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To solve $y^2 + 2 = x^3$ you can factor $(y - \sqrt{-2})(y + \sqrt{-2}) = x^3$ and then check that they are relatively prime and by unique factorization both must be cubes then you can solve it.

What about $y^2 + 5 = x^3$ which does not have unique factoring?

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Often what matters is not whether the number ring in question is a UFD but whether its class number is relatively prime to a certain key number. In the case of Fermat's Last Theorem, these factoring methods work well for $x^p + y^p = z^p$ for regular primes, i.e., primes for which $p$ does not divide the class number of the ring of integers of $\mathbb{Q}(\zeta_p)$.

For the Bachet-Mordell equation $y^2 + k = x^3$ (I suppose $k$ is squarefree), the favorable case is $k \equiv 1,2 \pmod 4$ (so that $\mathbb{Z}[\sqrt{-k}]$ is the full ring of integers of $\mathbb{Q}(\sqrt{-k})$) and that the class number of $\mathbb{Q}(\sqrt{-k})$ is prime to $3$. In the particular case $k = 5$, you're in luck: $5 \equiv 1 \pmod 4$ and the class number is $2$, which is as good as $1$!

What exactly do we win when this happens? By some happy coincidence I wrote up some notes on precisely this: please see Theorem 7 here. (Also see the "tables" on page 9.)

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    Wow! I did not expect that at all, thanks !2011-01-31
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    the link seems to be broken.2011-02-01
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    @Zev: actually, the link is fine but the entire UGA math server is down (!!). I'm hoping it comes back up soon, obviously.2011-02-01
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No factorization in rings without unique factorization is necessary. Look at the proof of Theorem 2.2 in http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf.

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    @K: It's good to know that there is a more elementary approach here, but this is particular to $k = 5$, right? That is, did we just "get lucky", or is there something more systematic behind the success of congruence arguments here? (Anyway, your notes are certainly an excellence reference for this topic: for instance, I make reference to them in my notes, as one will be able to see once normal service is resumed on the UGA math server.)2011-02-01
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    Yes, that is specific to k = 5.2011-02-12