Integral solutions to $y^{2}=x^{3}-1$

How to prove that the only integral solutions to the equation $$y^{2}=x^{3}-1$$ is $x=1, y=0$. I rewrote the equation as $y^{2}+1=x^{3}$ and then we can factorize $y^{2}+1$ as $$y^{2}+1 = (y+i) \cdot (y-i)$$ in $\mathbb{Z}[i]$. Next i claim that the factor's $y+i$ and $y-i$ are co-prime. But i am not able to show this. Any help would be useful. Moreover, i would also like to see different proofs of this question.

Extending Consider the equation $$y^{a}=x^{b}-1$$ where $a,b \in \mathbb{Z}$ and $(a,b)=1$ and $a < b$. Then is there any result which states about the nature of the solution to this equation.

Solution 1:

Alon Amit's answer is not right, I believe.

$2$ is not prime in the Gaussian integers! $(1+i)$ divides $2$.

The argument upto the fact that the prime $p$ divides $y+i$, $y-i$ and $2i$ is correct.

Since $p|2i$ we consider $p = 1+i$. Now any multiple of $1+i$ is of the form $2x + i2y$ or $2x+1 + i(2y+1)$. Since $y$ is even, $y+i$ cannot be divisible by $1+i$ and thus $y+i$ and $y-i$ are co-prime.

Setting $y+i$ to be a perfect cube (upto units), i.e. $y+i = (xi)^3$, easily gives us $y=0$ and so $x=1, y=0$ is the only solution.

Solution 2:

If $x$ is even then $x^3$ is divisible by 4, so $x^3-1 \equiv 3 \pmod 4$ and this cannot be a square. Thus $x$ is odd and $y$ is even.

Now, if a prime $p$ divides both $y+i$ and $y-i$ then it divides their difference $2i$. Thus $p=2$ (up to units), but then $p$ divides $y$. That's impossible since it divides $y+i$ as well.

EDIT: I was too hasty in writing this, as pointed out by Moron. The prime 2 ramifies in $\mathbb{Z}[i]$, and is (up to units) the square of $1+i$. Sorry for the confusion.

Solution 3:

Let $\alpha\in\mathbb{Z}[i]$ be a divisor of $y+i$ and $y-i$. Then $\alpha|2=i(y-i-(y+i))$ and $\alpha|(y-i)(y+i)=x^3$. Since $x$ is odd then $x^3$ is odd and therefore, by Bezout there exist $A,B\in\mathbb{Z}$ such that $Ax^3+2B=1$ and therefore $\alpha|1$ implying $\alpha\in\mathbb{Z}[i]^{\times}$. We conclude that $y+i$ and $y-i$ are coprime.