1. Introduction
In the first part of this paper, we focus on the study of the so-called Pythagoras number in real biquadratic fields. For that, let us consider some commutative ring 
$\mathcal O$ and its subset 
$\sum \mathcal O^2$, which contains those 
$\alpha\in \mathcal O$ which can be written as the sum of squares of elements in 
$\mathcal O$. Then the Pythagoras number 
$\mathcal P(\mathcal O)$ of 
$\mathcal O$ is the minimal number in 
$\mathbb N\cup\{\infty\}$ such that every 
$\alpha\in\sum \mathcal O^2$ can be written as the sum of at most 
$\mathcal P(\mathcal O)$ squares of elements in 
$\mathcal O$.
The exceptional case when 
$\mathcal O$ is a field is well known and richly studied. However, in this paper, we will discuss the case when 
$\mathcal O\subseteq \mathcal O_K$ is an order in a totally real number field K, and 
$\mathcal O_K$ is the ring of algebraic integers of K. In comparison with results for fields, we do not have much information about such orders.
In that case, i.e., if 
$\mathcal O\subseteq \mathcal O_K$, we have 
$\mathcal P(\mathcal O) \lt \infty$. However, it can also attain arbitrarily large values by the result of Scharlau [Reference Scharlau27]. In his proof, Scharlau efficiently uses properties of multiquadratic number fields to get an increasing lower bound for some particular family of these fields. Regarding the upper bound, we have 
$\mathcal P(\mathcal O)\leq f(d)$ where the function f depends only on the degree d of the field K (see, e.g., [Reference Kala and Yatsyna17, Corollary 3.3]). Furthermore, we have 
$\mathcal P(\mathcal O)\leq d+3$ whenever 
$2\leq d\leq 5$.
Except for these general statements, we only have some results for several families of number fields. Peters [Reference Peters26] investigated the case of orders 
$\mathcal O\subseteq\mathcal O_K$ where K is a real quadratic field. He proved that 
$\mathcal P(\mathcal O)=5$ with the exception of the following orders: 
$\mathcal P(\mathbb Z[\sqrt{5}])=\mathcal P(\mathbb Z[\sqrt{6}])=\mathcal P(\mathbb Z[\sqrt{7}])=4$ and 
$\mathcal P(\mathbb Z[\sqrt{2}])=\mathcal P(\mathbb Z[\sqrt{3}])=\mathcal P\big(\mathbb Z[\frac{1+\sqrt{5}}{2}]\big)=3$, which were mostly resolved before. Thus, in quadratic fields, the upper bound 5 from the above result is attained in all but finitely many cases.
The case of the simplest cubic fields, which are generated by a root ρ of the polynomial 
$x^3-ax^2-(a+3)x-1$ with 
$a\in\mathbb Z$, 
$a\geq-1$, was discussed in [Reference Tinková29]. In that paper, the present author proves that 
$\mathcal P(\mathbb Z[\rho])=6$ if 
$a\geq 3$, for which she used properties and knowledge of additively indecomposable integers in these fields. The fact that 
$P(\mathbb Z[\rho])=P(\mathcal O_K)=4$ for 
$a=-1$ was consequently shown in [Reference Krásenský20]. The method developed in [Reference Tinková29] was also applied to other families of totally real cubic fields [Reference Gil Muñoz and Tinková5, Reference Kala, Sgallová and Tinková14].
For biquadratic fields, the above upper bound is 7. In [Reference Krásenský, Raška and Sgallová21], the authors proved that for an infinite subfamily of biquadratic fields K, we indeed have 
$\mathcal P(\mathcal O_K)=7$. On the other hand, they also discuss one concrete infinite subfamily of biquadratic fields (containing 
$\sqrt{5}$), for which 
$P(\mathcal O_K)=5$. Note that this was later extended to quartic fields containing 
$\sqrt{2}$ in [Reference He and Hu6]. Moreover, in [Reference Krásenský, Raška and Sgallová21], one can find more results regarding the upper and lower bounds on 
$\mathcal P(\mathcal O_K)$, and we will refine some of them. In particular, their Theorem 1.2(2) says that if 
$n\neq 2,3,5,6,7$ is a square-free positive integer, then there can exist only finitely many real biquadratic fields K such that 
$\sqrt{n}\in K$ and 
$\mathcal P(\mathcal O_K) \lt 6$. Under a similar restriction, we will show that 
$\mathcal P(\mathcal O_K)\geq 6$ is always satisfied by proving the following theorem:
Theorem 1.1. Let K be a real biquadratic field not containing 
$\sqrt{2}$, 
$\sqrt{3}$, 
$\sqrt{5}$, 
$\sqrt{6}$, 
$\sqrt{7}$ and 
$\sqrt{13}$. Then 
$\mathcal P(\mathcal O_K)\geq 6$.
Note that in [Reference Krásenský, Raška and Sgallová21, proof of Theorem 8.1], which covers almost the whole proof of [Reference Krásenský, Raška and Sgallová21, proof of Theorem 1.2(2)], the authors needed to discuss 12 different cases of biquadratic fields. In the proof of Theorem 1.1, we will distinguish much less cases by using some properties related to integral bases of biquadratic fields.
Moreover, we will prove that the following subfamily of biquadratic fields has the maximal possible Pythagoras number for degree 4:
Theorem 1.2. Let 
$n\geq 6$ be a rational integer such that 
$p=(2n-1)(2n+1)$, 
$q=(2n-1)(2n+3)$ and 
$r=(2n+1)(2n+3)$ are square-free integers, and let 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$. Then 
$\mathcal P(\mathcal O_K)=7$.
As mentioned, the authors in [Reference Krásenský, Raška and Sgallová21] also show one infinity family of biquadratic fields with Pythagoras number 7. However, our family has a different type of integral basis (see § 2), and, moreover, in our case, 
$\gcd(p,q)\neq 1$. Therefore, it further confirms that 
$\mathcal P(\mathcal O_K)=7$ is not so rare for biquadratic fields.
The second part of this paper is devoted to a deeper look at additively indecomposable integers in real biquadratic fields. Let 
$\mathcal O_K^{+}\subset\mathcal O_K$ be the set of totally positive algebraic integers in 
$\mathcal O_K$. Recall that 
$\alpha\in K$ is totally positive if all its Galois conjugates are positive. Then 
$\alpha\in\mathcal O_K^{+}$ is indecomposable in 
$\mathcal O_K$ if α cannot be written as a sum of totally positive algebraic integers.
Especially in the last few years, indecomposable integers (also called indecomposables) were heavily used in the study of universal quadratic forms over number fields [Reference Blomer and Kala1, Reference Kala9, Reference Kala11, Reference Kala and Svoboda15, Reference Kala and Tinková16, Reference Yatsyna32], which is one of the classical topics in the number theory. For more details in this direction, see also the survey of Kala [Reference Kala12]. Moreover, as indicated before, these elements have a nice application in the determination of the Pythagoras number [Reference Gil Muñoz and Tinková5, Reference Kala, Sgallová and Tinková14, Reference Tinková29].
On the other hand, we have more information about these elements only for degrees 2, 3 and 4. Let us describe these results in detail. The structure of indecomposable integers in real quadratic fields can be determined from the continued fraction of concrete quadratic irrationalities as proved by Perron [Reference Perron25] and Dress and Scharlau [Reference Dress and Scharlau4]. Note that at the moment, we do not know a similar relation between indecomposable integers and, e.g., multidimensional continued fractions, although in [Reference Kala, Sgallová and Tinková14], the authors started some investigation in this direction. Then, the structure of indecomposable integers was determined for some families of monogenic [Reference Kala and Tinková16, Reference Tinková28] and non-mogenic [Reference Gil Muñoz and Tinková5] cubic fields.
Of course, the most relevant case for this paper is the case of real biquadratic fields. The research in this direction started with [Reference Čech, Lachman, Svoboda, Tinková and Zemková3], where the authors were trying to answer the following question: Are all indecomposable integers from a quadratic field still indecomposable in a biquadratic field containing this quadratic field? In that paper, we can find some sufficient conditions when this is true. However, in [Reference Krásenský, Tinková and Zemková22], the authors provide a counterexample. This problem was recently almost resolved by Man [Reference Man23], who proved that indecomposable integers from at least two quadratic subfields cannot decompose. Moreover, Man was able to use techniques developed in [Reference Kala and Tinková16] to find the structure of indecomposables for some subfamilies of biquadratic fields. In the following, we will benefit from this result. See also [Reference Kala and Man13] for some recent results.
What is known about indecomposable integers in general is that their algebraic norm is bounded by some constant depending only on K [Reference Brunotte2]. This result was later refined by Kala and Yatsyna [Reference Kala and Yatsyna18] by proving that we can take the discriminant of K as this upper bound. However, even that is not always the best possible. For quadratic fields, where we can use properties of the continued fractions, there appeared a whole series of articles improving this bound [Reference Dress and Scharlau4, Reference Jang and Kim7, Reference Kala10, Reference Tinková and Voutier30]. Similar results for several families of cubic fields followed [Reference Gil Muñoz and Tinková5, Reference Tinková28]. In this spirit, we prove the following theorem:
Theorem 1.3. Let 
$n\geq 6$ be a rational integer such that 
$p=(2n-1)(2n+1)$, 
$q=(2n-1)(2n+3)$ and 
$r=(2n+1)(2n+3)$ are square-free integers. If 
$\alpha\in\mathcal O_K^{+}$ is indecomposable in 
$\mathcal O_K$ for 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$, then 
$N(\alpha)\leq 16n^4+64n^3+16n^2-96n+36$.
Here, by 
$N(\alpha)$, we mean the norm of 
$\alpha\in K$. Since the discriminant of these fields is 
$16pqr\sim 1024n^6$, there is a great discrepancy between the above bound and the bound derived by Kala and Yatsyna [Reference Kala and Yatsyna18, Theorem 5]. Moreover, our result is sharp since there is an indecomposable integer in 
$\mathcal O_K$ with norm 
$16n^4+64n^3+16n^2-96n+36$. We also provide a similar result for two more families of biquadratic fields.
Moreover, we will also study the so-called minimal (codifferent) traces of indecomposables α. By that, we mean the minimum of traces of αδ where δ runs over all totally positive elements of the codifferent 
$\mathcal O_K^{\vee}$ (for more details, see § 2). If 
$\min_{\delta\in\mathcal O_K^{\vee,+}}\text{Tr}\left(\alpha\delta\right)=1$ for 
$\alpha\in\mathcal O_K^{+}$, then, easily, α is indecomposable in 
$\mathcal O_K$. The opposite implication is not true in general, although it is valid for real quadratic fields [Reference Kala and Tinková16]. On the other hand, we know families of cubic fields with this minimal trace equal to 1 or 2 [Reference Kala and Tinková16], or a family of cubic orders, for which it can attain arbitrarily large values [Reference Tinková29]. In this paper, we prove the following:
Theorem 1.4. Let 
$n\geq 6$ be a rational integer such that 
$p=(2n-1)(2n+1)$, 
$q=(2n-1)(2n+3)$ and 
$r=(2n+1)(2n+3)$ are square-free integers. If 
$\alpha\in\mathcal O_K^{+}$ is indecomposable in 
$\mathcal O_K$ for 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$, then 
$\min_{\delta\in\mathcal O_K^{\vee,+}}\textrm{Tr}\left(\alpha\delta\right)\leq 2$.
Therefore, this family is an example of biquadratic fields for which these minimal traces are small. Moreover, we show subfamilies of such fields for which this minimal trace is 1 for all indecomposable integers; such examples were known before only for degree 2. Note that these kinds of results can serve as an alternative proof of the indecomposability of these elements. Using this, we are also able to find a lower bound on the minimal number of variables of universal quadratic forms. Note that indecomposable integers of small codifferent traces can also be used to resolve the existence of universal 
$\mathbb Z$-forms over 
$\mathcal O_K$ – see, e.g., [Reference Kim and Lee19] for biquadratic case.
In § 2, we provide the necessary facts, notation and tools we use in the paper. Section 3 is devoted to the proof of Theorem 1.1, for which we discuss several different cases separately, and we will also prove Theorem 1.2. Finally, in § 4, we give some new results for three families of biquadratic fields, including proofs of Theorems 1.3 and 1.4.
2. Preliminaries
Let K be a totally real number field of degree d and denote by 
$\mathcal O_K$ its ring of algebraic integers. We say that an element 
$\alpha\in K$ is totally smaller than 
$\beta\in K$ if 
$\sigma(\alpha) \lt \sigma(\beta)$ for every embedding σ of K into complex numbers, for which we use the symbol 
$\alpha \prec \beta$. An element 
$\alpha\in K$ is totally positive if 
$\alpha\succ 0$. Moreover, we denote the set of totally positive algebraic integers by 
$\mathcal O_K^{+}$. An element 
$\alpha\in\mathcal O_K^{+}$ is indecomposable if 
$\alpha\neq \beta+\gamma$ for any 
$\beta,\gamma\in\mathcal O_K^{+}$. We define the trace and norm of 
$\alpha\in K$ as 
$\text{Tr}(\alpha)=\sum_{\sigma}\sigma(\alpha)$ and 
$N(\alpha)=\prod_{\sigma}\sigma(\alpha)$, respectively, where σ runs over all embedding of K into the field 
$\mathbb C$ of complex numbers. Moreover, we say that elements 
$\alpha,\beta\in\mathcal O_K$ are associated if there is a unit 
$\varepsilon\in\mathcal O_K$ such that 
$\alpha=\beta\varepsilon$.
By the codifferent of K, we understand the set of the form
\begin{equation*}
\mathcal O_K^{\vee}=\{\delta\in K: \text{Tr}(\alpha\delta)\in\mathbb Z\,\text{for all }\alpha\in\mathcal O_K\}.
\end{equation*} If 
$\mathcal O_K=\mathbb Z[\gamma]$ for some algebraic integer γ with the minimal polynomial f, then 
$\mathcal O_K^{\vee}=\frac{\mathbb Z[\gamma]}{f'(\gamma)}$ where f ʹ is the derivative of f. However, in the other cases, we are able to determine the form of the codifferent using [Reference Narkiewicz24, Proposition 4.14]. Let 
${\gamma_1,\ldots,\gamma_d}$ be an integral basis of 
$\mathcal O_K$. Then if 
$\varphi_1,\ldots,\varphi_d\in K$ satisfy 
$\text{Tr}(\gamma_i\varphi_j)=\delta_{ij}$ where δij is the Kronecker delta, then 
${\varphi_1,\ldots,\varphi_d}$ is a 
$\mathbb Z$-basis of 
$\mathcal O_K^{\vee}$. Using this, we can easily compute the form of 
${\varphi_1,\ldots,\varphi_d}$, and, moreover, we have
\begin{equation}
\text{Tr}\left(\Bigg(\sum_{i=1}^d a_i\gamma_i\Bigg)\Bigg(\sum_{j=1}^d b_j\varphi_j\Bigg)\right)=\sum_{i=1}^d a_ib_i
\end{equation} for 
$a_i,b_j\in\mathbb Z$. By 
$\mathcal O_K^{\vee,+}$, we will mean the subset of totally positive elements in 
$\mathcal O_K^{\vee}$. We will also denote 
$\text{minTr}(\alpha)=\min_{\delta\in\mathcal O_K^{\vee,+}}\text{Tr}(\alpha\delta)$ for 
$\alpha\in\mathcal O_K^{+}$ and call it the minimal (codifferent) trace of α.
By a quadratic form over 
$\mathcal O_K$ in n variables, we mean 
$Q(x_1,\ldots,x_n)=\sum_{1\leq i,j\leq n}a_{ij}x_ix_j$ where 
$a_{ij}\in\mathcal O_K$. We say that a quadratic form Q represents an element 
$\beta\in\mathcal O_K$ if there exist elements 
$\alpha_1,\ldots,\alpha_n\in\mathcal O_K$ such that 
$Q(\alpha_1,\ldots,\alpha_n)=\beta$. The form Q is totally positive definite if 
$Q(\alpha_1,\ldots,\alpha_n)\in\mathcal O_K^{+}$ for all 
$\alpha_i\in\mathcal O_K$ such that 
$(\alpha_1,\ldots,\alpha_n)\neq (0,\ldots,0)$. In this paper, we will always assume that our quadratic forms are totally positive definite. We say that Q is classical if 
$2|a_{ij}$ for all i ≠ j, and diagonal if 
$a_{ij}=0$ for all i ≠ j. Moreover, Q is universal if every element in 
$\mathcal O_K^{+}$ can be represented by Q.
In this paper, we will benefit from the following result:
Proposition 2.1 ([Reference Kala and Tinková16, Section 7])
Let K be a totally real number field of degree 4. Assume that there exist 
$\delta_1\in\mathcal O_K^{\vee,+}$ and 
$\alpha_1,\ldots, \alpha_n\in\mathcal O_K^{+}$ such that 
$\textrm{Tr}(\alpha_i\delta_1)=1$ for all 
$1\leq i\leq n$. Moreover, assume that there exist 
$\delta_2\in\mathcal O_K^{\vee,+}$ and 
$\beta_1,\ldots, \beta_m\in\mathcal O_K^{+}$ such that 
$\textrm{Tr}(\beta_i\delta_2)=2$ for all 
$1\leq i\leq m$. Then every classical universal quadratic form over 
$\mathcal O_K$ has at least 
$\frac{n}{4}$ variables, and every diagonal universal quadratic form over 
$\mathcal O_K$ has at least 
$\max\left(\frac{n}{4},\frac{m}{12}\right)$ variables.
Now, let us introduce the Pythagoras number of a ring. Let 
$\mathcal O$ be a commutative ring. Let us denote by 
$\sum \mathcal O^2$ the set of 
$\alpha\in\mathcal O$ such that α can be expressed as the sum of elements in 
$\mathcal O^2$, and let 
$\sum^m \mathcal O^2\subseteq\sum \mathcal O^2$ be the subset of those elements, for which we need at most m squares. Then, by the Pythagoras number of 
$\mathcal O$, we mean
\begin{equation*}
\mathcal P(\mathcal O)=\inf\Big\{m\in\mathbb N\cup\{+\infty\}:\; \sum \mathcal O^2=\sum^m \mathcal O^2\Big\}.
\end{equation*} Moreover, we will focus on the case when 
$\mathcal O = \mathcal O_K$ and K is a real biquadratic field. In that case, we know that 
$\mathcal P(\mathcal O_K)\leq 7$, see, e.g., [Reference Kala and Yatsyna17, Corollary 3.3].
In this paper, we study real biquadratic fields 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$, where 
$p,q \gt 1$ are distinct square-free integers. We will denote by r the remaining number whose square root is contained in K, i.e., 
$r=\frac{pq}{\gcd(p,q)^2}$. Let 
$\alpha=x+y\sqrt{p}+z\sqrt{q}+w\sqrt{r}\in\mathbb Q(\sqrt{p},\sqrt{q})$. The embeddings of 
$\mathbb Q(\sqrt{p},\sqrt{q})$ into 
$\mathbb C$ are given as
\begin{align*}
\sigma_1(\alpha)&=x+y\sqrt{p}+z\sqrt{q}+w\sqrt{r},\\
\sigma_2(\alpha)&=x-y\sqrt{p}+z\sqrt{q}-w\sqrt{r},\\
\sigma_3(\alpha)&=x+y\sqrt{p}-z\sqrt{q}-w\sqrt{r},\\
\sigma_4(\alpha)&=x-y\sqrt{p}-z\sqrt{q}+w\sqrt{r}.
\end{align*}Depending on the values of p and q (after possible interchanging of p, q and r), we can distinguish the following cases of integral bases of K [Reference Jarvis8, Reference Williams31]:
(1) if
$p\equiv 2\;(\text{mod }4)$ and
$q\equiv 3\;(\text{mod }4)$, then
$\mathcal O_K=\mathbb Z\Big[1,\sqrt{p},\sqrt{q},\frac{\sqrt{p}+\sqrt{r}}{2}\Big]$,(2) if
$p\equiv 2\;(\text{mod }4)$ and
$q\equiv 1\;(\text{mod }4)$, then
$\mathcal O_K=\mathbb Z\Big[1,\sqrt{p},\frac{1+\sqrt{q}}{2},\frac{\sqrt{p}+\sqrt{r}}{2}\Big]$,(3) if
$p\equiv 3\;(\text{mod }4)$ and
$q\equiv 1\;(\text{mod }4)$, then
$\mathcal O_K=\mathbb Z\Big[1,\sqrt{p},\frac{1+\sqrt{q}}{2},\frac{\sqrt{p}+\sqrt{r}}{2}\Big]$,(4) if
$p,q\equiv 1\;(\text{mod }4)$, then(1)
$\mathcal O_K=\mathbb Z\Big[1,\frac{1+\sqrt{p}}{2},\frac{1+\sqrt{q}}{2},\frac{1+\sqrt{p}+\sqrt{q}+\sqrt{r}}{4}\Big]$ if
$\frac{p}{\gcd(p,q)}\equiv \frac{q}{\gcd(p,q)}\equiv 1\;(\text{mod }4)$,(2)
$\mathcal O_K=\mathbb Z\Big[1,\frac{1+\sqrt{p}}{2},\frac{1+\sqrt{q}}{2},\frac{1-\sqrt{p}+\sqrt{q}+\sqrt{r}}{4}\Big]$ if
$\frac{p}{\gcd(p,q)}\equiv \frac{q}{\gcd(p,q)}\equiv 3\;(\text{mod }4)$.
Moreover, if we set 
$p_0=\gcd(q,r)$, 
$q_0=\gcd(p,r)$ and 
$r_0=\gcd(p,q)$, we can write p, q and r as 
$p=q_0r_0$, 
$q=p_0r_0$ and 
$r=p_0q_0$.
Note that in totally real fields of degree 4, an element 
$\alpha\in K$ is totally positive if and only if it is a root of a polynomial of the form
\begin{equation*}
x^4-Ax^3+Bx^2-Cx+D
\end{equation*} where 
$A,B,C,D\in\mathbb Q$, 
$A,B,C,D \gt 0$. In biquadratic fields, it is an easy computational task to find coefficients of the polynomial above, and for that, we can use Vieta’s formulas.
3. Pythagoras number in biquadratic fields
In this part, our goal is to prove Theorem 1.1, and for that, we will solve several different cases separately. In the following proofs, we will repeatedly use the fact that we need at least four squares of elements in 
$\mathcal O_K$ to express the element 7 if our biquadratic field does not contain 
$\sqrt{2}$, 
$\sqrt{3}$, 
$\sqrt{5}$, 
$\sqrt{6}$, 
$\sqrt{7}$ and 
$\sqrt{13}$. If 
$7=\sum_{i=1}^N\alpha_i^2$ for some 
$\alpha_i\in\mathcal O_K$ and 
$N\in\mathbb N$, the elements 
$0\prec \alpha_i^2$ are totally smaller than 7. Lemma 4.5(1) in [Reference Krásenský, Tinková and Zemková22] implies that in biquadratic fields not containing the above square roots, only the squares 1, 4, 
$\frac{9+\sqrt{17}}{2}$, 
$\frac{9-\sqrt{17}}{2}$, 
$\frac{11+\sqrt{21}}{2}$ and 
$\frac{11-\sqrt{21}}{2}$ can be totally smaller than 8 (of course, under the condition that they lie in the field). Since 
$7\prec 8$, these are also the only candidates for elements 
$\alpha_i^2$. From this list, we can exclude 
$\frac{11+\sqrt{21}}{2}$ and 
$\frac{11-\sqrt{21}}{2}$ since 
$\frac{11+\sqrt{21}}{2}\sim 7.79$, i.e., these two algebraic integers are not totally smaller than 7. If 
$\alpha_i^2=\frac{9+\sqrt{17}}{2}$ for some i, then, necessarily, 
$\alpha_j^2=\frac{9-\sqrt{17}}{2}$ for some j ≠ i. However, that gives
\begin{equation*}
\alpha_i^2+\alpha_j^2=9\succ 7,
\end{equation*} which is a contradiction. Therefore, 
$\alpha_i^2\in\{1,4\}$ for all 
$1\leq i\leq N$, and it is well known that we need at least four squares of rational integers to express 7.
Note that, e.g., if 
$\sqrt{7}\in K$, we only need one square to express 7, if 
$\sqrt{6}\in K$, two squares are sufficient, or if 
$\sqrt{5}\in K$, it is enough to consider three squares.
First of all, we will focus on the integral basis of type (1) as introduced above in § 2.
Proposition 3.1. Let 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$ where 
$p,q \gt 1$ are square-free positive integers such that 
$p\equiv 2\;(\textrm{mod }4)$, 
$q\equiv 3\;(\textrm{mod }4)$ and 
$p,q,r\geq 10$. Then 
$\mathcal P(\mathcal O_K)\geq 6$.
Proof. Recall that, in this case, we have
\begin{equation*}
\mathcal O_K=\mathbb Z\Bigg[1,\sqrt{p},\sqrt{q},\frac{\sqrt{p}+\sqrt{r}}{2}\Bigg].
\end{equation*} Clearly, 
$r=\frac{pq}{\gcd(p,q)^2}\equiv 2\;(\textrm{mod }4)$. Therefore, the form of the integral basis implies that the role of p and r is interchangeable. Thus, without loss of generality, we can assume p < r. First of all, suppose q < r, i.e., we get 
$r_0 \lt p_0,q_0$. To obtain the lower bound, let us consider the following element in 
$\sum\mathcal O_K^2$:
\begin{equation*}
\alpha = 7+(1+\sqrt{p})^2+(1+\sqrt{q})^2=9+p+q+2\sqrt{p}+2\sqrt{q}.
\end{equation*} In the following, we will derive the minimal number of squares of elements in 
$\mathcal O_K$ that we need to express α.
Let
\begin{equation*}
\alpha=\sum_{i=1}^N \Big(a_i+\frac{b_i}{2}\sqrt{p}+c_i\sqrt{q}+\frac{d_i}{2}\frac{\sqrt{pq}}{r_0}\Big)^2
\end{equation*} where 
$a_i,b_i,c_i,d_i\in\mathbb Z$ and 
$b_i\equiv d_i\;(\textrm{mod }2)$ for all 
$1\leq i\leq N$, and 
$N\in\mathbb N$. Note that the condition for coefficients bi and di follows from the form of the integral basis. Thus, we get the following equalities:
\begin{align}
9+p+q&=\sum_{i=1}^N\Big(a_i^2+\frac{b_i^2p}{4}+c_i^2q+\frac{d_i^2pq}{4r_0^2}\Big),
\end{align}
\begin{align}
2&=\sum_{i=1}^N( a_ib_i+c_id_ip_0),
\end{align}
\begin{align}
2&=\sum_{i=1}^N\Big( 2a_ic_i+\frac{1}{2}b_id_iq_0\Big),
\end{align}
\begin{align}
0&=\sum_{i=1}^N( b_ic_ir_0+a_id_i).
\end{align} If all the coefficients bi were equal to 0, the equality (3.2) would imply 
$2=\sum_{i=1}^N c_id_ip_0$, which is not possible as 
$p_0\geq 3$ (since 
$1\leq r_0 \lt p_0$ and p 0 is odd). Thus, we have at least one non-zero coefficient bi. Using (3.2), we can make the same conclusion for coefficients ai. On the other hand, if all the coefficients ci were zeros, the equality (3.3) would give 
$2=\frac{q_0}{2}\sum_{i=1}^N b_id_i$, which is again not possible since 
$p=q_0r_0\geq 10$ and 
$r_0 \lt q_0$, implying 
$\frac{q_0}{2}\geq 3$.
Let us now focus on the case when 
$d_j\neq 0$ for some 
$j\in\{1,\ldots,N\}$. If dj were even, the right side of (3.1) would be at least 
$1+\frac{p}{4}+q+\frac{pq}{r_0^2}$, and the necessary inequality
\begin{equation*}
1+\frac{p}{4}+q+\frac{pq}{r_0^2}\leq 9+p+q
\end{equation*} is equivalent to 
$q_0(4p_0-3r_0)\leq 32$. This cannot be satisfied since 
$q_0\geq 6$ and 
$4p_0-3r_0\geq 4(r_0+2)-3r_0=r_0+8\geq 9$. On the other hand, if dj were odd, (3.3) would imply the existence of k ≠ j such that 
$d_k\neq 0$, and in addition, 
$b_k\neq 0$ (note that in this case, clearly, 
$b_j\neq 0$). Therefore, as before, we get
\begin{equation*}
1+\frac{p}{2}+q+\frac{pq}{2r_0^2}\leq 9+p+q,
\end{equation*} which can be rewritten as 
$q_0(p_0-r_0)\leq 16$. Since 
$q_0\geq 6$ and 
$p_0-r_0\geq 2$, our inequality can be fulfilled only if 
$q_0=6$ and 
$p_0=r_0+2$. However, the only possible case satisfying these conditions is when p = 30 and q = 35. Using a computer program, we can easily check that 
$\mathcal P(\mathcal O_K)\geq 6$ even in this concrete case (we can consider, e.g., the element 
$44+\sqrt{30}+\sqrt{42}\in\sum\mathcal O_K^2$, for which we need at least six squares).
Thus, 
$d_i=0$ for all 
$1\leq i\leq N$. Since bi is congruent to di modulo 2, we obtain 
$b_i=2b'_i$ where 
$b_i'\in\mathbb Z$ for all 
$1\leq i\leq N$. Moreover, if we had two non-zero coefficients 
$b'_i$, we would obtain 
$1+2p+q\leq 9+p+q$, which is impossible for 
$p\geq 10$. The same must be true for coefficients ci. Thus, we have exactly one non-zero coefficient 
$b'_j$, and exactly one non-zero coefficient ck, and, clearly, 
$|b'_j|=|c_k|=1$.
Under this condition, 
$2=\sum_{i=1}^N 2a_ib'_i$ (i.e., (3.2)) implies 
$|a_j|=1$ and 
$\text{sgn}(a_j)=\text{sgn}(b'_j)$ where sgn is the sign function. Moreover, from 
$2=\sum_{i=1}^N 2a_ic_i$ (i.e., (3.3)), we have 
$|a_k|=1$ and 
$\text{sgn}(a_k)=\text{sgn}(c_k)$. From 
$0=\sum_{i=1}^N 2b'_ic_ir_0$ (i.e., (3.4)), we obtain 
$c_j=b'_k=0$, i.e., j ≠ k. Thus, without loss of generality, we can put 
$a_j=b'_j=1$ and 
$a_k=c_k=1$. It follows that every relevant decomposition of α must be of the form
\begin{equation*}
\alpha=\sum_{i\neq j,k}a_i^2+(1+\sqrt{p})^2+(1+\sqrt{q})^2,
\end{equation*} where 
$7=\sum_{i\neq j,k}a_i^2$. From the discussion at the beginning of this section, we know that to express 7, we need at least four squares for 
$p,q,r\geq 10$. Thus, we indeed get 
$\mathcal P(\mathcal O_K)\geq 6$ in this case.
Now, let us focus on the case when r < q, i.e., 
$q_0 \lt r_0 \lt p_0$. Using a similar procedure as before, we will consider the element 
$\beta\in\sum \mathcal O_K^2$ of the form
\begin{equation*}
\beta=7+\Bigg(1+\frac{\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2+\Bigg(\frac{\sqrt{p}-\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2=8+\frac{p}{2}+\frac{pq}{2r_0^2}+\sqrt{p}+\sqrt{r}.
\end{equation*}Using an analogous decomposition, we get equalities of the form
\begin{align}
8+\frac{p}{2}+\frac{pq}{2r_0^2}&=\sum_{i=1}^N\Big(a_i^2+\frac{b_i^2p}{4}+c_i^2q+\frac{d_i^2pq}{4r_0^2}\Big),
\end{align}
\begin{align}
1&=\sum_{i=1}^N( a_ib_i+c_id_ip_0),
\end{align}
\begin{align}
0&=\sum_{i=1}^N\Big( 2a_ic_i+\frac{1}{2}b_id_iq_0\Big),
\end{align}
\begin{align}
1&=\sum_{i=1}^N( b_ic_ir_0+a_id_i)
\end{align} with 
$b_i\equiv d_i\;(\text{mod }2)$. If all the coefficients bi (or ai) were equal to zero, we would have 
$1=p_0\sum_{i=1}^N c_id_i$, which is impossible for 
$p_0\geq 3$. Similarly, all zero coefficients di imply 
$1=r_0\sum_{i=1}^N b_ic_i$, which contradicts with 
$r_0\geq 3$. Thus, we have at least one non-zero ai, bi and di.
Now, let us assume that at least one ci is non-zero. Under this assumption, we have
\begin{equation*}
1+\frac{p}{4}+q+\frac{pq}{4r_0^2}\leq 8+\frac{p}{2}+\frac{pq}{2r_0^2},
\end{equation*} which can be rewritten as 
$4q\leq 28+p+r$. However, 
$p \lt r \lt q$, and, moreover, 
$10\leq q\leq 14$ only if q = 11. Nevertheless, for q = 11, we cannot have 
$10\leq p \lt r \lt q=11$. Thus, it implies that 
$c_i=0$ for all 
$1\leq i\leq N$.
If 
$|d_i|\geq 2$ for some i, we get
\begin{equation*}
1+\frac{p}{4}+\frac{pq}{r_0^2}\leq 8+\frac{p}{2}+\frac{pq}{2r_0^2},
\end{equation*} which leads to 
$28\geq 2r-p = r-p+r$. That can be satisfied only for 
$10 \lt r\leq 24$, which gives square-free r only if 
$r=14,22$. However, only the pair p = 10 and r = 14 fulfils 
$ r-p+r\leq 28$. For it, we can check by a computer program that for the element 
$\beta=20+\sqrt{10}+\sqrt{14}\in\sum\mathcal O_K^2$, we need at least six squares. Thus, we may assume that 
$|d_i|\leq 1$ for all 
$1\leq i\leq N$.
The equality (3.7) gives 
$\sum_{i=1}^N b_id_i=0$. Since we have at least one 
$|d_i|=1$ with bi odd as the congruence condition on bi and di requires, there must exist j ≠ i such that 
$|d_j|=1$, and thus bj is also odd. Moreover, an even number of non-zero coefficients di exists. Assume that there are more than two non-zero coefficients di. It implies
\begin{equation*}
1+p+r\leq 8+\frac{p}{2}+\frac{pq}{2r_0^2},
\end{equation*} i.e., 
$p+r\leq 14$, which is impossible for 
$p,r\geq 10$. Thus, there are exactly two pairs of odd bi and di, let us say 
$(b_j,d_j)$ and 
$(b_k,d_k)$. Moreover, similarly as before, it can be easily shown that 
$|b_j|=|b_k|=1$ and 
$b_i=0$ for 
$i\neq j,k$.
Since 
$\sum_{i=1}^N b_id_i=0$, without loss of generality, we can suppose 
$b_j=1$, 
$d_j=1$, 
$b_k=1$ and 
$d_k=-1$. The equalities 
$1=\sum_{i=1}^Na_ib_i=a_j+a_k$ and 
$1=\sum_{i=1}^Na_id_i=a_j-a_k$ then give 
$a_j=1$ and 
$a_k=0$. Thus, every decomposition of β must be of the form
\begin{equation*}
\beta=\sum_{i\neq j,k}a_i^2+\Bigg(1+\frac{\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2+\Bigg(\frac{\sqrt{p}-\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2,
\end{equation*} where 
$7=\sum_{i\neq j,k}a_i^2$. As before, for 7, we need at least four squares. Thus, 
$\mathcal P(\mathcal O_K)\geq 6$ even in this case.
We will proceed with bases of types (2) and (3).
Proposition 3.2. Let 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$ where 
$p,q \gt 1$ are square-free positive integers such that 
$p\equiv 2,3\;(\textrm{mod }4)$, 
$q\equiv 1\;(\textrm{mod }4)$, 
$p,r\geq 10$ and 
$q\geq 17$. Then 
$\mathcal P(\mathcal O_K)\geq 6$.
Proof. In this case, an integral basis of our field is of the form
\begin{equation*}
\mathcal O_K=\mathbb Z\Bigg[1,\sqrt{p},\frac{1+\sqrt{q}}{2},\frac{\sqrt{p}+\sqrt{r}}{2}\Bigg].
\end{equation*} It is not hard to see that 
$p\equiv r\;(\text{mod }4)$. Therefore, without loss of generality, we can suppose p < r implying 
$r_0 \lt p_0$. Moreover, since the case 
$r_0=1$ was already resolved in [Reference Krásenský, Raška and Sgallová21, Proposition 5.4], we can also assume 
$r_0\geq 3$.
Moreover, in the first part of this proof, we will suppose 
$q_0\geq 2$, i.e., that p and r are not coprime. Similarly, as before, let us focus on the element 
$\alpha\in\sum\mathcal O_K^2$ of the form
\begin{equation*}
\alpha = 7 + \Bigg(1+\frac{\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2+\Bigg(\frac{1-\sqrt{q}}{2}\Bigg)^2=\frac{33}{4}+\frac{p}{4}+\frac{q}{4}+\frac{pq}{4r_0^2}+\sqrt{p}+\frac{q_0-1}{2}\sqrt{q}+\sqrt{r}.
\end{equation*}Suppose that we can express α as
\begin{equation*}
\alpha=\sum_{i=1}^N\left(\frac{a_i}{2}+\frac{b_i}{2}\sqrt{p}+\frac{c_i}{2}\sqrt{q}+\frac{d_i}{2}\sqrt{r}\right)^2
\end{equation*} where 
$a_i,b_i,c_i,d_i\in\mathbb Z$, 
$a_i\equiv c_i\;(\text{mod }2)$, 
$b_i\equiv d_i\;(\text{mod }2)$ for all 
$1\leq i\leq N$, and 
$N\in\mathbb N$. The congruence conditions follow from the form of the above integral basis. That gives the following equalities:
\begin{align}
\frac{33}{4}+\frac{p}{4}+\frac{q}{4}+\frac{pq}{4r_0^2}&=\sum_{i=1}^N\Big(\frac{a_i^2}{4}+\frac{b_i^2p}{4}+\frac{c_i^2q}{4}+\frac{d_i^2pq}{4r_0^2}\Big),
\end{align}
\begin{align}
1&=\sum_{i=1}^N\Big(\frac{1}{2} a_ib_i+\frac{1}{2}c_id_ip_0\Big),
\end{align}
\begin{align}
\frac{q_0-1}{2}&=\sum_{i=1}^N\Big( \frac{1}{2}a_ic_i+\frac{1}{2}b_id_iq_0\Big),
\end{align}
\begin{align}
1&=\sum_{i=1}^N\Big(\frac{1}{2} b_ic_ir_0+\frac{1}{2}a_id_i\Big).
\end{align} As in the previous case, since 
$r_0,p_0\geq 3$ and 
$q_0\geq 2$, we must have at least one non-zero coefficient ai, bi, ci and di.
If this non-zero di were even, the right side of (3.9) would be at least
\begin{equation*}
\frac{1}{4}+\frac{p}{4}+\frac{q}{4}+\frac{pq}{r_0^2},
\end{equation*} which is lower than the left side only if 
$3r\leq 32$. That is impossible for 
$r\geq 14$. On the other hand, if there existed even non-zero bi, the corresponding inequality
\begin{equation*}
\frac{1}{4}+\frac{5p}{4}+\frac{q}{4}+\frac{pq}{4r_0^2}\leq \frac{33}{4}+\frac{p}{4}+\frac{q}{4}+\frac{pq}{4r_0^2}
\end{equation*} would be satisfied only when 
$p\leq 8$, which is excluded in the statement of this proposition. Thus, we have only non-zero pairs of odd coefficients bi and di. Furthermore, if there were more than one such pair, the inequality needed in (3.9) would lead to 
$p+r\leq 32$, which is not true for any case of type (2) or (3) if 
$q_0\geq 2$. Therefore, there exists exactly one 
$j\in\{1,2,\ldots,N\}$ such that 
$b_j,d_j\neq 0$ and, clearly, 
$|b_j|=|d_j|=1$.
Similarly, for 
$q\geq 17$, we can exclude the case when 
$c_i\neq 0$ is even. Furthermore, using the previous part, we see that (3.11) can be rewritten as
\begin{equation*}
\frac{q_0}{2}(1- b_jd_j) -\frac{1}{2}=\sum_{i=1}^N \frac{1}{2}a_ic_i.
\end{equation*} Since 
$1- b_jd_j$ is even, it implies that if there were more than one non-zero (odd) ci, then there would be an odd number of coefficients 
$c_i\neq 0$ (since, in that case, corresponding ai is also odd). That gives at least
\begin{equation*}
\frac{3}{4}+\frac{p}{4}+\frac{3q}{4}+\frac{pq}{4r_0^2}
\end{equation*} on the right side of (3.9), which can be satisfied only for 
$q\leq 15$, which we exclude in the statement. Thus, there is only one non-zero ci, let us say ck, which satisfies 
$|c_k|=1$.
Now, we will discuss the case when j = k. Without loss of generality, set 
$b_j=1$. Then:
(1) If
$d_j=1$, then (3.10) and (3.12) can be expressed as
$2=a_j+c_jp_0$ and
$2=a_j+c_jr_0$. However, these equalities can be both satisfied only if
$0=c_j(p_0-r_0)$, which is impossible for
$c_j\neq 0$ and
$p_0\neq r_0$.(2) If
$d_j=-1$, in a similar way, we get equalities
$2=a_j-c_jp_0$ and
$2=-a_j+c_jr_0$, which leads to
$p_0=r_0+4$,
$c_j=-1$ and
$a_j=2-p_0$. When we insert this into (3.11), we obtain
$p_0=2q_0+1$. Moreover, (3.9) implies
$(p_0-2)^2\leq 33$. This leads to
$p_0=7$,
$q_0=3$ and
$r_0=3$, which cannot be satisfied for square-free p, q and r.
Thus, j ≠ k. If we put 
$c_k=-1$ and 
$b_j=1$, without loss of generality, then one may verify from (3.9) to (3.11) that 
$a_j=2$, 
$d_j=1$ and 
$a_k=1$. That gives the decomposition
\begin{equation*}
\alpha =\sum_{i\neq j,k}a_i^2+ \Bigg(1+\frac{\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2+\Bigg(\frac{1-\sqrt{q}}{2}\Bigg)^2,
\end{equation*} where 
$7=\sum_{i\neq j,k}a_i^2$. We know that for 7, we need at least four squares. By this, the proof for the case when 
$q_0\geq 2$ is completed.
For 
$q_0=1$, likewise, we can use the element 
$\beta\in\sum\mathcal O_K^2$ from the second part of the proof of Proposition 3.1, namely
\begin{equation*}
\beta=7+\Bigg(1+\frac{\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2+\Bigg(\frac{\sqrt{p}-\frac{\sqrt{pq}}{r_0}}{2}\Bigg)^2=8+\frac{p}{2}+\frac{pq}{2r_0^2}+\sqrt{p}+\sqrt{r}.
\end{equation*}And finally, we will focus on bases of type (4).
Proposition 3.3. Let 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$ where 
$p,q \gt 1$ are square-free positive integers such that 
$p,q\equiv 1\;(\textrm{mod }4)$ and 
$p,q,r\geq 17$. Then 
$\mathcal P(\mathcal O_K)\geq 6$.
Proof. Recall that, in this case, an integral basis of 
$\mathcal O_K$ is of the form
\begin{equation*}\mathcal O_K=\mathbb Z\Big[1,\frac{1+\sqrt{p}}{2},\frac{1+\sqrt{q}}{2},\frac{1+\sqrt{p}+\sqrt{q}+\sqrt{r}}{4}\Big]
\end{equation*} if 
$\frac{p}{\gcd(p,q)}\equiv \frac{q}{\gcd(p,q)}\equiv 1\;(\text{mod }4)$, or
\begin{equation*}
\mathcal O_K=\mathbb Z\Big[1,\frac{1+\sqrt{p}}{2},\frac{1+\sqrt{q}}{2},\frac{1-\sqrt{p}+\sqrt{q}+\sqrt{r}}{4}\Big]
\end{equation*} if 
$\frac{p}{\gcd(p,q)}\equiv \frac{q}{\gcd(p,q)}\equiv 3\;(\text{mod }4)$. In both cases, the roles of p, q and r are interchangeable. Therefore, without loss of generality, we can assume 
$p \lt q \lt r$, which gives 
$r_0 \lt q_0 \lt p_0$. Clearly, 
$r_0,q_0$ and p 0 are odd and 
$r_0\equiv q_0\equiv p_0\;(\text{mod }4)$, which implies 
$q_0-r_0,p_0-q_0\geq 4$ and 
$p_0-r_0\geq 8$. Moreover, since the case when 
$r_0=1$ was already resolved in [Reference Krásenský, Raška and Sgallová21], we can assume 
$r_0\geq 3$.
For this basis, we will discuss the element 
$\alpha\in\sum\mathcal O_K^2$ of the form
\begin{multline*}
\alpha=7+\Bigg(\frac{1+\sqrt{p}}{2}\Bigg)^2+\Bigg(\frac{1+\sqrt{p}+\sqrt{q}+\sqrt{r}}{4}\Bigg)^2\\=\frac{117}{16}+\frac{5p}{16}+\frac{q}{16}+\frac{r}{16}+\frac{p_0+5}{8}\sqrt{p}+\frac{q_0+1}{8}\sqrt{q}+\frac{r_0+1}{8}\sqrt{r}.
\end{multline*}As before, suppose that
\begin{equation*}
\alpha=\sum_{i=1}^N \left(\frac{a_i}{4}+\frac{b_i}{4}\sqrt{p}+\frac{c_i}{4}\sqrt{q}+\frac{d_i}{4}\sqrt{r}\right)^2
\end{equation*} where 
$a_i,b_i,c_i,d_i\in\mathbb Z$, 
$a_i\equiv b_i\equiv c_i\equiv d_i\;(\text{mod }2)$ for all 
$1\leq i\leq N$, and 
$N\in\mathbb N$. Therefore, we get equalities
\begin{align}
\frac{117}{16}+\frac{5p}{16}+\frac{q}{16}+\frac{r}{16}&=\sum_{i=1}^N\Big(\frac{a_i^2}{16}+\frac{b_i^2p}{16}+\frac{c_i^2q}{16}+\frac{d_i^2pq}{16r_0^2}\Big),
\end{align}
\begin{align}
\frac{p_0+5}{8}&=\sum_{i=1}^N\Big(\frac{1}{8} a_ib_i+\frac{1}{8}c_id_ip_0\Big),
\end{align}
\begin{align}
\frac{q_0+1}{8}&=\sum_{i=1}^N\Big( \frac{1}{8}a_ic_i+\frac{1}{8}b_id_iq_0\Big),
\end{align}
\begin{align}
\frac{r_0+1}{8}&=\sum_{i=1}^N\Big(\frac{1}{8} b_ic_ir_0+\frac{1}{8}a_id_i\Big).
\end{align} Since 
$r_0\geq 3$, 
$q_0\geq 7$ and 
$p_0\geq 11$, there must be at least one non-zero coefficient ai, bi, ci and di. Moreover, we can easily check that every non-zero di must satisfy 
$|d_i|=1$; for that, we need to use the condition 
$a_i\equiv b_i\equiv c_i\equiv d_i\;(\text{mod }2)$ for all 
$1\leq i\leq N$.
Now, let us assume that there exists an index 
$k\in\{1,\ldots,N\}$ such that 
$c_k\neq 0$ is even. Under this assumption, we can derive from (3.13) that 
$|c_k|=2$, 
$b_k=d_k=0$, and there is only one index 
$j\in\{1,\ldots,N\}$, j ≠ k, with 
$|b_j|=|c_j|=|d_j|=1$, and 
$b_i=c_i=d_i=0$ for 
$i\neq j,k$. Without loss of generality, put 
$d_j=1$. Then we get the following two cases:
(1) If
$b_j=1$, then from (3.14) and (3.16), we obtain
$p_0+5=a_j+c_jp_0$ and
$r_0+1=a_j+c_jr_0$. These two equations are both satisfied only if
$4=(c_j-1)(p_0-r_0)$, which is impossible for
$p_0-r_0\geq 8$.(2) If
$b_j=-1$, we similarly obtain
$p_0+5=-a_j+c_jp_0$ and
$r_0+1=a_j-c_jr_0$. This leads to
$p_0+r_0+6=c_j(p_0-r_0)$. If
$c_j=-1$, the right side is negative. If
$c_j=1$, we can deduce that
$2r_0+6=0$, which is impossible for
$r_0\geq 3$.
Therefore, every non-zero coefficient ci is odd, and, moreover, we can easily check that 
$|c_i|\leq 1$ for all i.
Now, suppose that there exist two indices j ≠ k such that bj, cj, dj, bk, ck and dk are all odd. In this case, easily, if 
$i\neq j,k$, then 
$b_i=c_i=d_i=0$ and 
$|b_j|=|c_j|=|d_j|=|b_k|=|c_k|=|d_k|=1$. Without loss of generality, set 
$d_j=d_k=1$. Then we get:
(1) If
$b_j=b_k=1$, then the relations (3.14) and (3.16) give
$p_0+5=a_j+a_k+(c_j+c_k)p_0$ and
$r_0+1=a_j+a_k+(c_j+c_k)r_0$. That implies
$4=(c_j+c_k-1)(p_0-r_0)$, which is impossible for
$p_0-r_0\geq 8$.(2) If
$b_j=b_k=-1$, we similarly obtain
$p_0+r_0+6=(p_0-r_0)(c_j+c_k)$. Clearly,
$c_j+c_k\in\{-2,0,2\}$, but
$c_j+c_k=-2,0$ can be immediately excluded as
$p_0+r_0+6 \gt 0$. If
$c_j+c_k=2$, we get
$c_j=c_k=1$,
$p_0=3r_0+6$ and
$a_j+a_k=3r_0+1$. Having this, (3.15) gives
$q_0+1=a_j+a_k-2q_0=3r_0+1-2q_0$, i.e.,
$3q_0+1=3r_0+1$, which is impossible for
$q_0\neq r_0$.(3) If
$b_j=1$ and
$b_k=-1$ (the case
$b_j=-1$ and
$b_k=1$ is analogous), we obtain
$p_0+5=a_j-a_k+(c_j+c_k)p_0$ and
$r_0+1=a_j+a_k+(c_j-c_k)r_0$.(1) If
$c_j=c_k=1$, then
$a_j+a_k=r_0+1$ and (3.15) gives
$q_0+1=a_j+a_k+(1-1)q_0=r_0+1$, which is impossible for
$q_0\neq r_0$.(2) If
$c_j=1$ and
$c_k=-1$, we obtain
$p_0+5=a_j-a_k$, which leads to
$q_0+1=a_j-a_k=p_0+5$. That is impossible for
$q_0 \lt p_0$.(3) If
$c_j=-1$ and
$c_k=1$, we again get
$p_0+5=a_j-a_k$, which implies
$q_0+1=-a_j+a_k=-p_0-5$, which is impossible.(4) If
$c_j=c_k=-1$, then
$r_0+1=a_j+a_k$ and
$q_0+1=-a_j-a_k=-r_0-1$. However, that is impossible for
$q_0,r_0 \gt 0$.
Thus, we have exactly one triple of odd coefficients bj, cj and dj with 
$|c_j|=|d_j|=1$ and 
$c_i=d_i=0$ for i ≠ j.
Easily, 
$|b_j|\leq 3$. Set 
$d_j=1$. If 
$c_j=-1$, Equations (3.15) and (3.16) imply 
$q_0+r_0+2=b_j(q_0-r_0)$, which can be satisfied only for 
$b_j=1,3$. The case 
$b_j=1$ gives 
$2r_0+2=0$, which is impossible. On the other hand, 
$b_j=3$ leads to 
$a_j=4r_0+1$. Then the right side of (3.13) is at least
\begin{equation*}
\frac{(4r_0+1)^2}{16}+\frac{9p}{16}+\frac{q}{16}+\frac{r}{16},
\end{equation*} and it is lower than the left side only if 
$(4r_0+1)^2+4p\leq 117$, which is impossible for 
$r_0\geq 3$. Thus, 
$c_j=1$.
Since 
$d_j=c_j=1$, (3.15) and (3.16) give 
$a_j=b_j=1$. Then (3.14) leads to 
$p_0+5=\sum_{i\neq j}a_ib_i+1+p_0$, i.e., 
$4=\sum_{i\neq j}a_ib_i=4\sum_{i\neq j}a'_ib'_i$ where 
$a_i=2a'_i$ and 
$b_i=2b'_i$. Moreover, 
$a_i'\equiv b_i'\;(\text{mod }2)$, which follows from the form of the above integral basis. Therefore, 
$1=\sum_{i\neq j} a'_ib'_i$, which implies that there exist at least one index k ≠ j such that 
$a'_k$ and 
$b'_k$ are odd, and there is an odd number of such indices. If there were three or more such indices, the right side of (3.13) would be too large. Moreover, if there existed 
$i\neq j,k$ such that 
$b_i'\neq 0$ were even, the right side of (3.13) would be again too large. Thus, 
$b_i'=0$ for all 
$i\neq j,k$, and, moreover, 
$|b_k'|=1$.
Without loss of generality, we can set 
$b_k'=1$, and then 
$a_k'=1$. Thus, the only possible relevant decomposition of α is of the form
\begin{equation*}
\alpha=\sum_{i\neq j,k}a_i^2+\Bigg(\frac{1+\sqrt{p}}{2}\Bigg)^2+\Bigg(\frac{1+\sqrt{p}+\sqrt{q}+\sqrt{r}}{4}\Bigg)^2,
\end{equation*} where 
$7=\sum_{i\neq j,k}a_i^2$. As before, for 7, we need at least four squares. Therefore, 
$\mathcal P(\mathcal O_K)\geq 6$ even in this case.
In a similar fashion, we will prove Theorem 1.2.
Proof of Theorem 1.2.
Recall that in this case, 
$p=(2n-1)(2n+1)$, 
$q=(2n-1)(2n+3)$, 
$r=(2n+1)(2n+3)$ and 
$n\geq 6$. Moreover, 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$ has an integral basis of type (3), i.e.,
\begin{equation*}
\mathcal O_K=\mathbb Z\Bigg[1,\sqrt{p},\frac{1+\sqrt{q}}{2},\frac{\sqrt{p}+\sqrt{r}}{2}\Bigg].
\end{equation*} To give our assertion, let us consider the element 
$\alpha\in\sum\mathcal O_K^2$ of the form
\begin{align*}
\alpha&=7+\left(\frac{1-\sqrt{q}}{2}\right)^2+\left(1+\frac{\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\right)^2+\left(2+\frac{-\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\right)^2\\
&=\frac{49}{4}+\frac{p}{2}+\frac{q}{4}+\frac{pq}{2r_0^2}-\sqrt{p}-\frac{1}{2}\sqrt{q}+3\frac{\sqrt{pq}}{r_0}.
\end{align*}Similarly as in the proof of Proposition 3.2, we consider decompositions leading to the equations of the form
\begin{align}
\frac{49}{4}+\frac{p}{2}+\frac{q}{4}+\frac{pq}{2r_0^2}&=\sum_{i=1}^N\Big(\frac{a_i^2}{4}+\frac{b_i^2p}{4}+\frac{c_i^2q}{4}+\frac{d_i^2pq}{4r_0^2}\Big),
\end{align}
\begin{align}
-1&=\sum_{i=1}^N\Big(\frac{1}{2} a_ib_i+\frac{1}{2}c_id_ip_0\Big),
\end{align}
\begin{align}
-\frac{1}{2}&=\sum_{i=1}^N\Big( \frac{1}{2}a_ic_i+\frac{1}{2}b_id_iq_0\Big),
\end{align}
\begin{align}
3&=\sum_{i=1}^N\Big(\frac{1}{2} b_ic_ir_0+\frac{1}{2}a_id_i\Big)
\end{align} where the coefficients 
$a_i,b_i,c_i,d_i\in\mathbb Z$ satisfy 
$a_i\equiv c_i\;(\text{mod }2)$ and 
$b_i\equiv d_i\;(\text{mod }2)$. Since 
$r_0\geq 11$, 
$q_0\geq 13$ and 
$p_0\geq 15$, we can again conclude the existence of at least one non-zero coefficient ai, bi, ci and di.
If 
$|d_i|\geq 2$ for some i, then (3.17) implies
\begin{equation*}
\frac{1}{4}+\frac{p}{4}+\frac{q}{4}+\frac{pq}{r_0^2}\leq \frac{49}{4}+\frac{p}{2}+\frac{q}{4}+\frac{pq}{2r_0^2},
\end{equation*} i.e., 
$r+(r-p)\leq 48$, which is not true for 
$r=p_0q_0\geq 195$. Thus, 
$|d_i|\leq 1$ for all i. Using similar arguments and explicit expressions for p, q and r, we can show 
$|b_i|\leq 1$ and 
$|c_i|\leq 1$ for all i. Moreover, in the same manner, there exist at most two non-zero coefficients di, and since 
$|b_i|\leq 1$ and 
$b_i\equiv d_i\;(\text{mod }2)$, the same is also true for coefficients bi.
First of all, let us assume that there is only one pair of non-zero coefficients bi and di, let us say 
$(b_j,d_j)\in\{(1,1),(-1,1)\}$ where, without loss of generality, we can suppose 
$d_j=1$. Then (3.18) and (3.20) give 
$-2=a_jb_j+c_jd_jp_0$ and 
$6=b_jc_jr_0+a_jd_j$. We obtain:
(1) If
$b_j=1$, then
$8=c_j(r_0-p_0)=-4c_j$, which is a contradiction with
$|c_j|\leq 1$.(2) If
$b_j=-1$, then
$4=c_j(p_0-r_0)=4c_j$, which gives
$c_j=1$ and
$a_j=p_0+2=2n+5$. However, (3.19) implies
\begin{equation*}
-1=\sum_{i\neq j}a_ic_i+a_jc_j+b_jd_jq_0=\sum_{i\neq j}a_ic_i+p_0+2-q_0=\sum_{i\neq j}a_ic_i+4.
\end{equation*}Thus, there exists at least one index i ≠ j such that
$a_i\neq 0$ and
$c_i\neq 0$. Then (3.17) gives
\begin{equation*}
\frac{(2n+5)^2}{4}+\frac{p}{4}+\frac{q}{4}+\frac{r}{4}+\frac{1}{4}+\frac{q}{4}\leq \frac{49}{4}+\frac{p}{2}+\frac{q}{4}+\frac{r}{2},
\end{equation*}which can be rewritten as
$(2n+5)^2\leq 48+r+p-q \lt 48+r$, which is never true for
$n\geq 6$.
Therefore, we have exactly two pairs 
$(b_j,d_j),(b_k,d_k)\in\{(1,1),(-1,1)\}$ with j ≠ k, for which, without loss of generality, we can suppose 
$d_j=d_k=1$. In that case, (3.17) implies that there exists exactly one non-zero ci, let us say 
$c_l\in\{-1,1\}$. Moreover, (3.19) implies 
$-1=a_lc_l+(b_jd_j+b_kd_k)q_0$. If 
$b_jd_j+b_kd_k=\pm 2$, then 
$|a_l|\geq 2q_0-1=4n+1$. However, for such coefficients al, the right side of (3.17) is too large. Thus, 
$b_jd_j+b_kd_k=0$, and without loss of generality, 
$b_j=1$ and 
$b_k=-1$. Moreover, 
$a_lc_l=-1$.
If 
$l\in\{j,k\}$, then, similarly as before, (3.18) and (3.20) lead to a contradiction for both cases of the pair 
$(a_l,c_l)\in\{(1,-1),(-1,1)\}$. Thus, 
$l\neq j,k$, and, without loss of generality, 
$a_l=1$ and 
$c_l=-1$. Furthermore, from (3.18) and (3.20), we also have 
$a_j=2$ and 
$a_k=4$. It follows that our decomposition must be of the form
\begin{equation*}
\alpha=\sum_{i\neq j,k,l}a_i^2+\left(\frac{1-\sqrt{q}}{2}\right)^2+\left(1+\frac{\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\right)^2+\left(2+\frac{-\sqrt{p}+\frac{\sqrt{pq}}{r_0}}{2}\right)^2
\end{equation*} where 
$7=\sum_{i\neq j,k,l}a_i^2$. Since we need at least four squares to express 7, the proof is completed.
4. Indecomposable integers in biquadratic fields
Now, we will focus our attention on indecomposable integers in biquadratic fields. Recall that these algebraic integers are totally positive and cannot be expressed as a sum of two elements from 
$\mathcal O_K^{+}$. In the following, we will study three subfamilies of biquadratic fields from [Reference Man23]. For them, we will find an upper bound on the norm of indecomposable integers, and we will also study the minimal traces of their indecomposables. For that, we need to know a form of the codifferent 
$\mathcal O_K^{\vee}$.
All of these families are of type (3). As indicated in § 2, it can be easily checked that a 
$\mathbb Z$-basis of 
$\mathcal O_K^{\vee}$ for type (3) can consist of the elements
\begin{align*}
\varphi_1&=\frac{1}{4}+\frac{1}{4p_0r_0}-\frac{1}{2p_0r_0}\frac{1+\sqrt{q}}{2},\\
\varphi_2&=\left(\frac{1}{4p_0q_0}+\frac{1}{4q_0r_0}\right)\sqrt{p}-\frac{1}{2p_0q_0}\frac{\sqrt{p}+\sqrt{r}}{2},\\
\varphi_3&=-\frac{1}{2p_0r_0}+\frac{1}{p_0r_0}\frac{1+\sqrt{q}}{2},\\
\varphi_4&=-\frac{1}{2p_0q_0}\sqrt{p}+\frac{1}{p_0q_0}\frac{\sqrt{p}+\sqrt{r}}{2}.
\end{align*}4.1. First family
The first family is the one from Theorem 1.2, which is defined by
\begin{equation*}
p=(2n-1)(2n+1), \quad q=(2n-1)(2n+3), \quad r=(2n+1)(2n+3)
\end{equation*} where 
$n\geq 6$ is such that 
$p,q,r$ are all square-free. The fundamental units of the corresponding quadratic subfields 
$\mathbb Q(\sqrt{p})$, 
$\mathbb Q(\sqrt{q})$ and 
$\mathbb Q(\sqrt{r})$ are 
$\varepsilon_p=2n+\sqrt{p}$, 
$\varepsilon_q=\frac{2n+1+\sqrt{q}}{2}$ and 
$\varepsilon_r=2n+2+\sqrt{r}$, and they are all totally positive. Note that in these quadratic subfields, only totally positive units are indecomposable. On the other hand, the situation in 
$\mathbb Q(\sqrt{p},\sqrt{q})$ is much more complicated as proved by Man [Reference Man23] in the following theorem:
Theorem 4.1 ([Reference Man23, Theorem 1.6])
Let 
$p,q,r$ be as above. Then up to multiplication by totally positive units, the indecomposable integers in 
$\mathbb Q(\sqrt{p},\sqrt{q})$ are 1, 
$\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}$, 
$\mu=n+\frac{3}{2}+\frac{1}{2}\sqrt{p}+\frac{1}{2}\sqrt{q}+\frac{1}{2}\sqrt{r}$,
\begin{equation*}
\alpha_t=1+\varepsilon_p+t(\mu-1), \quad \beta_t=\frac{1+\varepsilon_p\varepsilon_r}{2}+\varepsilon_p+t(\mu-1), \quad 3\leq t\leq 2n-2,
\end{equation*}
\begin{equation*}
\gamma_t=1+\varepsilon_q^{-1}+t(\mu-1), \quad \delta_t=\frac{1+\varepsilon_p\varepsilon_r}{2}+\varepsilon_q^{-1}+t(\mu-1), \quad 4\leq t\leq 2n-1,
\end{equation*}
\begin{equation*}
\omega_t=\frac{\varepsilon_r^{-1}+\varepsilon_p}{2}+t(\mu-2), \quad 2\leq t\leq 2n-1.
\end{equation*}Moreover, we have the following relations among these indecomposable integers:
\begin{align*}
\delta_{2n-1-(t-3)}&=\sigma_3(\alpha_t)\varepsilon_r,\\
\beta_{2n-2-(t-3)}&=\sigma_3(\beta_t)\varepsilon_r,\\
\gamma_{t+1}&=\sigma_4(\beta_t)\varepsilon_p,\\
\omega_{2n-1-(t-2)}&=\sigma_2(\omega_t)\varepsilon_p
\end{align*} where maps σi are the embeddings introduced in § 2. Trivially, 
$N(\sigma_i(\alpha)\varepsilon)=N(\alpha)$ and 
$\text{Tr}(\alpha\delta)=\text{Tr}((\sigma_i(\alpha)\varepsilon)(\varepsilon^{-1}\sigma_i(\delta)))$ for every 
$\alpha\in\mathcal O_K$, 
$\varepsilon\in\mathcal O_K$ a unit and 
$\delta\in\mathcal O_K^{\vee}$. Therefore, regarding norms and minimal traces, it is enough to consider the elements 1, 
$\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}$, µ, αt with 
$3\leq t\leq 2n-2$, βt with 
$3\leq t\leq n$ and ωt with 
$2\leq t\leq n$.
Now, we will prove Theorem 1.3:
Proof of Theorem 1.3.
In biquadratic fields, which are Galois, it is not so hard to compute norms of elements using 
$N(\alpha)=\sigma_1(\alpha)\sigma_2(\alpha)\sigma_3(\alpha)\sigma_4(\alpha)$. In particular, we obtain
\begin{align*}
N\left(\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}\right)&=(2n+1)^2, \\
N(\mu)&=4, \\
N(\alpha_t)&=((4n+2)(t+1)-t^2)^2,\\
N(\beta_t)&=t^4-(4n+2)t^3-(4n^2+12n+3)t^2+(16n^3+40n^2+24n+4)t\\ & \,\,\,\,+16n^3+48n^2+44n+13,\\
N(\omega_t)&=(2n+1+(4n+2)t-2t^2)^2.
\end{align*} Moreover, we can immediately exclude µ as its norm 4 is too small in comparison with, e.g., 
$(2n+1)^2$ for 
$n\geq 6$.
We will start with elements αt. Let us set 
$g_1(t)=(4n+2)(t+1)-t^2$. Since derivative of 
$g_1(t)$, which is 
$4n+2-2t$, has zero in 
$t=2n+1$, the polynomial 
$g_1(t)$ is a monotonic function on 
$[3,2n-2]$. Therefore, the largest value among norms of αt is attained for either t = 3 or 
$t=2n-2$. In our case, the latter is true, and we obtain the bound 
$N(\alpha_{2n-2})=16 n^4+64n^3+16n^2-96n+36$.
We will proceed with βt. Let 
$g_2(t)=N(\beta_t)$. Then g 2 is a quartic polynomial in t with a positive leading coefficient, 
$g_2(-1)=g_2(2n+2)=4n^2+12n+9 \gt 0$ and 
$g_2(-2)=g_2(2n+3)=-16n^3-48n^2-20n+25 \lt 0$ for 
$n\geq 6$. Moreover, we have 
$N(\beta_{2n-2-(t-3)})=N(\beta_t)$ following from 
$\beta_{2n-2-(t-3)}=\sigma_3(\beta_t)\varepsilon_r$. Therefore, the maximum of 
$g_2(t)$ on the interval 
$[3,n]$ is attained in t = n and equals 
$g_2(n)=9n^4+42n^3+69n+48n+13$.
Similarly, for ωt, the derivative of 
$2n+1+(4n+2)t-2t^2$ is 
$2+4n-4t$, which is zero for 
$t=n+\frac{1}{2}$. Thus, 
$2n+1+(4n+2)t-2t^2$ is monotonic on the interval 
$[2,n]$. The largest norm is attained in some of the border points, and, finally, we get the bound 
$N(\omega_n)=4n^4+16n^3+20n^2+8n+1$.
Thus, we are left with the norms 
$(2n+1)^2$, 
$16n^4+64n^3+16n^2-96n+36$, 
$9n^4+42n^3+69n+48n+13$ and 
$4n^4+16n^3+20n^2+8n+1$, from which 
$16n^4+64n^3+16n^2-96n+36$ is the largest for 
$n\geq 6$.
Now, we will discuss minimal traces of our elements. Recall that for 
$\alpha\in\mathcal O_K^+$,
\begin{equation*}\text{minTr}(\alpha)=\min_{\delta\in\mathcal O_K^{\vee,+}}\text{Tr}(\alpha\delta).\end{equation*}Proposition 4.2. Let 
$p,q,r$ be as above. Then we have:
(1)
$\textrm{minTr}\left(\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}\right)=\textrm{minTr}\left(\mu\right)=1$,(2)
$\textrm{minTr}\left(\alpha_t\right)=\textrm{minTr}\left(\beta_t\right)=2$ for all
$3\leq t\leq 2n-2$,(3)
$\textrm{minTr}\left(\gamma_t\right)=\textrm{minTr}\left(\delta_t\right)=2$ for all
$4\leq t\leq 2n-1$,(4)
$\textrm{minTr}\left(\omega_t\right)=2$ for all
$2\leq t\leq 2n-1$.
Proof. Again, we can restrict to elements 
$\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}$, µ, αt, 
$3\leq t\leq 2n-2$, βt, 
$3\leq t\leq n$ and ωt, 
$2\leq t\leq n$; the rest is a consequence of the above relations.
In the first part, we see that 
$\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}=2n+1-\sqrt{p}+\frac{\sqrt{p}+\sqrt{r}}{2}$. If we take the element of the codifferent 
$\varphi_1+(2n-1)\varphi_2-(n-1)\varphi_3-\varphi_4$, it gives trace 1 with 
$\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}$. Note that for this computation, we can use the relation (2.1). Thus, it suffices to prove that this element of the codifferent is totally positive. It is a straightforward computational task to verify that 
$\varphi_1+(2n-1)\varphi_2-(n-1)\varphi_3-\varphi_4$ is a root of
\begin{equation*}
g(x)=x^4-x^3+\frac{4n+3}{8n^2+16n+6}x^2-\frac{1}{16n^2+32n+12}x+\frac{1}{16(4n^2+8n+3)^2},
\end{equation*} which has only positive roots as indicated in § 2. That completes our proof for element 
$\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}$.
We will proceed with the element 
$\mu=n+1+\frac{1+\sqrt{q}}{2}+\frac{\sqrt{p}+\sqrt{r}}{2}$. To get trace 1, we can use the element of the codifferent 
$\varphi_1-(2n-1)\varphi_2+n\varphi_3-2n\varphi_4$, which is also a root of g and, thus, totally positive.
For 
$\alpha_t=(t+2)n+1+\sqrt{p}+t\frac{1+\sqrt{q}}{2}+t\frac{\sqrt{p}+\sqrt{r}}{2}$, we have
\begin{equation*}
\text{Tr}(\alpha_t(\varphi_1-(2n-1)\varphi_2+n\varphi_3-2n\varphi_4))=2,
\end{equation*} which gives an upper bound on 
$\text{minTr}(\alpha_t)$. We will show that 2 is, in fact, 
$\text{minTr}(\alpha_t)$. Let us suppose that there is an element 
$a\varphi_1+b\varphi_2+c\varphi_3+d\varphi_4\in\mathcal O_K^{\vee,+}$ such that
\begin{equation*}
\text{Tr}(\alpha_t(a\varphi_1+b\varphi_2+c\varphi_3+d\varphi_4))=((t+2)n+1)a+b+tc+td=1.
\end{equation*} That gives 
$b=1-((t+2)n+1)a-t(c+d)$. If we multiply 
$a\varphi_1+b\varphi_2+c\varphi_3+d\varphi_4$ by totally positive elements 
$\nu\in\mathcal O_K$, we need to get
\begin{equation*}
\text{Tr}(\nu(a\varphi_1+b\varphi_2+c\varphi_3+d\varphi_4)) \gt 0.
\end{equation*} Therefore, we will choose suitable elements 
$\nu\in\mathcal O_K^{+}$ to get restrictions on a, b, c and d. For ν = 1, we obtain the trace a, which implies a > 0. Moreover,
\begin{align*}
\text{Tr}(\alpha_{t-1}(a\varphi_1+(1-((t+2)n+1)a-t(c+d))\varphi_2+c\varphi_3+d\varphi_4))&=1-an-c-d \gt 0,\\
\text{Tr}(\alpha_{t+1}(a\varphi_1+(1-((t+2)n+1)a-t(c+d))\varphi_2+c\varphi_3+d\varphi_4))&=1+an+c+d \gt 0.
\end{align*} (Note that both α 2 and 
$\alpha_{2n-1}$ are totally positive; see [Reference Man23, Subsection 5.3]). That gives 
$an+c+d=0$, i.e., our element of the codifferent is of the form
\begin{equation*}
a\varphi_1+(1-(2n+1)a)\varphi_2+c\varphi_3-(an+c)\varphi_4
\end{equation*} for some 
$a,c\in\mathbb Z$. Then we further have
\begin{equation*}
\text{Tr}(\varepsilon_p(a\varphi_1+(1-(2n+1)a)\varphi_2+c\varphi_3-(an+c)\varphi_4))=1-a \gt 0.
\end{equation*} However, that is a contradiction with a > 0. Thus, we obtain 
$\text{minTr}(\alpha_t)=2$ for all 
$3\leq t\leq 2n-2$.
The indecomposables
\begin{equation*}
\beta_t=2n^2+3n+tn+2\sqrt{p}+(2n+1+t)\frac{1+\sqrt{q}}{2}+(2n+t)\frac{\sqrt{p}+\sqrt{r}}{2}
\end{equation*} also give trace 2 with the element 
$\varphi_1-(2n-1)\varphi_2+n\varphi_3-2n\varphi_4$. Similarly, as before, if there were an element 
$\delta\in\mathcal O_K^{\vee,+}$ such that 
$\text{Tr}(\beta_t\delta)=1$, it would have to be of the form
\begin{equation*}
\delta=a\varphi_1+\frac{1}{2}(1-(2n^2+3n+tn)a-(2n+t)(c+d)-c)\varphi_2+c\varphi_3+d\varphi_4
\end{equation*} for suitable 
$a,c,d\in\mathbb Z$ and a > 0. Multiplying by 
$\beta_{t-1}$ and 
$\beta_{t+1}$, respectively, implies 
$an+c+d=0$, i.e., 
$\delta=a\varphi+\frac{1}{2}(1-3an-c)\varphi_2+c\varphi_3-(an+c)\varphi_4$. Then
\begin{align*}
\text{Tr}(\sigma_3(\varepsilon_q)\delta)&=\text{Tr}\left(\Big(n+1-\frac{1+\sqrt{q}}{2}\Big)\delta\right)=(n+1)a-c \gt 0,\\
\text{Tr}(\varepsilon_p\varepsilon_r\delta)&=1-(n+1)a+c \gt 0.
\end{align*} However, the above inequalities cannot be both satisfied, giving 
$\text{minTr}(\beta_t)=2$.
Finally, for 
$\omega_t=2n+1+t(n-1)+\sqrt{p}+t\frac{1+\sqrt{q}}{2}+(t-1)\frac{\sqrt{p}+\sqrt{r}}{2}$, we will consider the element 
$\varphi_1-(2n-1)\varphi_2-(n-1)\varphi_3$. This element is a root of
\begin{equation*}
x^4-x^3+\frac{3}{4n+6}x^2-\frac{1}{8n^2+16n+6}x+\frac{1}{4(4n^2+8n+3)^2};
\end{equation*} thus, it is totally positive and gives trace 2 with ωt. Let us suppose that some 
$\delta=a\varphi+b\varphi_2+c\varphi_3+d\varphi_4\in\mathcal O_K^{\vee,+}$ satisfies 
$\text{Tr}(\omega_t\delta)=1$, which gives 
$b=1-(2n+1+t(n-1))a-(t-1)(c+d)-c$ and a > 0. Then, again, by multiplying by 
$\omega_{t-1}$ and 
$\omega_{t+1}$, we obtain the condition 
$a(n-1)+c+d=0$, i.e., 
$\delta=a\varphi_1+(1-3an-c)\varphi_2+c\varphi_3-(a(n-1)+c)\varphi_4$. (Note that ω 1 and 
$\omega_{n+1}$ are both totally positive). Moreover,
\begin{equation*}
\text{Tr}(\varepsilon_p\varepsilon_q\delta)=\text{Tr}\left(\Big(2n^2+\sqrt{p}+2n\frac{1+\sqrt{q}}{2}+(2n-1)\frac{\sqrt{p}+\sqrt{r}}{2}\Big)\delta\right)=1-a \gt 0,
\end{equation*}a contradiction. By this, we finished the proof.
Proof of Theorem 1.4
As we have proved, every indecomposable integer in 
$\mathbb Q(\sqrt{p},\sqrt{q})$ has minimal trace 1 or 2. That gives our assertion.
Moreover, we are able to provide a lower bound on the number of variables of universal diagonal quadratic forms.
Proposition 4.3. Let 
$p,q,r$ be as above. Then every universal diagonal quadratic form over 
$\mathcal O_K$ must have at least 
$\frac{2n-4}{3}$ variables.
Proof. We will apply Proposition 2.1. If we consider 
$\delta=\varphi_1-(2n-1)\varphi+n\varphi_3-2n\varphi_4$, we know that 
$\text{Tr}(\alpha_t\delta)=\text{Tr}(\beta_t\delta)=2$ for all 
$3\leq t\leq 2n-2$, and, moreover, we can easily show that 
$\text{Tr}(\gamma_t\delta)=\text{Tr}(\delta_t\delta)=2$ for all 
$4\leq t\leq 2n-1$. That gives the bound 
$\frac{4(2n-4)}{12}=\frac{2n-4}{3}$.
4.2. Second family
We will proceed with the family of the form 
$p=(2n-1)(2n+1)$, 
$q=(4n-3)(4n+1)$ and r = pq where we assume 
$n\geq 9$ and that p and q are coprime and square-free. Moreover, similarly, as before, 
$\varepsilon_p=2n+\sqrt{p}$, 
$\varepsilon_q=\frac{4n-1+\sqrt{q}}{2}$ and 
$\varepsilon_r=8n^2-2n-2+\sqrt{r}$ are fundamental units of quadratic subfields. This family was also considered in [Reference Man23] and has the following structure of indecomposable integers:
Theorem 4.4 ([Reference Man23, Theorem 5.3])
Let 
$p,q,r$ be as above. Then up to multiplication by totally positive units, the indecomposable integers in 
$\mathbb Q(\sqrt{p},\sqrt{q})$ are 1,
\begin{equation*}
\alpha_t=\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}+t(\varepsilon_q-\varepsilon_p^{-1}), \quad 0\leq t\leq 2n-2,
\end{equation*}
\begin{equation*}
\beta_t=\varepsilon_p^{-1}-\varepsilon_q+t(\varepsilon_p\varepsilon_q-1), \quad 1\leq t\leq 2n-1.
\end{equation*} As in the previous family, some of the indecomposables are associated with conjugates of other ones. In particular, we have 
$\beta_{2n-1-(t-1)}=\sigma_2(\alpha_t)\varepsilon_r$ for 
$1\leq t\leq 2n-2$. Thus, it is enough to discuss the elements αt and β 1.
For these biquadratic fields, we prove the following:
Proposition 4.5. Let 
$p,q,r$ be as above. If 
$\alpha\in\mathcal O_K^{+}$ is indecomposable in 
$\mathcal O_K$ for 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$, then 
$N(\alpha)\leq 16n^4-8n^2+1$.
Proof. Again, the norms of our elements are
\begin{align*}
N(\alpha_t)&= (t^2-4n^2+1)^2,\\
N(\beta_1)&=16n^2-16n+4.
\end{align*} It is not hard to see that the maximum of 
$N(\alpha_t)$ on the interval 
$[0,2n-2]$ must be attained in one of the border points, and, moreover, 
$64n^2-80n+25=N(\alpha_{2n-2}) \lt N(\alpha_0)=16n^4-8n^2+1$ for 
$n\geq 9$. Furthermore, 
$16n^4-8n^2+1 \gt 16n^2-16n+4$ for 
$n\geq 9$, which completes the proof.
Now, we will discuss the minimal traces.
Proposition 4.6. Let 
$p,q,r$ be as above. Let α be an indecomposable integer in 
$\mathcal O_K$. Then 
$\textrm{minTr}(\alpha)=1$.
Proof. First, let us consider the elements 
$\alpha_t=4n^2-1-t+(t-1)\sqrt{p}+t\frac{1+\sqrt{q}}{2}+\frac{\sqrt{p}+\sqrt{r}}{2}$. The element of the codifferent 
$\varphi_1+(2n-1)\varphi_2-(2n-2)\varphi_3-(4n^2-2n-1)\varphi_4$ is a root of the polynomial
\begin{multline*}
x^4-x^3+\frac{8n^2-1}{32n^3+8n^2-8n-2}x^2-\frac{1}{4(16n^3+4n^2-4n-1)}x\\+\frac{1}{16(2n-1)^2(2n+1)^2(4n+1)^2}.
\end{multline*} Therefore, it is totally positive and gives trace 1 with αt. Moreover, its trace with 1 and 
$\beta_1=4n^2-2n+(2n-2)\sqrt{p}+(2n-1)\frac{1+\sqrt{q}}{2}+\frac{\sqrt{p}+\sqrt{r}}{2}$ is also 1, which finishes the proof.
We can apply the above result to get a lower bound on the minimal number of variables of universal quadratic forms.
Proposition 4.7. Let 
$p,q,r$ be as above. Then every universal classical quadratic form over 
$\mathcal O_K$ must have at least 
$\frac{2n+3}{2}$ variables.
Proof. We see that 
$\varphi_1+(2n-1)\varphi_2-(2n-2)\varphi_3-(4n^2-2n-1)\varphi_4\in\mathcal O_K^{\vee,+}$ produces trace 1 with αt, 
$0\leq t\leq 2n-2$, and, moreover, with 1 and β 1. Besides that, that is also true for the following elements:
\begin{align*}
\sigma_4(\beta_t)&=4n^2t-t+(1-2nt)\sqrt{p}+(1-2nt)\frac{1+\sqrt{q}}{2}+t\frac{\sqrt{p}+\sqrt{r}}{2}\quad (1\leq t\leq 2n-1),\\
\varepsilon_p^{-1}&=2n-\sqrt{p},\\
\varepsilon_q&=2n-1+\frac{1+\sqrt{q}}{2},\\
\varepsilon_r&=8n^2-2n-2-\sqrt{p}+2\frac{\sqrt{p}+\sqrt{r}}{2},\\
\sigma_4(\varepsilon_p\varepsilon_q)&=4n^2-2n\sqrt{p}-2n\frac{1+\sqrt{q}}{2}+\frac{\sqrt{p}+\sqrt{r}}{2},\\
\sigma_4(\varepsilon_p\varepsilon_r)&=16n^3-4n-1-(8n^2-2)\sqrt{p}-(8n^2-2)\frac{1+\sqrt{q}}{2}+4n\frac{\sqrt{p}+\sqrt{r}}{2},\\
\sigma_4(\varepsilon_q\varepsilon_r)&=16n^3-4n^2-4n-(8n^2-2n-2)\sqrt{p}-(8n^2-2n-2)\frac{1+\sqrt{q}}{2}\\
&+(4n-1)\frac{\sqrt{p}+\sqrt{r}}{2}.
\end{align*}Applying Proposition 2.1, we get the lower bound.
4.3. Third family
The last family from [Reference Man23], which we will examine, is of the form 
$p=(2n-1)(2n+1)$, 
$q=(4n-1)(4n+3)$ and r = pq where p and q are assumed to be coprime and square-free, and 
$n\geq 2$. The corresponding fundamental units of quadratic subfields are 
$\varepsilon_p=2n+\sqrt{p}$, 
$\varepsilon_q=\frac{4n+1+\sqrt{q}}{2}$ and 
$\varepsilon_r=8n^2+2n-2+\sqrt{r}$. Note that these units are also totally positive. The structure of indecomposable integers is the following:
Theorem 4.8 ([Reference Man23, Theorem 5.4])
Let 
$p,q,r$ be as above. Then up to multiplication by totally positive units, the indecomposable integers in 
$\mathbb Q(\sqrt{p},\sqrt{q})$ are 1, 
$\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}$,
\begin{equation*}
\alpha_t=\frac{\varepsilon_p+\varepsilon_r}{2}+t(\varepsilon_p-\varepsilon_q^{-1}), \quad 0\leq t\leq 2n-2,
\end{equation*}
\begin{equation*}
\beta_t=\varepsilon_q^{-1}-\varepsilon_p+t(\varepsilon_p\varepsilon_q-1), \quad 1\leq t\leq 2n-1.
\end{equation*} Moreover, 
$\alpha_0=\sigma_3\left(\frac{\varepsilon_p^{-1}+\varepsilon_r}{2}\right)\varepsilon_p\varepsilon_r$ and 
$\beta_{2n-1-(t-1)}=\sigma_3(\alpha_t)\varepsilon_r$ for 
$1\leq t\leq 2n-2$. Thus, it suffices to examine elements αt and β 1. Moreover, the situation is similar to the second family.
Proposition 4.9. Let 
$p,q,r$ be as above. If 
$\alpha\in\mathcal O_K^{+}$ is indecomposable in 
$\mathcal O_K$ for 
$K=\mathbb Q(\sqrt{p},\sqrt{q})$, then 
$N(\alpha)\leq 16n^4+16n^3-4n^2-4n+1$.
Proof. In this case, we have
\begin{align*}
N(\alpha_t)&= (t^2+t -4n^2-2n+1)^2,\\
N(\beta_1)&= 16n^2-8n+1.
\end{align*} Moreover, similarly, as before, 
$N(\alpha_t)$ attains its maximum in one of the border points of the interval 
$[0,2n-2]$. Since 
$N(\alpha_{2n-2})=64n^2-48n+9 \lt 16n^4+16n^3-4n^2-4n+1=N(\alpha_{0})$, and, clearly 
$N(\beta_1) \lt N(\alpha_{0})$ for 
$n\geq 2$, 
$N(\alpha_{0})$ bounds the norm of indecomposable integers in these fields.
As in the previous case, all indecomposables have minimal trace 1.
Proposition 4.10. Let 
$p,q,r$ be as above. Let α be an indecomposable integer in 
$\mathcal O_K$. Then 
$\textrm{minTr}(\alpha)=1$.
Proof. In this case, we consider the element of the codifferent 
$\varphi_1-(2n-1)\varphi_2+2n\varphi_3-(4n^2+2n-2)\varphi_4$, which is a root of the polynomial
\begin{multline*}
x^4-x^3+\frac{16 n^2+ 8 n-1}{64 n^3+ 64 n^2+ 4 n-6}x^2-\frac{1}{2 (32 n^3+ 32 n^2+ 2 n-3)}x\\+\frac{1}{4 (4 n-1)^2 (2 n+1)^2 (4 n+3)^2}.
\end{multline*} This totally positive element produces trace 1 both with algebraic integers 
$\alpha_t=4n^2+2n-1-t+t\sqrt{p}+t\frac{1+\sqrt{q}}{2}+\frac{\sqrt{p}+\sqrt{r}}{2}$ and 
$\beta_1=4n^2+(2n-1)\sqrt{p}+(2n-1)\frac{1+\sqrt{q}}{2}+\frac{\sqrt{p}+\sqrt{r}}{2}$, and, moreover, with 1.
This result also enables us to make the following conclusion.
Proposition 4.11. Let 
$p,q,r$ be as above. Then every universal classical quadratic form over 
$\mathcal O_K$ must have at least 
$\frac{2n+3}{2}$ variables.
Proof. We will consider the element of the codifferent as above. Besides αt with 
$0\leq t\leq 2n-2$, β 1 and 1, we can use the following elements:
\begin{align*}
\sigma_4(\beta_t)&=4n^2t+2nt-t-(2nt+t-1)\sqrt{p}-(2nt-1)\frac{1+\sqrt{q}}{2}+t\frac{\sqrt{p}+\sqrt{r}}{2}\quad (1\leq t\leq 2n-1),\\
\varepsilon_p&=2n+\sqrt{p},\\
\varepsilon_q^{-1}&=2n+1-\frac{1+\sqrt{q}}{2},\\
\varepsilon_r&=8n^2+2n-2-\sqrt{p}+2\frac{\sqrt{p}+\sqrt{r}}{2},\\
\sigma_4(\varepsilon_p\varepsilon_q)&=4n^2+2n-(2n+1)\sqrt{p}-2n\frac{1+\sqrt{q}}{2}+\frac{\sqrt{p}+\sqrt{r}}{2},\\
\sigma_4(\varepsilon_p\varepsilon_r)&=16n^3+8n^2-4n-1-(8n^2+4n-2)\sqrt{p}-(8n^2-2)\frac{1+\sqrt{q}}{2}+4n\frac{\sqrt{p}+\sqrt{r}}{2},\\
\sigma_4(\varepsilon_q\varepsilon_r)&=16n^3+12n^2-2n-2-(8n^2+6n-1)\sqrt{p}-(8n^2+2n-2)\frac{1+\sqrt{q}}{2}\\
& \quad +(4n+1)\frac{\sqrt{p}+\sqrt{r}}{2}.
\end{align*}Then the statement follows from Proposition 2.1.
Acknowledgements
The author was supported by Czech Science Foundation GAČR, grants 21-00420M and 22-11563O. I also thank the anonymous referee for helpful suggestions and corrections.
