Conjecture on Pythagoras numbers of maximal orders in biquadratic fields

Let $K$ be a totally real biquadratic field, and let $\mathcal{O}_K$ be its ring of integers. The Pythagoras number $\mathsf{P}(\mathcal{O}_K)$ is the least number of squares needed to represent every element of $\sum \mathcal{O}_K^2$. The biquadratic Pythagoras-number conjecture.

1. If $K$ contains none of $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$, then $\mathsf{P}(\mathcal{O}_K)\geq 6$ holds with finitely many exceptions.
2. If $K$ contains $\sqrt{2}$, $\sqrt{3}$, or $\sqrt{5}$, then $\mathsf{P}(\mathcal{O}_K)\leq 5$.
3. The inequality in the second assertion is strict only for the following seven fields: $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$, $\mathbb{Q}(\sqrt{2},\sqrt{5})$, and $\mathbb{Q}(\sqrt{3},\sqrt{5})$, where $\mathsf{P}(\mathcal{O}_K)=3$; and $K=\mathbb{Q}(\sqrt{2},\sqrt{7})$, $\mathbb{Q}(\sqrt{3},\sqrt{7})$, $\mathbb{Q}(\sqrt{5},\sqrt{6})$, and $\mathbb{Q}(\sqrt{5},\sqrt{7})$, where $\mathsf{P}(\mathcal{O}_K)=4$.

The conjecture synthesizes the paper's results on maximal orders: the bound in the $\sqrt{5}$ case is proved, while the analogous claims for $\sqrt{2}$ and $\sqrt{3}$ and the exceptional-field assertions are supported only by computations and remain open.