= Solution
Write $x=r+s\sqrt d$ with $r,s\in\mathbb Q$. The <field trace> and <field norm> are
$$
T=2r,\qquad N=r^2-ds^2.
$$
The element is an <algebraic integer> if and only if $T,N\in\mathbb Z$: necessity follows from its two conjugates; sufficiency follows from the <monic polynomial> $X^2-TX+N$.
Suppose $x$ is an <algebraic integer>. Set $a=2r\in\mathbb Z$. Then
$$
d(2s)^2=T^2-4N\in\mathbb Z.
$$
Write $2s=h/k$ in lowest terms, with $k>0$. Then $k^2\mid d$, and the hypothesis that $d$ is a <square-free integer> forces $k=1$. Thus $b=2s\in\mathbb Z$, and the remaining condition is
$$
a^2-db^2\equiv0\pmod4.
$$
The only squares modulo four are zero and one. For $d\equiv2\pmod4$, an odd $b$ is impossible, after which $a$ is even. For $d\equiv3\pmod4$, an odd $b$ would require $a^2\equiv3\pmod4$, again impossible. For $d\equiv1\pmod4$, the condition is precisely $a\equiv b\pmod2$.
Conversely, $\sqrt d$ satisfies $X^2-d$, and when $d\equiv1\pmod4$, the element $\omega=(1+\sqrt d)/2$ satisfies
$$
X^2-X+\frac{1-d}{4}\in\mathbb Z[X].
$$
Their integer spans therefore consist of <algebraic integers>. In the last case,
$$
\frac{a+b\sqrt d}{2}=\frac{a-b}{2}+b\omega.
$$
This proves the <ring of integers of a quadratic field> formula. The required <integral basis> is
$$
\boxed{\begin{cases}
\{1,\sqrt d\},&d\equiv2,3\pmod4,\\
\{1,(1+\sqrt d)/2\},&d\equiv1\pmod4.
\end{cases}}
$$
Their rational <linear independence> follows from $\sqrt d\notin\mathbb Q$.
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