1. ## Gaussian integers

Use the Gaussian integers to find the solutions in rational integers of the diophantine equation x^2+1=y^3

a) Show that if x and y are integers such that x^2+1=y^3, then x - i and x + i are relatively prime
b) Show that there are integers r and s such that x= r^3 - 3r*(s^2) and 3*(r^2)*s-s^3=1
c) Find all solutions in integers x^2+1=y^3 by analyzing the equations for r and s in part (b)

2. Originally Posted by totally-confused
a) Show that if x and y are integers such that x^2+1=y^3, then x - i and x + i are relatively prime
Let $d$ be a greatest common divisor between $x-i$ and $x+i$. Then $d|(x-i)$ and $d|(x+i)$. But since $d| (x-i)$ it means $d|[(x+i) - 2i]$ but $d|(x+i)$ thus $d|2i$. Thus, $d|i$ or $d|2$. If $d|i$ then $d$ is a unit, and this is what we want. Thus, it suffices to show $d|2$ is impossible, and the (non-unit) divisors of $2$ are $\pm 2,\pm i2$ because they are all associate it sufficies to show $d=2$ is impossible. If $2|(x-i)$ and $2|(x+i)$ then $4|(x^2+1)$. But simple trial and error on the congruences mod 4 shows it is impossible to have a square congruent to 1 mod 4. Thus, $4\not | (x^2+1)$, a contradiction. Thus, $d\not = 2$. And so $d$ must be a unit.
b) Show that there are integers r and s such that x= r^3 - 3r*(s^2) and 3*(r^2)*s-s^3=1
Here we use the fact that $\mathbb{Z}[i]$ is a unique factorization domain. As a consequence it means if $(x-i)(x+i) = y^3$ and $(x-i),(x+i)$ are relatively prime then each factor must be a cube itself, so, $x+i = \alpha^3 = (r+is)^3 = (r^3 -3rs^2) + i(3r^2s - s^3)$ for some $r,s\in \mathbb{Z}$. Thus, $x = r^3 - 3rs^2$ and $3r^2s - s^3 = 1$.
c) Find all solutions in integers x^2+1=y^3 by analyzing the equations for r and s in part (b)[/FONT][/COLOR]
Since $3r^2s - s^3 = 1 \implies s(3r^2 - s^2) = 1$. There are two possibilities $s=\pm 1$ and so $3r^2 - 1 = \pm 1$ so $r=0$ which means $x=0$ which leads to $y=0$. Thus, $x=0,y=0$ is the only solution.