# Prove this equation has only two integer solutions [duplicate]

The problem:

Prove that the only integer solutions of the equation $$x^3=y^2+2$$ are $$x=3$$ and $$y=\pm 5$$.

I have tried techinques similar to the ones used in those posts, however I have not been able to solve it yet.

Edit: The first comment gave me the hint of working in $$\mathbb{Z}[\sqrt{-2}]$$.

$$x^3=y^2-(-2)=(y+\sqrt{-2})(y-\sqrt{-2})$$ And we define $$a:=y+\sqrt{-2}$$ and $$b:=y-\sqrt{-2}$$. Now I need to prove that $$a$$ is a cube in $$\mathbb{Z}[\sqrt{-2}]$$. As a side note, that would imply that $$b$$ is also a cube in $$\mathbb{Z}[\sqrt{-2}]$$. Once I prove that, I can use the answers in the post titled "The only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $\mathbb{Z}[\sqrt{-2}]$ is $x=5$", and solve the problem.

How can I prove that $$a$$ is a cube in $$\mathbb{Z}[\sqrt{-2}]$$ ?

• factor in the Euclidean domain $\mathbb{Z}[\sqrt{-2}]$, etc. – user10354138 Jun 25 '19 at 0:52
• @user10354138 thanks for your comment! I edited my question to show what I did thanks to your hint. – evaristegd Jun 25 '19 at 1:20
• perhaps try solving for $c$ and $d$: $$\left (c + d \sqrt -2 \right)^3 = y + \sqrt -2$$ – jonan Jun 25 '19 at 3:23
• @jonan You are assuming that what we want to prove is true. You cannot do that in mathematics. We can only use the fact that the product $ab$ is a cube, because that is all we know. – evaristegd Jun 25 '19 at 16:54
• @evaristegd if we want to prove that 25 is a square, we are looking for an integer solution to the equation: $x^2 = 25$. If such a solution exists, we say 25 is a square. If no such solution exists, we say it is not a square. Am I correct in stating this? If I am, then I do not see how my prior comment is invalid as you are suggesting, with all due respect. – jonan Jun 26 '19 at 3:49