$$

Case 2: The imaginary part of $(1- i)(a + bi)^3$ is $-a^3+3a^2b+3ab^2-b^3$.

Since, $u + i = (1- i)(a + bi)^3$, so, $-a^3+3a^2b+3ab^2-b^3=1\implies (a+b)(-a^2+4ab-b^2)=1$, from this we get two systems, they are-

$$(a+b)=1; (-a^2+4ab-b^2)=1 \cdots \text{(System 3)}$$

and

$$(a+b)=-1; (-a^2+4ab-b^2)=-1 \cdots \text{(System 4)}$$

From system $(3), (4)$, we get $a+b= \pm 1$, this implies one of $a, b$ is negative integer (if both were negative, then $|a+b|> 1$). If $a<0$ then, $-a^3+3a^2b+3ab^2-b^3$ becomes

$$a^3+3a^2b-3ab^2-b^3$$

which is exactly same as case 1.

If $b<0$ then, $-a^3+3a^2b+3ab^2-b^3$ becomes

$$b^3+3b^2a-3ba^2b-a^3$$

which needs to follow the same procedure as case 1.

So, it is sufficient to solve one of the two cases, i.e. it is sufficient to solve $u + i = (1+ i)(a + bi)^3$.