# $y^{2}=x^{3}-2$

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 $\mathbbmss{Z}[\sqrt{-2}]$ 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 units (http://planetmath.org/UnitsOfQuadraticFields) on $\mathbbmss{Z}[\sqrt{-2}]$ are $1,-1$, we have that each factor must be a cube.

So let us write

 $(y+i\sqrt{2})=(a+bi\sqrt{2})^{3}=(a^{3}-6ab^{2})+i(3a^{2}b-2b^{3})\sqrt{2}$

Then $y=a^{3}-6ab^{2}$ and $1=3a^{2}b-2b^{3}=b(3a^{2}-2b^{2})$. These two equations imply $b=\pm 1$ and thus $a=\pm 1$, from where the only possible solutions are $x=3,y=\pm 5$.

## References

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Title $y^{2}=x^{3}-2$ Y2X32 2013-03-22 14:52:05 2013-03-22 14:52:05 CWoo (3771) CWoo (3771) 10 CWoo (3771) Application msc 12D05 msc 11R04 $y^{2}+2=x^{3}$ finding integer solutions to $y^{2}+2=x^{3}$ UFD