Proof of Theorem 1

(i) Suppose that some four distinct algebraic conjugates of an algebraic number $\alpha $ of degree $d\in \{4,5,6,7\}$ satisfy the relation $\alpha _{1}+\alpha _{2}+\alpha _{3}+\alpha _{4}=0$ . The case $d=4$ is trivial in view of $tr(\alpha )=\alpha _{1}+\alpha _{2}+\alpha _{3}+\alpha _{4}$ . By Lemma 5, d cannot be 5 or 7. Let $d=6$ . Lemma 8 implies that $tr(\alpha )=0$ . Then, $\alpha _5+\alpha _6 = tr(\alpha )-(\alpha _{1}+\alpha _{2}+\alpha _{3}+\alpha _{4})=0$ . Hence, $\alpha _6=-\alpha _5$ . Let $p(x)$ be the minimal polynomial of $\alpha $ over $\mathbb {Q}$ . We have that $p(\alpha _6)=p(-\alpha _5)=0$ . Hence, $\alpha _5$ is a root of $p(-x)$ . Thus, $p(x)$ divides the polynomial $p(-x)$ . Since both polynomials $p(x)$ and $p(-x)$ are of the same degree and their constant terms coincide, we have that $p(-x)=p(x)$ . So $p(x)$ is of the form

$$\begin{align*}x^{6}+ax^{4}+bx^{2}+ c \in\mathbb{Q}[x]. \end{align*}$$

The converse is clear, since the roots $\alpha _1,\alpha _2,\dotsc , \alpha _6$ of such polynomial satisfy

$$\begin{align*}\alpha_{1}=-\alpha_{2},\quad \alpha_{3}=-\alpha_{4},\quad \alpha_{5}=-\alpha_{6}. \end{align*}$$

Thus, $\alpha _{1}+\alpha _{2}+\alpha _{3}+\alpha _{4}=0$ .

(ii) Suppose that some four distinct algebraic conjugates of an algebraic number $\alpha $ of degree $d\in \{4,5,6,7\}$ satisfy the relation $\alpha _{1}+\alpha _{2}+\alpha _{3}=\alpha _{4}$ . If $d=4$ , then, by Lemma 8, $tr(\alpha )=\alpha _{1}+\alpha _{2}+\alpha _{3}+\alpha _{4} = 0$ and we obtain $\alpha _4 + \alpha _4 = (\alpha _{1}+\alpha _{2}+\alpha _{3}) + \alpha _4 =0$ . A contradiction. By Lemma 5, d cannot be 5 or 7. Hence, $d=6$ . By Lemma 8, $tr(\alpha )=\alpha _{1}+\dotsb +\alpha _{6} = 0$ . Since $\alpha _{1}+\alpha _{2}+\alpha _{3}=\alpha _{4}$ , we have that

$$\begin{align*}0=\alpha_{1}+\dotsb+\alpha_{6} = 2\alpha_4+\alpha_5+\alpha_6, \end{align*}$$

which is impossible in view of Lemma 7.

Proof of Proposition 10

Let $\alpha $ be an algebraic number of degree $d=6$ such that $tr(\alpha )=0$ and some four distinct conjugates of $\alpha $ satisfy the relation

$$\begin{align*}\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}=:\beta. \end{align*}$$

Let

$$\begin{align*}\mathcal{S}:=\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}, \alpha_{6}\} \end{align*}$$

be the full set of algebraic conjugates of $\alpha $ . Then, in view of $tr(\alpha )=0$ , we have

$$\begin{align*}\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}=\beta,\quad \alpha_{5}+\alpha_{6} = -2\beta. \end{align*}$$

Let G be the Galois group of the normal closure of $\mathbb {Q}(\alpha _{1})$ over $\mathbb {Q}$ . The group G is determined (in a unique way) by its action on $\mathcal {S}$ : it corresponds to some transitive subgroup of the full symmetric group $S_{6}$ . First, consider the trivial case:

$$\begin{align*}\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}=a,\quad \alpha_{5}+\alpha_{6} = -2a, \end{align*}$$

where $a\in \mathbb {Q}$ . Select an automorphism $\phi \in G$ that maps $\alpha _{1}$ to $\alpha _{5}$ . Setting $\phi (\alpha _{2}) = \alpha _{k}$ , we obtain $\alpha _{5} + \alpha _{k} = a$ . We claim that $k=6$ . Indeed, if $1\leqslant k\leqslant 2$ , then $\alpha _{1}+\alpha _{2}=a$ together with $\alpha _{5} + \alpha _{k} = a$ imply $\alpha _{5}=\alpha _{1}$ or $\alpha _{5}=\alpha _{2}$ , which is impossible. Similarly, if $3\leqslant k\leqslant 4$ , then $\alpha _{3}+\alpha _{4}=a$ together with $\alpha _{5} + \alpha _{k} = a$ imply $\alpha _{5}=\alpha _{3}$ or $\alpha _{5}=\alpha _{4}$ , and we get another contradiction. Clearly, $k\neq 5$ , so the only option is $\alpha _{5} + \alpha _{6} = a$ . But we already know that $\alpha _{5}+\alpha _{6} = -2a$ . Thus, $a=-2a$ , meaning that $a=0$ .

Now assume that

$$\begin{align*}\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}=\beta,\quad \alpha_{5}+\alpha_{6} = -2\beta, \end{align*}$$

where $\beta \notin \mathbb {Q}$ . We will prove that $\beta _{1} := \beta $ is a cubic algebraic number.

Let us write all possible distinct expressions of $\beta _{1}$ in terms of $\alpha _{i}+\alpha _{j}$ (sum of two distinct $\alpha $ conjugates). Assume that there are exactly l distinct expressions (two expressions $\alpha _i+\alpha _j$ and $\alpha _u+\alpha _v$ are distinct if $\{i,j\}\neq \{u,v\}$ ):

$$ \begin{align*} \beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4} =\alpha_u+\alpha_v= \dots. \end{align*} $$

Notice that $l\geqslant 2$ , since the equality $\alpha _{1}+\alpha _{2} = \alpha _{3}+\alpha _{4}$ provides at least two distinct expressions of $\beta _{1}$ . We will show that $l=2$ . Indeed, assume that $l\geqslant 3$ . Then, we have at least three distinct expressions of $\beta _1$ as a sum of two distinct conjugates of $\alpha $ :

(6) $$ \begin{align} \beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4} =\alpha_u+\alpha_v. \end{align} $$

Then, $\{u,v\} =\{5,6\}$ . Indeed, if $u\in \{1,2,3,4\}$ , then (6) implies that $\alpha _v$ coincides with one of the conjugates $\alpha _1$ , $\alpha _2$ , $\alpha _3$ , $\alpha _4$ , which is impossible, since all three expressions in (6) are distinct. Similarly, $v\in \{1,2,3,4\}$ also leads to a contradiction. Hence, ${\{u,v\}=\{5,6\}}$ . So (6) becomes

$$\begin{align*}\beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4} =\alpha_5+\alpha_6. \end{align*}$$

Adding all these expressions of $\beta _1$ , we obtain

$$\begin{align*}3\beta_{1} = \alpha_{1}+\alpha_{2} + \alpha_{3}+\alpha_{4} + \alpha_{5}+\alpha_{6} = 0 , \end{align*}$$

which is impossible in view of $\beta _{1}\notin \mathbb {Q}$ .

Now, we have that $l=2$ and

(7) $$ \begin{align} \beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4}. \end{align} $$

Next, we will obtain an upper bound for $\deg (\beta _{1})$ . Note that by acting on (7) with an appropriate automorphism from $G,$ we can obtain expressions of the form (7) for every algebraic conjugate of $\beta _1$ :

(8) $$ \begin{align} \beta_{1} &= \alpha_{1}\phantom{a}+\alpha_{2}\;\; = \alpha_{3}\phantom{a}+\alpha_{4}, \nonumber\\ \beta_{2} &= \alpha_{i_{21}}+\alpha_{i_{22}} = \alpha_{i_{23}}+\alpha_{i_{24}}, \\ &= \dotsb\;\;\;\dotsb \nonumber\\ \beta_t &= \alpha_{i_{t1}}+\alpha_{i_{t2}} = \alpha_{i_{t3}}+\alpha_{i_{t4}}.\nonumber \end{align} $$

Here, t is the degree of $\beta _1$ and $\beta _1,\beta _2,\dotsc , \beta _t$ are the algebraic conjugates of $\beta _1$ . We have precisely $2\cdot t$ distinct expressions of the form $\alpha _i+\alpha _j$ in (8), since there are exactly t algebraic conjugates of $\beta _1$ and every such conjugate has exactly two expressions. On the other hand, since $\deg (\alpha )=6$ , we have at most $\binom {6}{2} = 15$ possible pairs of indices for distinct expressions $\alpha _{i}+\alpha _{j}$ . Hence, $ 2t\leqslant 15$ and $t=\deg (\beta _1)\leqslant 7$ .

Next, we will show that, in fact, $\deg (\beta _{1})$ is divisible by $3$ . Indeed, in (8), there are $2t$ distinct expressions of conjugates of $\beta _{1}$ as sums $\alpha _{i}+\alpha _{j}$ . Each such sum contains two conjugates of $\alpha $ . Hence, there are exactly $2\cdot 2t$ appearances of conjugates of $\alpha $ in (8). On the other hand, since G is transitive on the set of algebraic conjugates of $\alpha $ , each $\alpha _{i}$ must appear the same number of times in (8). Suppose that every $\alpha _{i}$ appears exactly k times in (8). So we have exactly $k\cdot \deg (\alpha )=6k$ appearances of conjugates of $\alpha $ in (8). Hence, $4t=6k$ , and therefore t is divisible by 3. Recall that $t=\deg (\beta _1)\leqslant 7$ . So $\deg (\beta _1)=3$ or 6.

Finally, we will show that $\deg (\beta _{1})\neq 6$ . Indeed, assume that $\deg (\beta _{1})=6$ . Since each conjugate of $\beta _{1}$ has exactly two distinct expressions of the form $\alpha _{i}+\alpha _{j}$ , we obtain $6\cdot 2=12$ distinct expressions. Recall that there are at most $\binom {6}{2} = 15$ possible pairs of indices for distinct expressions $\alpha _{i}+\alpha _{j}$ and also

$$\begin{align*}-2\beta_{1} = -(\alpha_{1}+\alpha_{2}) - (\alpha_{3}+\alpha_{4}) = \alpha_{5}+\alpha_{6}. \end{align*}$$

By applying all automorphisms from G to $-2\beta _{1} = \alpha _{5}+\alpha _{6}$ , we get at least $\deg (\beta _{1})=6$ expressions of the form $\alpha _{i}+\alpha _{j}$ for algebraic conjugates of $-2\beta _{1}$ . These expressions must be distinct from each other and from the $12$ expressions that we already have. Note that there is no pair of indices $(i, j)$ such that $\beta _{i} = -2\beta _{j}$ . Indeed, let $q(x)$ be the minimal polynomial of the $\beta _1,\beta _2,\dotsc , \beta _t$ and assume that $\beta _{i} = -2\beta _{j}$ . In such a case, $q(x)$ and $(-2)^{t}q(-\frac {x}{2})$ would be the same polynomial. This implies that either $t=0$ or $q(x)=x^{t}$ , which are both impossible. But in that case, there would be

$$\begin{align*}12+6 = 18 \end{align*}$$

distinct expressions $\alpha _{i}+\alpha _{j}$ . A contradiction, since there are at most $15$ distinct such expressions. Hence, $\deg (\beta _{1})\neq 6$ , and the only possibility is $\deg (\beta _{1})=3$ .

Proof of Proposition 11

Let $\alpha $ be an algebraic number of degree $d=6$ and ${tr(\alpha )=0}$ . Assume that some four distinct algebraic conjugates of $\alpha $ satisfy the relation

(9) $$ \begin{align} \alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}=:\beta\not\in\mathbb{Q}. \end{align} $$

Then, by Proposition 10, $\beta $ is a cubic algebraic number and $tr(\beta )=0$ . Let $\beta _1=\beta ,\beta _2,\beta _3$ be the algebraic conjugates of $\beta $ . Let G be the Galois group of the normal closure of $\mathbb {Q}(\alpha )$ over $\mathbb {Q}$ . Consider G as a subgroup of $S_6$ , acting on the indices of the conjugates $\alpha _1,\alpha _2,\dotsc ,\alpha _6$ of $\alpha $ . Take two automorphisms of G such that one maps $\beta _1$ to $\beta _2$ and another maps $\beta _1$ to $\beta _3$ . Acting with these automorphisms on (9), we obtain

(10) $$ \begin{align} \begin{cases} \beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4},\\ \beta_{2} = \alpha_{i_{21}}+\alpha_{i_{22}} = \alpha_{i_{23}}+\alpha_{i_{24}},\\ \beta_{3} = \alpha_{i_{31}}+\alpha_{i_{32}} = \alpha_{i_{33}}+\alpha_{i_{34}}, \end{cases} \end{align} $$

where each $\alpha _{i_{kl}}$ is an algebraic conjugate of $\alpha $ . In view of $tr(\alpha )=0$ , we also obtain corresponding relations

(11) $$ \begin{align} \begin{cases} -2\beta_{1} = \alpha_{5}+\alpha_{6},\\ -2\beta_{2} = \alpha_{i_{25}}+\alpha_{i_{26}},\\ -2\beta_{3} = \alpha_{i_{35}}+\alpha_{i_{36}}.\\ \end{cases} \end{align} $$

Note that for every $k=2,3$ the numbers $i_{k1},i_{k2}, i_{k3}, i_{k4}, i_{k5}, i_{k6}$ are distinct. We will specify the indices in (10) and (11) by relabeling the conjugates $\alpha _1,\alpha _2,\dotsc ,\alpha _6$ , if necessary. First, we will prove that each $-2\beta _{k}$ has a unique expression in terms of $\alpha _i+\alpha _j$ (recall that two expressions $\alpha _i+\alpha _j$ and $\alpha _u+\alpha _v$ are distinct if $\{i,j\}\neq \{u,v\}$ ). Indeed, say, $-2\beta _{1}$ has two distinct expressions:

(12) $$ \begin{align} -2\beta_{1} = \alpha_{5}+\alpha_{6} = \alpha_{u}+\alpha_{v}. \end{align} $$

If $u=5$ (or $u=6$ ), then by (12), $v=6$ (or $v=5$ , respectively). In this scenario, the expressions $\alpha _{u}+\alpha _{v}$ and $\alpha _{5}+\alpha _{6}$ become identical. This implies that $u\notin \{5,6\}$ . A similar argument shows that $v\notin \{5,6\}$ . Consequently, $u, v\in \{1, 2, 3, 4\}$ . Without loss of generality, let $u=1$ . We then examine the following cases for $(u,v)$ :

1. Case 1: If $(u,v)=(1,1)$ , then $\alpha _{5}+\alpha _{6} = 2\alpha _{1}$ , which contradicts Lemma 6.

2. Case 2: If $(u,v)=(1,2)$ , then equations (10) and (12) yield $-2\beta _{1}=\beta _{1}$ implying $\beta _{1}=0$ . This contradicts the condition that $\beta _{1}\notin \mathbb {Q}$ .

3. Case 3: If $(u,v)=(1,3)$ , then we have $\beta _{1} = \alpha _{1}+\alpha _{2}$ and $-2\beta _{1} = \alpha _{1} + \alpha _{3}$ . Substituting the first into the second gives $3\alpha _1+2\alpha _2+\alpha _3=0$ , which contradicts Lemma 7.

4. Case 4: Similarly, $(u,v)\neq (1,4)$ .

These cases imply that $u\notin \{1, 2, 3, 4\}$ . Therefore, $-2\beta _{1}$ has a unique expression in terms of $\alpha _i+\alpha _j$ . Since $\beta _1,\beta _2,\beta _3$ are algebraic conjugates of $\beta _1$ , it follows that every $-2\beta _{k}$ (for $k=1,2,3$ ) also has a unique expression in terms of $\alpha _i+\alpha _j$ .

Now, we will prove that all the $\alpha $ ’s appear exactly once in (11). Indeed, note that each conjugate of $\alpha $ appears exactly three times counting both (10) and (11): the set $\{\alpha _1,\dotsc ,\alpha _6\}, \{\alpha _{i_{21}},\dotsc , \alpha _{i_{26}}\},$ and $\{\alpha _{i_{31}},\dotsc , \alpha _{i_{36}}\}$ make three copies of the set of conjugates of $\alpha $ , and since each appears exactly twice in (10) (as was proven in the proof of Proposition 10), there must be one full set of conjugates in (11).

We have that $\{\alpha _{i_{25}}, \alpha _{i_{26}}, \alpha _{i_{35}}, \alpha _{i_{36}}\}=\{\alpha _{1}, \alpha _{2}, \alpha _{3}, \alpha _{4}\}$ . Without loss of generality, we can assume that $\alpha _{i_{25}} = \alpha _{1}$ . If $\alpha _{i_{26}} = \alpha _{2}$ , then $-2\beta _2=\beta _1$ . Substituting this expression of $\beta _1$ into $tr(\beta )=\beta _1+\beta _2+\beta _3=0$ yields $\beta _2 = \beta _3$ , which is impossible. Hence, $\alpha _{i_{26}}\in \{\alpha _{3}, \alpha _{4}\}$ . Note that $\alpha _{3}$ and $\alpha _{4}$ appear symmetrically in the first equation of (10). Without loss of generality, by relabeling $\alpha _{3}$ and $\alpha _{4}$ , if necessary, we can assume that $\alpha _{i_{26}} = \alpha _{4}$ . From this, we immediately derive that $\{\alpha _{i_{21}}, \alpha _{i_{22}}, \alpha _{i_{23}}, \alpha _{i_{24}}\}=\{\alpha _{2}, \alpha _{3}, \alpha _{5}, \alpha _{6}\}$ . Note that $\beta _{2}\neq \alpha _{5}+\alpha _{6}$ . Indeed, if $\beta _{2}=\alpha _{5}+\alpha _{6}$ , then $\beta _{2}=-2\beta _{1}$ . Substituting this expression of $\beta _2$ into $tr(\beta )=\beta _1+\beta _2+\beta _3=0$ yields $\beta _1 = \beta _3$ , which is impossible. Thus, $\alpha _{5}$ and $ \alpha _{6}$ appear in distinct expressions of $\beta _{2}$ in (10), as well as $\alpha _{2}$ and $\alpha _{3}$ . Without loss of generality, we can assume that $\alpha _{i_{21}} = \alpha _{2}$ and $\alpha _{i_{23}} = \alpha _{3}$ . Since $\alpha _{5}$ and $\alpha _{6}$ appear symmetrically in the first equation of (11), by relabeling $\alpha _{5}$ and $\alpha _{6}$ , if necessary, we can assume that $\alpha _{i_{22}} = \alpha _{5}$ and $\alpha _{i_{24}} = \alpha _{6}$ . So far, we have obtained

$$ \begin{align*} \begin{cases} \beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4},\\ \beta_{2} = \alpha_{2}+\alpha_{5} = \alpha_{3}+\alpha_{6},\\ \beta_{3} = \alpha_{i_{31}}+\alpha_{i_{32}} = \alpha_{i_{33}}+\alpha_{i_{34}}, \end{cases} \begin{cases} -2\beta_{1} = \alpha_{5}+\alpha_{6},\\ -2\beta_{2} = \alpha_{1}+\alpha_{4},\\ -2\beta_{3} = \alpha_{i_{35}}+\alpha_{i_{36}}.\\ \end{cases} \end{align*} $$

Since $\{\alpha _{i_{35}}, \alpha _{i_{36}}\}=\{\alpha _{2}, \alpha _{3}\}$ , without loss of generality, we assume that $\alpha _{i_{35}} = \alpha _{2}$ and $\alpha _{i_{36}} = \alpha _{3}$ . Then, $\{\alpha _{i_{31}}, \alpha _{i_{32}}, \alpha _{i_{33}}, \alpha _{i_{34}}\}=\{\alpha _{1}, \alpha _{4}, \alpha _{5}, \alpha _{6}\}$ . Note that $\beta _{3}\neq \alpha _{5}+\alpha _{6}$ . Indeed, if $\beta _{3}=\alpha _{5}+\alpha _{6}$ , then $\beta _{3}=-2\beta _{1}$ . Substituting this expression of $\beta _3$ into $tr(\beta )=\beta _1+\beta _2+\beta _3=0$ yields $\beta _1 = \beta _2$ , which is impossible. Therefore, $\alpha _{5}$ and $\alpha _{6}$ appear in distinct expressions of $\beta _{3}$ in (10), as well as $\alpha _{1}$ and $\alpha _{4}$ . Thus, without loss of generality, we can assume that $\alpha _{i_{31}} = \alpha _{1}$ and $\alpha _{i_{33}} = \alpha _{4}$ . Now, we have two possible cases:

$$\begin{align*}\beta_{3} = \alpha_{1}+\alpha_{5} = \alpha_{4}+\alpha_{6}\;\; \text{or}\;\; \beta_{3} = \alpha_{1}+\alpha_{6} = \alpha_{4}+\alpha_{5}. \end{align*}$$

The first case is impossible. Indeed, by adding $\beta _{1} = \alpha _{1}+\alpha _{2}, \beta _{2} = \alpha _{2}+\alpha _{5}$ and $\beta _{3} = \alpha _{1}+\alpha _{5}$ , we obtain

$$\begin{align*}0 = \beta_{1} + \beta_{2} + \beta_{3} = 2(\alpha_{1}+ \alpha_{2}+\alpha_{5})=2(\beta_{1}+\alpha_{5}), \end{align*}$$

and hence $\beta _{1}=-\alpha _{5}$ , which is impossible, since $6=\deg (-\alpha _{5})\neq \deg (\beta _{1})=3$ . Finally, we can rewrite equations (10) and (11) as follows:

(13) $$ \begin{align} \begin{cases} \beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4},\\ \beta_{2} = \alpha_{2}+\alpha_{5} = \alpha_{3}+\alpha_{6},\\ \beta_{3} = \alpha_{1}+\alpha_{6} = \alpha_{4}+\alpha_{5}, \end{cases} \end{align} $$

(14) $$ \begin{align} \begin{cases} -2\beta_{1} = \alpha_{5}+\alpha_{6},\\ -2\beta_{2} = \alpha_{1}+\alpha_{4},\\ -2\beta_{3} = \alpha_{2}+\alpha_{3}.\\ \end{cases} \end{align} $$

Now, we will prove that the Galois group G is of order $|G| = 6$ or $12$ . Since G is transitive, the Orbit–Stabilizer Theorem implies

$$\begin{align*}|G| = |\mathrm{Orb}(\alpha_{1})|\cdot |\mathrm{Stab}(\alpha_{1})| = 6\cdot |\mathrm{Stab}(\alpha_{1})|, \end{align*}$$

where $\mathrm{Stab}(\alpha _1) = \{\psi \in G:\psi (\alpha _1)=\alpha _1\}$ . Hence, if we can show that $|\mathrm{Stab}(\alpha _{1})|\leqslant 2$ , then it follows that $|G|=6$ or 12. Let $\psi \in \mathrm{Stab}(\alpha _{1})$ .

If $\psi $ stabilizes $\alpha _{2}$ , then, in view of $-2\beta _{2} = \alpha _{1}+\alpha _{4}$ and $-2\beta _{3} = \alpha _{2}+\alpha _{3}$ , $\psi $ stabilizes $\beta _2$ , $\alpha _3,$ and $\alpha _4$ . Since $\beta _2=\alpha _2+\alpha _5$ , it follows that $\psi $ stabilizes $\alpha _5$ . Hence, $\psi $ stabilizes every conjugate of $\alpha $ . Therefore, $\psi =id$ .

Suppose that $\psi (\alpha _2)\neq \alpha _2$ . Since $\psi (\alpha _1)=\alpha _1$ and every conjugate of $\alpha $ appears exactly once in (14), it follows that $\psi $ stabilizes $\beta _2$ and $\alpha _4$ . Moreover, from (13), we obtain that $\psi $ maps $\beta _1=\alpha _1+\alpha _2$ to $\beta _3=\alpha _1+\alpha _6$ and vice versa. Hence, $\psi (\alpha _2)=\alpha _6$ and $\psi (\alpha _6)=\alpha _2$ . Furthermore, $\psi $ maps $-2\beta _1=\alpha _5+\alpha _6$ to $-2\beta _3=\alpha _2+\alpha _3$ and vice versa. So $\psi (\alpha _3)=\alpha _5$ and $\psi (\alpha _5)=\alpha _3$ . Hence, $\psi =(2\,6)(3\,5)\in S_6$ .

We have proved that if every $\psi \in \mathrm{Stab}(\alpha _1)$ stabilizes $\alpha _2$ , then the stabilizer subgroup $\mathrm{Stab}(\alpha _1)$ is trivial and G has order 6. If there exists $\psi \in \mathrm{Stab}(\alpha _1)$ which does not stabilize $\alpha _2$ , then $\mathrm{Stab}(\alpha _1)=\{id,(2\,6)(3\,5)\}$ and accordingly G has order $6\cdot 2=12$ . Hence, if G has order 12, then necessarily $(2\,6)(3\,5)\in G$ .

Next, we will prove that the group G contains the permutation $\sigma := (1\ 3\ 5) (2\ 6\ 4) \in S_6$ . Indeed, since $\beta _{1}$ is a cubic algebraic number, the Galois group G contains an element, denote it by $\varphi $ , that permutes the conjugates of $\beta _{1}$ . Without loss of generality, we can assume that

$$\begin{align*}\varphi(\beta_{1})=\beta_{2},\quad \varphi(\beta_{2})=\beta_{3},\quad \varphi(\beta_{3})=\beta_{1} \end{align*}$$

(by exchanging $\varphi $ with $\varphi ^2$ , if necessary).

Relations in (14) imply that $\varphi $ maps $\{\alpha _{1}, \alpha _{4}\}$ to $\{\alpha _{2}, \alpha _{3}\}$ . Consider two possible cases: $\varphi (\alpha _{1})=\alpha _{2}$ and $\varphi (\alpha _{1})=\alpha _{3}$ .

If $\varphi (\alpha _{1})=\alpha _{2}$ , then $\varphi (\alpha _{4})=\alpha _{3}$ . The expressions of $\beta _{1}$ and $\beta _{2}$ in (13) imply that $\varphi $ maps $\{\alpha _{1}, \alpha _{2}\}$ to $\{\alpha _{2}, \alpha _{5}\}$ . Since $\varphi (\alpha _{1})=\alpha _{2}$ , we obtain $\varphi (\alpha _{2})=\alpha _{5}$ . Similarly, we see that $\varphi $ maps $\{\alpha _{3}, \alpha _{4}\}$ to $\{\alpha _{3}, \alpha _{6}\}$ . Since $\varphi (\alpha _{4})=\alpha _{3}$ , we derive $\varphi (\alpha _{3})=\alpha _{6}$ . The expressions of $\beta _{2}$ and $\beta _{3}$ in (13) imply that $\varphi $ maps $\{\alpha _{2}, \alpha _{5}\}$ to $\{\alpha _{4}, \alpha _{5}\}$ . Since $\varphi (\alpha _{2})=\alpha _{5}$ , we obtain $\varphi (\alpha _{5})=\alpha _{4}$ . We are left with only one option for $\varphi (\alpha _6)$ , i.e., $\varphi (\alpha _6)=\alpha _1$ . Hence, $\varphi = (1\ 2\ 5\ 4\ 3\ 6)$ . Note that $\varphi ^4=(1\ 3\ 5)(2\ 6\ 4)=\sigma $ . So that in this case ( $\varphi (\alpha _1)=\alpha _2$ ) the permutation $\sigma $ is contained in G.

If $\varphi (\alpha _{1})=\alpha _{3}$ , then $\varphi (\alpha _{4})=\alpha _{2}$ . The expressions of $\beta _{1}$ and $\beta _{2}$ in (13) imply that $\varphi $ maps $\{\alpha _{1}, \alpha _{2}\}$ to $\{\alpha _{3}, \alpha _{6}\}$ . Since $\varphi (\alpha _{1})=\alpha _{3}$ , we have $\varphi (\alpha _{2})=\alpha _{6}$ . Similarly, we see that $\varphi $ maps $\{\alpha _{3}, \alpha _{4}\}$ to $\{\alpha _{2}, \alpha _{5}\}$ . Since $\varphi (\alpha _{4})=\alpha _{2}$ , we get $\varphi (\alpha _{3})=\alpha _{5}$ . The expressions of $\beta _{2}$ and $\beta _{3}$ in (13) imply that $\varphi $ maps $\{\alpha _{2}, \alpha _{5}\}$ to $\{\alpha _{1}, \alpha _{6}\}$ . Since $\varphi (\alpha _{2})=\alpha _{6}$ , we obtain $\varphi (\alpha _{5})=\alpha _{1}$ . We are left with only one option for $\varphi (\alpha _6)$ , i.e., $\varphi (\alpha _6)=\alpha _4$ . Hence, $\varphi = (1\ 3\ 5)(2\ 6\ 4)=\sigma $ .

We have proved that the group G contains the permutation $\sigma = (1\ 3\ 5)(2\ 6\ 4)$ . Now, we are in a position to find all possible groups G.

A simple computation with SageMath [14] shows that there is a unique transitive subgroup of $S_6$ which has order 12 and contains permutations $(2\,6)(3\,5)$ and $\sigma =(1\ 3\ 5)(2\ 6\ 4)$ . This subgroup is generated by the permutations $\tau = (1\,2)(3\,4)(5\,6)$ and $\pi = (1\, 2\, 5\, 4\, 3\, 6)$ and is isomorphic to the dihedral group $D_6$ of order 12.

Similarly, there are exactly four transitive subgroups of $S_6$ which have order 6 and contain the permutation $\sigma =(1\ 3\ 5)(2\ 6\ 4)$ . These are

• • $G_1=\langle \sigma ,\tau \rangle $ isomorphic to $S_3$ , where $\tau = (1\,2)(3\,4)(5\,6)$ ;

• • $G_2=\langle \pi \rangle $ isomorphic to the cyclic group $C_6$ , where $\pi =(1\, 2\, 5\, 4\, 3\, 6)$ ;

• • $G_3=\langle (1\,6\,5\,2\,3\,4) \rangle $ isomorphic to the cyclic group $C_6$ ;

• • $G_4=\langle (1\,4\,5\,6\,3\,2) \rangle $ isomorphic to the cyclic group $C_6$ .

Note that the groups $G_3$ and $G_4$ do not preserve the relations in (14). Indeed, the generator of $G_3$ maps $\alpha _1+\alpha _4$ to $\alpha _6+\alpha _1$ while the generator of $G_4$ maps $\alpha _1+\alpha _4$ to $\alpha _4+\alpha _5$ . Hence, $G\neq G_3$ and $G\neq G_4$ .

We have proved that there are three options for the group G:

1. (1) $G=\langle \tau , \pi \rangle \cong D_{6}$ ;

2. (2) $G=\langle \sigma ,\tau \ \rangle \cong S_3$ ;

3. (3) $G = \langle \pi \ \rangle \cong C_6$ .

This completes the proof of Proposition 11.

Proof of Theorem 3

Necessity. Suppose that $\alpha $ is an algebraic number of degree 6 whose four distinct algebraic conjugates satisfy the relation $\alpha _{1}+\alpha _{2}=\alpha _{3}+\alpha _{4}=: \beta \notin \mathbb {Q}$ . Note that for any $r\in \mathbb {Q,}$ the number $\alpha -r$ will also have this property. Moreover, $\alpha $ equals the sum of a quadratic and a cubic algebraic number if and only if $\alpha -r$ has the same property. Hence, by taking $r=tr(\alpha )/6$ , we can assume that $\alpha $ has trace zero, $tr(\alpha )=0$ . Let G be the Galois group of the normal closure of $\mathbb {Q}(\alpha )$ over $\mathbb {Q}$ . Consider G as a subgroup of $S_6$ , acting on the indices of the conjugates $\alpha _1,\alpha _2,\dotsc ,\alpha _6$ of $\alpha $ . Then, by Proposition 11, we have the following:

1. (i) $\beta $ is a cubic algebraic number.

2. (ii) One can label the algebraic conjugates $\alpha _1,\alpha _2,\dotsc ,\alpha _6$ of $\alpha $ in such a way that these satisfy the relations

   (15) $$ \begin{align} \begin{cases} \beta_{1} = \alpha_{1}+\alpha_{2} = \alpha_{3}+\alpha_{4},\\ \beta_{2} = \alpha_{2}+\alpha_{5} = \alpha_{3}+\alpha_{6},\\ \beta_{3} = \alpha_{1}+\alpha_{6} = \alpha_{4}+\alpha_{5}, \end{cases} \end{align} $$
   where $\beta _1=\beta ,\beta _2,\beta _3$ are the algebraic conjugates of $\beta $ .

3. (iii) Given the relations (15), there are exactly three options for the Galois group G:

   1. (1) $G=\langle \tau , \pi \ |\ \tau ^{2}=\pi ^{6}=id,\ \tau \pi \tau = \pi ^{5}\rangle \cong D_{6}$ ;

   2. (2) $G=\langle \pi \ |\ \pi ^{6}=id\rangle \cong C_{6}$ ;

   3. (3) $G = \{id, \sigma , \sigma ^{2}, \tau , \tau \sigma , \tau \sigma ^{2}\}\cong S_3$ .

Here, $\pi =(1\, 2\, 5\, 4\, 3\, 6)$ , $\sigma =(1\ 3\ 5)(2\ 6\ 4)$ and $\tau = (1\,2)(3\,4)(5\,6)$ .

Consider the number $\alpha _1-\alpha _4$ . The expression of $\beta _1$ in (15) implies that $\alpha _2-\alpha _3=-(\alpha _1-\alpha _4)$ . Moreover,

$$ \begin{align*} \tau(\alpha_1-\alpha_4) &= \alpha_2-\alpha_3=-(\alpha_1-\alpha_4),\\ \pi(\alpha_1-\alpha_4) &= \alpha_2-\alpha_3=-(\alpha_1-\alpha_4),\\ \sigma(\alpha_1-\alpha_4) &= \alpha_3-\alpha_2=\alpha_1-\alpha_4. \end{align*} $$

Hence, $\alpha _1-\alpha _4$ is a quadratic algebraic number. Moreover, the relations $\beta _2=\alpha _{2}+\alpha _{5} = \alpha _{3}+\alpha _{6}$ together with $tr(\alpha )=0$ imply $-2\beta _2 = \alpha _1+\alpha _4$ , which is equivalent to

$$\begin{align*}\alpha_1 = \frac{\alpha_1-\alpha_4}{2} - \beta_2. \end{align*}$$

Consequently, $\alpha =\alpha _1$ is the sum of the quadratic algebraic number $(\alpha _1-\alpha _4)/2$ and the cubic algebraic number $-\beta _2$ .

Sufficiency. Assume that $\alpha $ is the sum of a quadratic algebraic number $\gamma $ and a cubic algebraic number $\delta $ . We will prove that $\alpha $ has degree 6 and some four distinct algebraic conjugates of $\alpha $ satisfy the relation $\alpha _{1}+\alpha _{2}=\alpha _{3}+\alpha _{4}\notin \mathbb {Q}$ . Indeed, let $\gamma _1=\gamma $ , $\gamma _2$ be the algebraic conjugates of $\gamma $ and let $\delta _1=\delta $ , $\delta _2$ , $\delta _3$ be the algebraic conjugates of $\delta $ . Since the compositum $\mathbb {Q}(\gamma ,\delta )$ contains $\gamma $ and $\delta $ of degree 2 and 3, respectively, it follows that the degree of $\mathbb {Q}(\gamma ,\delta )$ is divisible by $2\cdot 3=6$ . On the other hand, $[\mathbb {Q}(\gamma ,\delta ):\mathbb {Q}]\leqslant [\mathbb {Q}(\gamma ):\mathbb {Q}]\cdot [\mathbb {Q}(\delta ):\mathbb {Q}] = 2\cdot 3=6$ . Hence, $[\mathbb {Q}(\gamma ,\delta ):\mathbb {Q}]=6$ . By Lemma 9, we obtain that $\mathbb {Q}(\gamma ,\delta ) = \mathbb {Q}(\gamma +\delta )$ . Therefore, $\alpha =\gamma +\delta $ has degree 6. Hence, the numbers ${\gamma _i+\delta _j}$ , for $i=1,2$ and $j=1,2,3$ , are distinct algebraic conjugates of $\gamma +\delta $ . The identity

$$\begin{align*}(\gamma_1+\delta_1) + (\gamma_2+\delta_2) = (\gamma_1+\delta_2) + (\gamma_2+\delta_1) \end{align*}$$

implies that four distinct algebraic conjugates of $\alpha =\gamma +\delta $ satisfy

$$\begin{align*}\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}, \end{align*}$$

where $\alpha _1=\gamma _1+\delta _1$ , $\alpha _2=\gamma _2+\delta _2$ , $\alpha _3=\gamma _1+\delta _2,$ and $\alpha _4=\gamma _2+\delta _1$ . Finally, since ${tr(\gamma )=\gamma _1+\gamma _2}$ and $tr(\delta )=\delta _1+\delta _2+\delta _3$ are rational numbers, we obtain that

$$\begin{align*}\alpha_1+\alpha_2 = \gamma_1+\delta_1 + \gamma_2+\delta_2 = tr(\gamma) + tr(\delta) - \delta_3 \end{align*}$$

is a cubic algebraic number, and therefore $\alpha _1+\alpha _2\notin \mathbb {Q}$ .

Proof of Theorem 2

Let $\alpha $ be an algebraic number of degree $d\in \{4,5,6,7\}$ such that $tr(\alpha )=0$ . Suppose that four distinct conjugates of $\alpha _{1} := \alpha $ satisfy the relation

(16) $$ \begin{align} \alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}. \end{align} $$

By Lemma 5, $d\neq 5$ and $d\neq 7$ . Hence, $d=4$ or 6.

$(i)$ Suppose that $d=4$ . Then, (16) together with $tr(\alpha )=0$ imply $\alpha _{1}+\alpha _{2}=\alpha _{3}+\alpha _{4}=0$ . Hence, $\alpha _2=-\alpha _1$ , and therefore the minimal polynomial $p(x)$ of $\alpha $ is of the form $p(x)=x^4+ax^{2}+b\in \mathbb {Q}[x]$ .

Conversely, let $p(x)=x^4+ax^{2}+b\in \mathbb {Q}[x]$ be an irreducible polynomial. Let $\beta ,\gamma \in \mathbb {C}$ be two distinct roots of $p(x)$ such that $\gamma \neq -\beta $ . Then $\alpha _1 = \beta $ , $\alpha _2 = -\beta $ , $\alpha _3 = \gamma ,$ and $\alpha _4 = -\gamma $ are all the roots of $p(x)$ and the relation (16) holds.

$(ii)$ Suppose that $d=6$ and the sum $\alpha _{1}+\alpha _{2}$ in (16) is a rational number. Then, by Proposition 10, $\alpha _{1}+\alpha _{2} = \alpha _{3}+\alpha _{4}=0$ . This implies that $\alpha _2=-\alpha _1$ . Therefore, the minimal polynomial $p(x)$ of $\alpha $ is of the form $p(x)=x^6+ax^{4}+bx^2+c\in \mathbb {Q}[x]$ . Similarly, as in case $(ii)$ , we see that some four distinct roots of any such irreducible polynomial satisfy the relation (16).

$(iii)$ Suppose that $d=6$ and the sum $\alpha _{1}+\alpha _{2}$ in (16) is not a rational number. Then, by Theorem 3, $\alpha $ is a sum of a quadratic algebraic number $\gamma $ and a cubic algebraic number $\delta $ . Let $\gamma _1=\gamma $ , $\gamma _2$ be the algebraic conjugates of $\gamma $ and let $\delta _1=\delta $ , $\delta _2$ , $\delta _3$ be the algebraic conjugates of $\delta $ . Then, the numbers $\gamma _i+\delta _j$ , for $i=1,2$ and $j=1,2,3$ , are the algebraic conjugates of $\alpha =\gamma +\delta $ . We have that $tr(\alpha )=0$ . On the other hand, $tr(\alpha )$ equals the sum of all the numbers $\gamma _i+\delta _j$ , for $i=1,2$ and $j=1,2,3$ . The later sum equals $3(\gamma _1+\gamma _2)+2(\delta _1+\delta _2+\delta _3)$ . Hence, $0=tr(\alpha ) = 3tr(\gamma )+2tr(\delta )$ . Therefore, $tr(\gamma )/2+tr(\delta )/3=0$ and we can represent $\alpha $ as

$$\begin{align*}\alpha = \left(\gamma-\frac{tr(\gamma)}{2}\right) + \left(\delta-\frac{tr(\delta)}{3}\right). \end{align*}$$

Note that $\gamma -tr(\gamma )/2$ and $\delta -tr(\delta )/3$ are quadratic and cubic algebraic numbers, respectively, both having trace zero. Consequently, without loss of generality, we can assume that $tr(\gamma )=0$ and $tr(\delta )=0$ in the expression $\alpha =\gamma +\delta $ . Then, the minimal polynomial of $\gamma $ is of the form $x^2-a$ and the minimal polynomial of $\delta $ is of the form $R(x)=x^3+bx+c$ , where $a,b,c\in \mathbb {Q}$ . Moreover, in view of $tr(\gamma )=0$ and $\gamma _1\gamma _2=-a$ , we obtain that $\gamma _1=\pm \sqrt {a}$ and $\gamma _2=\mp \sqrt {a}$ . Now, the minimal polynomial $p(x)$ of $\alpha $ can be expressed as

(17) $$ \begin{align} p(x) &= \prod_{\substack{i=1,2\\j=1,2,3}} (x-\gamma_i-\delta_j) = \prod_{i=1,2} (x-\gamma_i-\delta_1)(x-\gamma_i-\delta_2)(x-\gamma_i-\delta_3)\nonumber\\ &= R(x-\gamma_1)R(x-\gamma_2) = R(x-\sqrt{a})R(x+\sqrt{a})\nonumber\\ &=x^{6} + (2 b - 3 a) x^{4} + 2 c x^{3} + (3 a^{2} + b^{2}) x^{2} + 2c(3 a + b) x\\ &\phantom{=}\,\,- a^{3} - 2 a^{2} b - a b^{2} + c^{2}.\nonumber \end{align} $$

Conversely, given an irreducible polynomial $p(x)$ of the form (17), we can factor it as $p(x)=R(x-\sqrt {a})R(x+\sqrt {a})$ , where $R(x)=x^3+bx+c$ . Note that $\sqrt {a}\notin \mathbb {Q}$ , since $p(x)$ is irreducible. Consequently, $\sqrt {a}$ is a quadratic algebraic number. Moreover, $R(x)$ is irreducible. Indeed, if $R(x)$ factors as $R(x)=P(x)Q(x)$ with some polynomials $P(x),Q(x)\in \mathbb {Q}[x]$ both of degree $\geqslant 1$ , then

$$\begin{align*}p(x) = R(x-\sqrt{a})R(x+\sqrt{a}) = P(x-\sqrt{a})P(x+\sqrt{a})Q(x-\sqrt{a})Q(x+\sqrt{a}) \end{align*}$$

with both polynomials $P(x-\sqrt {a})P(x+\sqrt {a})$ and $Q(x-\sqrt {a})Q(x+\sqrt {a})$ having rational coefficients. This contradicts the assumption that $p(x)$ is irreducible. Hence, $R(x)$ is irreducible. Finally, the factorization $p(x)=R(x-\sqrt {a})R(x+\sqrt {a})$ implies that every root of $p(x)$ is a sum of a quadratic algebraic number $\pm \sqrt {a}$ and a root of $R(x)$ , which is a cubic algebraic number. Therefore, by Theorem 3, some four distinct roots of $p(x)$ satisfy the relation (16) with $\alpha _1+\alpha _2\notin \mathbb {Q}$ .