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[parent] $y^2= x^3-2$ (Application)

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

\begin{displaymath}(y-i\sqrt{2})(y+i\sqrt{2})=x^3. \end{displaymath}

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

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

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

Bibliography

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



"$y^2= x^3-2$" is owned by drini. [ full author list (2) | owner history (1) ]
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See Also: UFD

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

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Cross-references: solutions, imply, cube, norm, even, divisor, relatively prime, odd, congruences, factor, integers, equation
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This is version 6 of $y^2= x^3-2$, born on 2004-11-30, modified 2005-05-16.
Object id is 6544, canonical name is Y2X32.
Accessed 3519 times total.

Classification:
AMS MSC: 11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers)
 12D05 (Field theory and polynomials :: Real and complex fields :: Polynomials: factorization)
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