 Solution

Write $x=r+s\sqrt{d}$ with $r,s\in \mathbb{Q}$. The field trace and field norm areThe 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 $a=2r\in \mathbb{Z}$. ThenWrite $2s=h/k$ 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 isThe only squares modulo four are zero and one. For $d\equiv 2(\mathrm{mod}4)$, an odd $b$ is impossible, after which $a$ is even. For $d\equiv 3(\mathrm{mod}4)$, an odd $b$ would require $a^2\equiv 3(\mathrm{mod}4)$, again impossible. For $d\equiv 1(\mathrm{mod}4)$, the condition is precisely $a\equiv b(\mathrm{mod}2)$.

Conversely, $\sqrt{d}$ satisfies $X^2-d$, and when $d\equiv 1(\mathrm{mod}4)$, the element $\omega =(1+\sqrt{d})/2$ satisfiesTheir integer spans therefore consist of algebraic integers. In the last case,This proves the ring of integers of a quadratic field formula. The required integral basis isTheir rational linear independence follows from $\sqrt{d}\notin \mathbb{Q}$.