# Exercise on Fundamentals of Divisibility

(This question is on page 236 of Falko Lorenz's Algebra Volume 1: Fields and Galois Theory, exercise 4.5)

Prove that $Y^2 = X^3 - 2$ has exactly one solution in the natural numbers.

(Hint: use the fact that $\mathbb{Z}\sqrt{-2}$ is a Euclidean domain, and that in any unique factorization domain $R$, if $\alpha_1,\dots ,\alpha_n$ are pairwise rel. prime in $R$ and their product is an m-th power in $R$, each $\alpha_i$ is associated to an m-th power in $R$.

Here's what I know. Since $\mathbb{Z}\sqrt{-2}$ is a Euclidean domain, then it is also a unique factorization domain. As well, $X^3=Y^2+2=Y^2-\sqrt{-2}^2=(Y-\sqrt{-2})(Y+\sqrt{-2})$. Clues for the clueless?

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An outline: The units are $\pm 1$. Show that $X$ and $Y$ must be odd. Then show $Y+\sqrt{-2}$ and $Y-\sqrt{-2}$ are relatively prime. So each is a cube.
Let $Y+\sqrt{-2}=(a+b\sqrt{-2})^3$. Expand. Compute in particular the coefficient of $\sqrt{-2}$. This must be $1$. That will tell you something very important about $b$. But then you will know what $a$ must be, more or less. That will give the only solutions, and the only positive one.
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me? – Chris Oct 3 '12 at 5:05
That $b$ divides $1$! – André Nicolas Oct 3 '12 at 5:06
Well I got it, $X=3$ and $Y=5$! – Chris Oct 3 '12 at 5:13
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+\sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd. – André Nicolas Oct 3 '12 at 5:20