${y}^{2}={x}^{3}-2$
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 $\mathbb{Z}[\sqrt{-2}]$ 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 $\mathbb{Z}[\sqrt{-2}]$ are $1,-1$, we have that each factor must be a cube.
So let us write
$$(y+i\sqrt{2})={(a+bi\sqrt{2})}^{3}=({a}^{3}-6a{b}^{2})+i(3{a}^{2}b-2{b}^{3})\sqrt{2}$$ |
Then $y={a}^{3}-6a{b}^{2}$ and $1=3{a}^{2}b-2{b}^{3}=b(3{a}^{2}-2{b}^{2})$. 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.
Title | ${y}^{2}={x}^{3}-2$ |
---|---|
Canonical name | Y2X32 |
Date of creation | 2013-03-22 14:52:05 |
Last modified on | 2013-03-22 14:52:05 |
Owner | CWoo (3771) |
Last modified by | CWoo (3771) |
Numerical id | 10 |
Author | CWoo (3771) |
Entry type | Application |
Classification | msc 12D05 |
Classification | msc 11R04 |
Synonym | ${y}^{2}+2={x}^{3}$ |
Synonym | finding integer solutions to ${y}^{2}+2={x}^{3}$ |
Related topic | UFD |