$\mathbb{Q}(\sqrt{d})$ with specific integral basis I would like some help with the following question.
Ireland and Rosen (ch.13#10)

For which $d$ does $\mathbb{Q}(\sqrt{d})$ have an integral basis of the form $\alpha, \alpha '$ where $\alpha '$ is the conjugate of $\alpha$?

As I understand this, {$a,b\sqrt{d}$} is a basis for $\mathbb{Q}(\sqrt{d})$, so we can set $a+b\sqrt{d}=a_{1}\alpha +b_{2}\alpha '$ where $a,b,a_{1},b_{1}\in \mathbb{Q}$ and $\alpha '$ is the conjugate of $\alpha$. However, I do not see where the restrictions on $d$ come from. Thanks. 
 A: The ring of integers of a number field $K$ is the set of elements of $K$ that are integral over $\mathbb{Z}$, i.e. they are roots of monic irreducible polynomials with coefficients in $\mathbb{Z}$. As the name implies, they form a subring of $K$. The ring of integers of $K$ is usually denoted $\mathcal{O}_K$.
An integral basis for a number field $K$ is a basis for $\mathcal{O}_K$ as a $\mathbb{Z}$-module (it is a theorem that $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $n$, where $n=[K:\mathbb{Q}]$). 
Proposition 13.1.1 in Ireland & Rosen tells you that, for $K=\mathbb{Q}(\sqrt{d})$ with $d$ a squarefree integer, 
$$\mathcal{O}_K=\begin{cases}\mathbb{Z}[\sqrt{d}] & \text{ if }d\equiv 2,3\bmod 4,\\\\\mathbb{Z}\left[\tfrac{-1+\sqrt{d}}{2}\right] & \text{ if }d\equiv 1\bmod 4.\end{cases}$$
So, given two $\alpha,\beta\in\mathcal{O}_K$, the set $\{\alpha,\beta\}$ is an integral basis for $K$ if every $\gamma\in \mathcal{O}_K$ can be uniquely represented as $\gamma=r\alpha+s\beta$ for some $r,s\in\mathbb{Z}$. Clearly, if $\{\alpha,\alpha'\}$ is to be an integral basis, $\alpha$ cannot be in $\mathbb{Z}$, so we must have that 
$$\alpha=\begin{cases}h+k\sqrt{d} \text{ for some }h,k\in\mathbb{Z}, k\neq 0 & \text{ if }d\equiv 2,3\bmod 4,\\\\h+k\left(\tfrac{-1+\sqrt{d}}{2}\right) \text{ for some }h,k\in\mathbb{Z}, k\neq 0 & \text{ if }d\equiv 1\bmod 4.\end{cases}$$
Suppose $d\equiv 2,3\bmod 4$, and that $\alpha=h+k\sqrt{d}$. Then $\alpha'=h-k\sqrt{d}$. Can you write every element of $\mathcal{O}_K$ as a $\mathbb{Z}$-linear combination of $\alpha$ and $\alpha'$?
Now try the case of $d\equiv 1\bmod 4$ for yourself :)
A: Hint $\rm\ (j\alpha + k\alpha'= 1)'\Rightarrow\: j\alpha'+k\alpha= 1\:\Rightarrow\: j = k\:\Rightarrow\:  k\:(\alpha+\alpha') = 1\:\Rightarrow\: tr\:\alpha = \pm 1 $
A: So, here's a hint: if $d$ is 1 mod 4, there's no problem, because we know the ring of integers has the basis
$$
1, \frac{1 + \sqrt{d}}{2},
$$
but that means that
$$
\frac{1 + \sqrt{d}}{2}, \frac{1 - \sqrt{d}}{2}
$$
is also a basis since the sum of those two elements is 1.
I claim there's at least sometimes a problem when $d$ is not 1 mod 4. Now the known basis is
$$
1, \sqrt{d}.
$$
It's clear that
$$
-\sqrt{d}, \sqrt{d}
$$
won't work (those aren't even linearly independent!), but of course we could try to work with something like
$$
a + b \sqrt{d}, a - b\sqrt{d},
$$
and what you need to show is that you can't always pick $a$ and $b$ to make this work. As a hint, think about the norms of all the elements in the $\mathbb{Z}$ span of this basis.
