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So the question is completely stated by the title. My own thoughts:

I can prove that $x^2 + 1 = y^3$ has no solutions for $x,y \in \mathbb{Z}$ by using the factorization: $$ y^3 = (x-i)(x+i) $$ in $\mathbb{Z}[i]$, using the fact that $\mathbb{Z}[i]$ is a UFD and that $(x-i)$ and $(x+i)$ are coprime we obtain that they are both cubes and it follows that this is impossible.

Now I would like to apply similar reasoning for this equation, we factor it as follows: $$ y^3 = (x-\sqrt{-5})(x+\sqrt{-5}) $$ in $\mathbb{Z}[\sqrt{-5}]$ but this time $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. However, since $\mathbb{Q}(\sqrt{-5})/\mathbb{Q}$ is a finite seperable field extension we find that the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{-5})$ is a Dedekind domain, this integral closure is exactly our ring $\mathbb{Z}[\sqrt{-5}]$ (which follows from the fact that $-5 = 3 \neq 1 \text{ mod } 4$). In a Dedekind domain we know that we write ideals uniquely as a product of maximal ideals, we can apply this to the ideal $(y)$, $(x-\sqrt{-5})$ and $(x+\sqrt{-5})$ for this to be useful we must first prove that the ideals $(x-\sqrt{-5})$ and $(x+\sqrt{-5})$ are coprime, for this it suffices to show that there is no prime ideal $P \subseteq \mathbb{Z}[\sqrt{-5}]$ that contains $(x-\sqrt{-5})$ and $(x+\sqrt{-5})$.

Let $P$ be a prime ideal s.t. $x-\sqrt{-5},x+\sqrt{-5} \in P$, then it follows that also: $$ 2\sqrt{-5} = x+\sqrt{-5} - (x-\sqrt{-5}) \in P $$ and thus by the given equality: $$ -20 = (2\sqrt{-5})^2 \in P^2 \subseteq (y)^3 $$ so we find an element $a+b \sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$ s.t. $-20 = (a+b\sqrt{-5}) \cdot y^3$ using the norm: $$ N(r + s \sqrt{-5}) := r^2 +s^2 5 $$ we obtain from this equality thath: $$ 400 = (a^2 + 5b^2) \cdot y^6 $$ but as $400 = 2^4 \cdot 5^2$ we find that $y$ must be equal to $1$ and thus we find that: $$ x^2 + 5 = 1 $$ which is clearly impossible as for $x \in \mathbb{Z}$ we have $x^2 \geq 0$.

From this we can conclude that $(x-\sqrt{-5})$ and $(x+\sqrt{-5})$ are indeed coprime so their decomposition in a product of maximal ideals is disjoint (i.e. we find maximal ideals $\mathfrak{p}_1,\dots,\mathfrak{p}_r$, $\mathfrak{q}_1,\dots,\mathfrak{q}_s$ and natural numbers $e_1,\dots,e_r,f_1,\dots,f_s$ s.t.: $$(x-\sqrt{-5}) = \mathfrak{p}_1^{e_1}\dots\mathfrak{p}_r^{e_r}, \qquad (x+\sqrt{-5}) = \mathfrak{q}_1^{f_1}\dots\mathfrak{q}_r^{f_r}$$ where all $\mathfrak{p}_i$ and $\mathfrak{q}_j$ are different. For $(y)$ we also find such a decomposition, which implies that the decompositions of $(x-\sqrt{-5})$ and $(x+\sqrt{-5})$ must be of the form: $$ (x-\sqrt{-5}) = \mathfrak{p}_1^{3e_1}\dots\mathfrak{p}_r^{3e_r}, \qquad (x+\sqrt{-5}) = \mathfrak{q}_1^{3f_1}\dots\mathfrak{q}_r^{3f_r} $$ which implies that there is some ideal $I$ in $\mathbb{Z}[\sqrt{-5}]$ s.t. $(x+\sqrt{-5}) = I^3$ so we find elements of $I$: $a_i+ b_i \sqrt{-5} \in I$ for which: $$ x+\sqrt{-5} = \prod_{i=1}^3(a_i+ b_i \sqrt{-5}) $$ and then I would like that this is impossible, but at this point I'm stuck.

I'm sorry if my own idea is absolutaly worthless but this is the first time I try to solve this kind of exercise.

HK Lee
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Holymonk
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    Mordell's equation $y^2=x^3-n$ has no solution for $n=3, 5, 6, 9, 10, 12, 14, 16, 17, 21, 22, \ldots $. See here. – Dietrich Burde Apr 17 '15 at 14:11
  • Sweet, thanks the proof is significantly simpler than what I had in mind :) – Holymonk Apr 17 '15 at 14:15
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  • $y=2l\Rightarrow x^2+5=8l^3\equiv 0\equiv 1+5\pmod{8}$, contr.
  • $y$ odd $\Rightarrow x=2k\Rightarrow 4(k^2+1)=(y-1)(y^2+y+1)\Rightarrow y\equiv 1\pmod{4}$, but $y^2+y+1\equiv 1+1+1\equiv 3\pmod {4}$, impossible, since $\exists p\in\mathbb P (p=4c+3\mid k^2+1)$.
    • – user26486 Apr 17 '15 at 14:17
  • When you say $y = 1$ from $y^6 = 1$, actually you can only say $y = \pm 1$. But you still get a contradiction after that, as before. – KCd Apr 17 '15 at 14:17
  • You are absolutaly right! – Holymonk Apr 17 '15 at 14:20
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    Do you know about ideal class groups? The ring $\mathbf Z[\sqrt{-5}]$ has class number $2$, which is relatively prime to the exponent $3$ in the equation, so even though $\mathbf Z[\sqrt{-5}]$ is not a UFD one can prove that $x+\sqrt{-5}$ must be a cube in $\mathbf Z[\sqrt{-5}]$, which is the kind of conclusion you'd expect if $\mathbf Z[\sqrt{-5}]$ were a UFD. Trying to solve $x + \sqrt{-5} = (a + b\sqrt{-5})^3$ in integers $a$ and $b$ leads to a contradiction by looking at the coefficient of $\sqrt{-5}$ on both sides. That's how you solve this with algebraic number theory. – KCd Apr 17 '15 at 14:22
  • I had heard of the ideal class group but I don't really know anything about them except their definition (I'm not at all experienced in algebraic number theory) – Holymonk Apr 17 '15 at 14:25
  • @Holymonk Do you understand my solution? I wanted to make the solution succinct, so I can clarify if anything's not clear. The solution checks 2 cases. Since in the first case $x$ is odd, we know $x^2\equiv 1\pmod {8}$. In the 2nd case, since $y^2+y+1$ is odd, we know $4\mid y-1$. Then I used the fact that $\left(\frac{-1}{p}\right)=1\iff p\equiv 1\pmod 4$ with $p$ odd prime. I used Legendre symbol here. – user26486 Apr 17 '15 at 19:06
  • This question is by no means a duplicate of the proposed one, and it is a rather shame that moderators here go around closing interesting questions without a good-enough reason!!! – barak manos Apr 18 '15 at 08:03
  • @user31415 I almost understand your answer, so I can follow it to the point when you say $y \equiv 1 (\text{mod }4)$, but then there are 2 things unclear:
    1. is there an easy proof to see that equivalence of the Legendre symbol?
    2. How can we use this to get a contradiction (I don't know the notation "$\mathbb{P}(p = 4c+3 \mid k^2 +1)"$). I'm sorry for the late response I was on a trip this weekend and thanks for the proof, I confirmed that my question was a duplicate because I found a proof using the link provided by Dietrich Burde. But I think the proof of user31415 is better!
     – Holymonk Apr 19 '15 at 14:56
  • @Holymonk I'll re-write it a bit: ..but $y^2+y+1\equiv 1+1+1\equiv 3\pmod {4}$, which means $\exists p=4c+3\in\mathbb P (p\mid k^2+1)$. But $p\mid k^2+1\Rightarrow k^2\equiv -1\pmod {p}\Rightarrow \text{ord}_p k=4\stackrel{\text{FLT}}\Rightarrow 4\mid p-1\Rightarrow p=4m+1$, so we have a contradiction. – user26486 Apr 19 '15 at 15:21
  • @user3141: I'm sorry but I still don't fully understand, why can you conclude that $\textup{ord}_p(k) = 4$ and is $FLT$ Fermat's little theorem? – Holymonk Apr 19 '15 at 15:46
  • @Holymonk \text{FLT} is Fermat's Little Theorem. Fact: If $\text{ord}_n a = s, a^k\equiv 1\pmod {n}$, then $s\mid k$. Proof: Let $k=sl+r, 0\le r<s$. Then $a^{sl+r}\equiv a^{sl}a^{r}\equiv \left(a^s\right)^la^r\equiv a^r\equiv 1\pmod {n}$. But $s$ is the least positive number such that $a^s\equiv 1\pmod {n}$, and $0\le r<s$, so $r$ can't be positive and is $0$. Then $k=sl$.$\ \ \ \square$ Since in our case $k^4\equiv 1\pmod {p}$, we know $\text{ord}_p k=1,2,4$. It's not $2$, because $k^2\equiv -1\pmod {p}$ and it's not $1$, since $k\equiv 1\pmod {p}$ would imply $k^2\equiv 1\pmod {p}$, contr. – user26486 Apr 19 '15 at 15:57
  • @Holymonk More generally, $a^{q^{n}}\equiv -1\pmod {m}\Rightarrow \text{ord}_m a=q^{n+1}$ (with $q$ prime). Proof: assume $\text{ord}_m a\neq q^{n+1}$. Then $\text{ord}_m a=q^{k}, 0<q^k<q^{n+1}$. But then $(a^{q^{k}})^{q^{n-k}}\equiv 1^{q^{n-k}}\equiv 1\not\equiv -1\pmod{m}$, contradiction. – user26486 Apr 19 '15 at 16:03
  • Ok, I got it, thanks man! – Holymonk Apr 22 '15 at 13:19
  • You are the number theory magician! :) – Holymonk Apr 22 '15 at 13:20
  • @KCd: I know what you mean now and that way is nice also :) – Holymonk May 18 '15 at 14:48