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The problem:

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

I found these related posts:
Prove that diophantine equation has only two solutions.
Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
Show that $x^2-dy^2 = -2$ with $d = m^2+2$ has infinitetly many integer solutions

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 $\mathbb{Z}[\sqrt{-2}]$ ?

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    $\begingroup$ factor in the Euclidean domain $\mathbb{Z}[\sqrt{-2}]$, etc. $\endgroup$ – user10354138 Jun 25 '19 at 0:52
  • $\begingroup$ @user10354138 thanks for your comment! I edited my question to show what I did thanks to your hint. $\endgroup$ – evaristegd Jun 25 '19 at 1:20
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    $\begingroup$ perhaps try solving for $c$ and $d$: $$ \left (c + d \sqrt -2 \right)^3 = y + \sqrt -2$$ $\endgroup$ – jonan Jun 25 '19 at 3:23
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    $\begingroup$ @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. $\endgroup$ – evaristegd Jun 25 '19 at 16:54
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    $\begingroup$ @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. $\endgroup$ – jonan Jun 26 '19 at 3:49