## You are here

Home$y^2= x^3-2$

## Primary tabs

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

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 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

- 1
Esmonde, Ram Murty;
*Problems in Algebraic Number Theory*. Springer.

Related:

UFD

Synonym:

$y^2+2=x^3$,finding integer solutions to $y^2+2=x^3$

Type of Math Object:

Application

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

12D05*no label found*11R04

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Corrections

Units of Z[RQ(-2)] by pahio ✓

title by mathcam ✓

x,y must be odd, by vmoraru ✘

factoring out b by curious ✓

title by mathcam ✓

x,y must be odd, by vmoraru ✘

factoring out b by curious ✓