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I'm trying to show that the Diophantine equation $x^2+5=y^3$ has no integer solutions using the fact that $\mathbb Z[ \sqrt{-5}]$ has class number two. I think I have the general idea, but I'm having a tough time establishing the statement in bold below. Any ideas?

The equation gives a factorization into ideals $$ (x+\sqrt{-5})(x-\sqrt{-5})=(y)^3 $$ in $\mathbb Z[ \sqrt{-5}]$. Assuming that $(x+\sqrt{-5})$ and $(x-\sqrt{-5})$ are relatively prime, we know that $(x+\sqrt{-5})=\mathfrak a^3$ and $(x-\sqrt{-5})=\mathfrak b^3$ for some ideals $\mathfrak a, \mathfrak b$ in $\mathbb Z[ \sqrt{-5}]$. Then the classes of $\mathfrak a$ and $\mathfrak b$ have order dividing 3 in the class group, hence they must both be principal, generated by some $\alpha,\beta\in \mathbb Z[\sqrt{-5}]$, respectively, as the class number is 2. Hence $\alpha^3=x+\sqrt{-5}$ up to units ($\pm 1$), and matching up real and imaginary parts then gives a contradiction.

How can I show that the ideals $(x+\sqrt{-5})$ and $(x-\sqrt{-5})$ are relatively prime? I was trying something along the lines of letting $\mathfrak p$ be a prime of $\mathbb Z[\sqrt{-5}]$ diving both ideals, in which case it also divides their sum (gcd). Then $$2\sqrt{-5}=x+\sqrt{-5}-(x-\sqrt{-5})\in (x+\sqrt{-5},x-\sqrt{-5})\subseteq \mathfrak p,$$ so $\mathfrak p\mid (2\sqrt{-5})$... But I'm not totally sure where I'm going with this.

Arbutus
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    Just realized this could be a duplicate of: https://math.stackexchange.com/questions/1239090/prove-that-x2-5-y3-has-no-solutions-for-x-y-in-mathbbz?rq=1. Will read this solution to see if it answers my question... – Arbutus Jan 25 '19 at 16:05
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    The ring $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. – egreg Jan 25 '19 at 16:12
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    Right, but it is a Dedekind domain, so we have unique factorization of ideals. – Arbutus Jan 25 '19 at 16:12
  • That's indeed the case. – egreg Jan 25 '19 at 16:13
  • So, I don't think my question is a duplicate: in the link above, @KCd's comment essentially explains the method outlined above, but does not shed light on how to show the ideals are relatively prime. – Arbutus Jan 25 '19 at 16:17

4 Answers4

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$\mathfrak p\mid(2\sqrt{-5})$, so $\mathfrak p\mid(2)$ or $\mathfrak p\mid(\sqrt{-5})$. Since $(2),(\sqrt{-5})$ are prime, $\mathfrak p$ is equal to one of them. We can't have $\mathfrak p=(2)$, since $2\nmid x+\sqrt{-5}$, so $\mathfrak p=(\sqrt{-5})$, so $\sqrt{-5}\mid x+\sqrt{-5}$. This implies $5\mid x$, and you can derive a contradiction from $x^2+5=y^3$.

EDIT: As Yong Hao Ng notes in the comment, $(2)$ is not a prime ideal - it factors as $(2,1+\sqrt{-5})^2$, so we get that $\mathfrak p=(2,1+\sqrt{-5})$. Then $x+\sqrt{-5}\in(2,1+\sqrt{-5})$, so $x-1\in(2,1+\sqrt{-5})$ and we get $x$ is odd. Now we get a contradiction from $x^2+5\equiv 2\pmod 4$, since $y^3$ can't be even but indivisible by $4$.

Wojowu
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  • @Servaes That isn't right - this is true in any ring as long as $\mathfrak p$ is prime, at least assuming we define $I\mid J$ as $J\subseteq I$. We don't need $(2),(\sqrt{-5})$ prime for this either. – Wojowu Jan 27 '19 at 13:34
  • (+1) but I'm wondering if $(2)$ isn't prime and $$(2)=(2,1-\sqrt 5)^2$$ instead. The ring mod $(2)$, $\mathbb Z[\sqrt 5]/(2)$, has the residue classes ${0,1,\sqrt 5,1+\sqrt 5}$ and there is a zero divisor since $$(1+\sqrt 5)^2 = 6+2\sqrt 5 \equiv 0\pmod 2$$ But the residue ring should be an integral domain. – Yong Hao Ng Jan 27 '19 at 17:33
  • @YongHaoNg I'm afraid you are right (except we are in $\mathbb Z[\sqrt{-5}]$, not $\mathbb Z[\sqrt{5}]$). I will edit accordingly. – Wojowu Jan 27 '19 at 18:48
  • @Wojowu I'm afraid I don't understand your argument on the odness of $x$. As I interpret it, you write $x-1$ under the form $2a+b(1-\sqrt{-5})$, but here $a, b \in \mathbf Z[\sqrt {-5}]$, so I think you can't conclude a priori. Did I miss something ? – nguyen quang do Jan 27 '19 at 22:09
  • @nguyenquangdo If $x-1$ was odd, say of the form $2k+1$, then $(2,1+\sqrt{-5})$ would also contain $x-1-2k=1$. In general, if you have a proper ideal $I$ in a number ring containing a rational prime $p$, then $I\cap\mathbb Z=(p)$. – Wojowu Jan 27 '19 at 22:11
  • OK. I misinterpreted your argument. – nguyen quang do Jan 27 '19 at 22:27
  • Cool, this is so much shorter. – Yong Hao Ng Jan 28 '19 at 02:43
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This has been answered many times on this site, so it is certainly a duplicate. On the other hand, it may be still worth to give the link to a very nice lecture note by Keith Conrad here, where he does proofs for many interesting cases of the Mordell equation $y^2=x^3-n$. For $n=5$ there are no integer solutions. More references here are found at OEIS A054504.

Dietrich Burde
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In the ideal factorization $(x+\sqrt {-5})(x-\sqrt {-5})=(y)^3$, "how can I show that the ideals $(x \pm \sqrt {-5})$ are relatively prime?" Suppose that $P$ is a common prime divisor, then $P^{2a}$ will appear in the LHS of the factorization. But $P^{3b}$ will appear in the RHS because of the cube, hence at least $P^6$ will divide both sides because of uniqueness. Following your argument, $P^3$ must divide $(2)(\sqrt {-5})$. In the quadratic field $\mathbf Q(\sqrt {-5})$, we know that the prime $5$ is totally ramified, and $(5)=(\sqrt {-5})^2$ shows that $(\sqrt {-5})$ is prime. Besides $(2)=Q^2$, with $Q=(2, 1+\sqrt {-5})$ (so the class of $Q$ is the unique class of order $2$ of the class group). It follows again by uniqueness of factorization that $P^3$ cannot divide $(\sqrt {-5})Q^2$, and we are done.

nguyen quang do
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If ideals $I,J$ are relatively prime then their sum generates the ring $R$, so we can show this. In particular we can show also that $$ (a + b\sqrt {- 5})(x+ \sqrt{- 5}) + (c + d\sqrt {- 5})(x - \sqrt {- 5}) = 1 $$ for some $a,b,c,d$.

This gives the equations $$ \begin{align} (a+c)x - 5(b-d) &= 1\\ (a-c) + (b+d)x &= 0 \end{align} $$ So a necessary condition is $5$ does not divide $x$. This can be checked to be true from $x^2 + 5 = y^3$, but otherwise the ideals are not necessarily relatively prime in general.

Setting $$ c = a + (b+d)x $$ gives us $$ 2ax = (1+5(b-d)) - (b+d)x^2 $$ Taking $\pmod 2$ we see that $x$ must be even, so to get an integer solution for $a$ we must have $1+5(b-d)$ divisible by $2x$.

The obvious way is to look at the linear diophantine solutions $$ xu - 5v = 1 $$ We can always choose $u$ to be even, so now we can set $d=0, v=b$ to get $$ \begin{align} a &= \frac{1+5(b-d)}{2x} - (b+d)\frac{x}{2}\\ &= \frac{1+5v}{2x} - v\frac{x}{2}\\ &= \frac{u}{2} - v\frac{x}{2} \end{align} $$ Since $u/2$ is integral we just need $vx$ to be even, which is true based on an earlier assumption: $x$ must be even.

It remains to show that $x$ is indeed even. If $x=2m+1$ is odd, then $y$ is even and $$ x^2 + 5 = 4m^2+4m+ 6 \equiv 2\pmod 4 $$ which contradicts $y^3 \equiv 0 \pmod 4$. Hence $x$ is even and we are done, finding a solution to the initial equation as $$ (u/2 -vx/2 + v\sqrt {-5})(x + \sqrt {-5}) + (u/2 + vx/2)(x - \sqrt {-5}) = 1 $$ (Recall that $xu-5v=1$ and $u$ chosen to be even.)

Since the two ideals generate $1$, they are relatively prime.

Yong Hao Ng
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