Solving the diophantine equation $y^{2}=x^{3}-2$

It is known that the diophantine equation $y^{2}=x^{3}-2$ has only one positive integer solution $(x,y)=(3,5)$. The proof of it can be read from the book “About Indeterminate Equation” (in Chinese, by Ke Zhao and Sun Qi). But the method I have known of solving this problem is algebraic number theory.

My question:

Does there exists an elementary method for solving the diophantine equation $y^{2}=x^{3}-2$?

I have gotten an elementary method by the assumption that “$x$” is a prime. That is, we can get a conclusion from the equation that $3\mid x$. Since $x$ is a prime we get the only solution $(x,y)=(3,5)$. But if $x$ is not a prime, the answer seems to be difficult.

Thank you for your help.

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The key is to factor this over $\mathbf{Z}[\sqrt{-2}]$ as $(y-\sqrt{-2})(y+\sqrt{-2})=x^3$. Working mod $8$ it is not too hard to see that $x$ and $y$ must be odd, and that $\gcd((y-\sqrt{-2}),(y+\sqrt{-2}))=1$; else every divisor would have an even norm, contradiction.

Therefore since $\mathbf{Z}[\sqrt{-2}]$ is a UFD and since the only units are $\pm 1$, each factor must be a cube. Thus we can write $$(y+\sqrt{-2})=(a+b\sqrt{-2})^3=(a^3-6ab^2)+(3a^2b-2b^4)\sqrt{-2}$$ Then, $y=a^3-6ab^2$ and $1=3a^2b-2b^3=b(3a^2-2b^2)$. From the two equations it is clear that $b=\pm 1\implies a=\pm 1$, so the only possible solutions are $(x,y)=(3,\pm 5)$.

This is a Mordell curve, with $n=-2$. From this fact alone you know that there are finitely many integer solutions.

There is a table of data for various $n$; here you have the information, that you already know, that there is only one solution in positive integers. I don’t know of any elementary way of proving this fact.