\classheader{2013-01-16}

We claim that the only solutions $x,y\in\Z$ to the equation $x^3=y^2+4$ are $(x,y)=(2,\pm 2)$ and $(x,y)=(5,\pm 11)$.

As we did last time, we factor the left side in $\Z[i]$ (which is a UFD):
\[x^3=(y+2i)(y-2i).\]
We then obtained last time that this implies $(y+2i)=\alpha^3$ for some $\alpha\in\Z[i]$. If $\alpha=a+bi$, then
\begin{align*}
y+2i &= (a+bi)^3\\
&=(a^3-3ab^2)+(3ab^2-b^3)i
\end{align*}
and therefore
\begin{align*}
y &= a^3-3ab^2\\
2 &= 3a^2b-b^3=(3a^2-b^2)b
\end{align*}
The second of these equations forces that $b\in\{1,-1,2,-2\}$.
\begin{align*}
b=1 &\implies 3a^2-1=2  &&\implies a=\pm 1\implies y=\mp 2\\
b=-1 &\implies 3a^2-1=-2 &&\text{(not OK)}\\
b=2 &\implies 3a^2-4=1 &&\text{(not OK)}\\
b=-2 &\implies 3a^2-4=-1 && \implies a=\pm 1\implies y=\mp 11
\end{align*}

Now let's go back to proving that the integral solutions to $x^3=y^2+20$ are $(x,y)=(6,\pm 14)$.

Factoring the equation in $\Z[\sqrt{-5}]$, we get
\[x^3=(y+2\sqrt{-5})(y-2\sqrt{-5}).\]
As before, this impies that there is an $\alpha\in \Z[\sqrt{-5}]$ such that $y+2\sqrt{-5}=\alpha$. If $\alpha=a+b\sqrt{-5}$, then we obtain that
\begin{align*}
y &= a^3-3\times 5ab^2\\
2 &= 3a^2b - 5b^3=(3a^2-5b^2)b
\end{align*}
Again, we break into cases for each possibility $b\in\{1,-1,2,-2\}$, and can conclude the result.

Let $K=\Q(\sqrt{-5})$. Then $\ringofintegersof{K}=\Z[\sqrt{-5}]$ is not a UFD, because
\[6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5}).\]
The beautiful world of ideals exists behind the ugly world of numbers.

In the world of ideals, this is fixed, because
\[(2)=\frak{p}^2,\quad (3)=\frak{q}\frak{q}',\quad (1+\sqrt{-5})=\frak{p}\frak{q},\quad (1-\sqrt{-5})=\frak{p}\frak{q}'\]
where
\[\frak{p} = (2,1+\sqrt{-5})=(2,1-\sqrt{-5}),\qquad \frak{q} = (3,1+\sqrt{-5}),\qquad \frak{q}' = (3,1-\sqrt{-5}).\]
For a maximal ideal $\frak{p}$ of $\Z[\sqrt{-5}]$ and $\beta\in\Z[\sqrt{-5}]$, recall that $\ord_\frak{p}(\beta)=e$ for that integer $e$ for which $(\beta)=\frak{p}^eI$ for some ideal $I$, but for which there is no ideal $I$ such that $(\beta)=\frak{p}^{e+1}I$.

Considering the equation $x^3=(y+2\sqrt{-5})(y-2\sqrt{-5})$, we can conclude that $\ord_\frak{p}(y+2\sqrt{-5})$ for every maximal ideal $\frak{p}$ of $\Z[\sqrt{-5}]$. This is because if $y+2\sqrt{-5}\in\frak{p}$ and $y-2\sqrt{-5}\notin\frak{p}$, then \[3\ord_\frak{p}(x)=\ord_\frak{p}(x^3)=\ord_\frak{p}(y+2i),\]
and if $y+2\sqrt{-5}\in \frak{p}$ and $y-2\sqrt{-5}\in\frak{p}$, then $4\sqrt{-5}\in\frak{p}$, which implies that either $\frak{p}=(2,1+\sqrt{-5})=(2,1-\sqrt{-5})$, or $\frak{p}=(\sqrt{-5})=(-\sqrt{-5})$. Thus, we have obtained that $(y+2\sqrt{-5})=I^3$ for some non-zero ideal $I$ of $\Z[\sqrt{-5}]$. Because the class number of $\Q(\sqrt{-5})$ is 2 and $\class(I)^3=1$, we can conclude that $\class(I)=1$, because the order of $\class(I)$ is either 1 or 2.

Therefore there is a principal ideal $I=(\alpha)$ such that $(y+2\sqrt{-5})=I^3$, so that $y+2\sqrt{-5}=\text{unit}\cdot\alpha^3$.  The units of $\Z[\sqrt{-5}]$ are just $\{\pm 1\}$, so $y+2\sqrt{-5}=\pm\alpha^3=(\pm \alpha)^3$.


This method does not work for solving $x^3=y^2+26$, because $\Q(\sqrt{-26})$ has class number 6, which is divisible by 3. Thus, we get everything up to the conclusion that $(y+\sqrt{-26})=I^3$ for some ideal $I$, but we do not then obtain that $y+\sqrt{-26}=\alpha^3$ for some element $\alpha$.  For example, $(x,y)=(3,1)$ is a solution, so
\[3^3=(1+\sqrt{-26})(1-\sqrt{-26}),\]
but $1+\sqrt{-26}$ is not a cube, i.e. $1+\sqrt{-26}\neq\alpha^{3}$ for any $\alpha\in\Z[\sqrt{-26}]$. However, $(1+\sqrt{-26})=\frak{p}^3$ where $\frak{p}=(3,1+\sqrt{-26})$.

\subsection*{Three big theorems in algebraic number theory in the 19th century}

Let $K$ be a number field.

\begin{enumerate}
\item Unique factorization in the world of ideals
\begin{quote}
If $I$ is a non-zero ideal of $\ringofintegersof{K}$, then $I=\frak{p}_1^{e_1}\cdots\frak{p}_r^{e_r}$ in a unique way, where the $\frak{p}_i$ are maximal ideals of $\ringofintegersof{K}$.
\end{quote}
\item Finiteness of class number
\begin{quote}
The ideal class group of $K$ is finite.
\end{quote}
\item Dirichlet's unit theorem
\begin{quote}
The unit group $(\ringofintegersof{K})^\times$ is isomorphic to
\[\Z^{r_1+r_2-1}\oplus \Z/w_K\Z\]
where $w_K$ is the number of roots of unity in $K$, $r_1$ is the number of distinct embeddings $K\hookrightarrow\R$, and $r_2$ is (one half of) the number of distinct embeddings $K\hookrightarrow\C$ whose image is not $\R$. Note that
\[r_1+2r_2=\#\{K\hookrightarrow \C\}=[K:\Q].\]
\end{quote}
\end{enumerate}

Let's look at some examples of Dirichlet's unit theorem.

If $K=\Q(i)$, then there are no embeddings of $K$ into $\R$, and two embeddings of $K$ into $\C$ (via $i\mapsto i$ and $i\mapsto -i$), so $r_1=0$ and $r_2=1$, and
\[\Z[i]^\times=\{\pm 1,\pm i\}\cong \Z/4\Z.\]
If $K=\Q(\sqrt{2})$, then $r_1=2$ and $r_2=0$, and
\[\Z[\sqrt{2}]^\times=\{\pm(1+\sqrt{2})^n\mid n\in\Z\}\cong\Z\oplus\Z/2\Z.\]
If $K=\Q(\sqrt[3]{2})$, then $r_1=1$, $r_2=1$, and
\[\Z[\sqrt[3]{2}]^\times=\{\pm(1-\sqrt[3]{2})^n\mid n\in\Z\}\cong\Z\times\Z/2\Z.\]
The finiteness of class number tells us that the difference between the world of ideals and the world of numbers is finite in some sense.

Recall that Kummer wanted to approach Fermat's Last Theorem by
\begin{align*}
x^n & = y^n-z^n\\
&=\prod_{a=1}^n(z-\zeta_n^a y)
\end{align*}

Kummer wanted to be able to say that
\[(z-\zeta_n^ay)=I^n\implies (z-\zeta_n^ay)=(\alpha^n)\]
for some $\alpha$, which we've seen that we can do when the class number of $\Q(\zeta_n)$ is prime to $n$. However, the next step requires us to understand the unit group because
\[(z-\zeta_n^ay)=(\alpha^n)\implies z-\zeta_n^ay=u\alpha^n\]
for a unit $u$.

















