TITLE: Finding integer solutions to $y^2=x^3-2$
QUESTION [5 upvotes]: I have the equation:
$$y^2=x^3-2$$
It seems to be deceivingly simple, yet I simply cannot crack it. It is obviously equivalent to finding a perfect cube that is two more than a perfect square, and a brute force check shows no solutions other than $y=5$ and $x=3$ under 10,000. 
However, I can't prove it. 
Are there other integer solutions to this equation? If so, how many? If not, can you prove that there aren't? 
Bonus: What about the more general equation"
$$y^2=x^3-c$$
Where $c$ is a positive integer?

REPLY [6 votes]: Fact : $\mathbb{Z}[\sqrt{-2}]$ is Unique factorization domain.
lemma : in every UFD, if the product of two numbers, which are relatively prime is a cube, then each of them must be a cube.
There is no any solution
$$x^3 = (y+\sqrt{-2})\times(y-\sqrt{-2})$$
the greatest common divisor of these factors will divide 2-times the $\sqrt{-2}$,
which is lead to only finitly many cases. (some cases can be shown impossible, only by the modular an congrunce arithmetic.)
finally we have:
$$y+\sqrt{-2}=(a+b\sqrt{-2})^3$$
which lead us to the system of equations as follows:
$$a^3-6ab^2=y$$ and
$$3a^2b-2b^3=1$$
then
$$b(3a^2-2b^2)=1$$
which implies the assertion.