= Solution
For a <square-free integer> $\delta\equiv2$ or $3\pmod4$, the ring of <integers> of the quadratic field is $\mathcal O_{\mathbb Q(\sqrt\delta)}=\mathbb Z[\sqrt\delta]$. Square-freeness is the usual convention underlying the first request: congruence alone would not suffice, for example $\delta=27$ gives the same field as $\delta=3$.
Now take the stipulated square-free $\delta\equiv3\pmod4$ and put $r=\sqrt2$, $s=\sqrt\delta$. For any integral $\alpha=a+br+cs+drs$, <relative traces> to the three quadratic subfields show $2a,2b,2c,2d\in\mathbb Z$. Those subfields all have integral bases $1$ and their <square root>, including $\mathbb Q(\sqrt{2\delta})$. Write $\alpha=(A+Br+Cs+Drs)/2$ with <integer> coefficients.
Its relative <field norm> to $\mathbb Q(s)$ is
$$
\frac{A^2+\delta C^2-2B^2-2\delta D^2}{4}+\frac{AC-2BD}{2}s\in\mathbb Z[s].
$$
Thus $AC$ is even, and the constant numerator is divisible by four. Since $\delta\equiv3\pmod4$, these conditions force $A,C$ both even: one odd makes the numerator odd, and both odd contradicts $AC$ even. They then force $B,D$ to have the same <integer parity>. Every <algebraic integer> consequently belongs to the lattice generated by
$$
\boxed{1,\quad\sqrt2,\quad\sqrt\delta,\quad\frac{\sqrt2+\sqrt{2\delta}}2.}
$$
Conversely the last generator $w$ satisfies $w^2=(1+\delta)/2+\sqrt\delta$ and the monic <integer> equation
$$
w^4-(1+\delta)w^2+\frac{(\delta-1)^2}{4}=0.
$$
All four generators are integral and <linearly independent> over $\mathbb Q$, so this lattice is exactly $\mathcal O_L$, proving this <integral basis of an odd biquadratic field containing square root of two> without an unproved index assumption.
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