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?
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$\begingroup$The only integral solutions to your first problem are $(3, \pm 5)$. The general class of equations are known as Mordell's equation. A fairly elaborate discussion and case by case analysis is provided here.
$\endgroup$ 7 $\begingroup$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.
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