$y^2= x^3-2$Y2X322004-11-30 12:41:012009-01-26 12:45:29ApplicationEuclidean domain\usepackage{graphicx}
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\newtheorem{dfn}{Definition}We want to solve the equation $y^2=x^3 - 2$ over the integers.
By writing $y^2+2=x^3$ we can factor on $\Z[\sqrt{-2}]$ as
\[(y-i\sqrt{2})(y+i\sqrt{2})=x^3.
\]
Using congruences modulo $8$, one can show that both $x,y$ must be odd, and it can also be shown that $(y-i\sqrt{2})$ and $(y+i\sqrt{2})$ are relatively prime (if it were not the case, any divisor would have even norm, which is not possible).
Therefore, by unique factorization, and using that the only \PMlinkname{units}{UnitsOfQuadraticFields} on $\Z[\sqrt{-2}]$ are $1,-1$, we have that each factor must be a cube.
So let us write
\[
(y+i\sqrt{2}) = (a+bi\sqrt{2})^3 = (a^3 - 6ab^2) + i(3a^2b-2b^3)\sqrt{2}
\]
Then $y=a^3 - 6ab^2$ and $1=3a^2b-2b^3=b(3a^2-2b^2)$. These two equations imply $b=\pm 1$ and thus $a=\pm 1$, from where the only possible solutions are $x=3, y=\pm 5$.
\begin{thebibliography}{9}
\bibitem{esm}
Esmonde, Ram Murty; \emph{Problems in Algebraic Number Theory}. Springer.
\end{thebibliography}