The product of a quartic and a sextic number cannot be octic

1 Introduction

In the study by Drungilas et al. [1] a curious version of an a b c -problem for sums and products of algebraic numbers has been introduced. (It has nothing to do with the a b c -conjecture of Masser and Oesterlé.) According to the definitions in [1], a triplet ( a , b , c ) ∈ N 3 is called

• sum-feasible if there exist algebraic numbers α and β of degrees a and b (over Q ), respectively, such that the degree of α + β is c ,

• product-feasible if there exist algebraic numbers α and β of degrees a and b (over Q ), respectively, such that the degree of α ⋅ β is c ,

• compositum-feasible if there exist number fields K and L of degrees a and b (over Q ), respectively, such that the degree of their compositum K L is c .

It is clear that the first two definitions are symmetric with respect to a , b , and c in the sense that the triplet ( a , b , c ) is sum-feasible (resp. product-feasible) if and only if so are six possible permutations of the triplet ( a , b , c ) . So, without restriction of generality, we may assume that a ≤ b ≤ c . The case of compositum-feasible triplets is less symmetric. Then, only the degrees a and b (but not c ) are involved in a symmetric way. However, since the field K L contains both K and L , we can also assume that a ≤ b ≤ c . Of course, if ( a , b , c ) ∈ N 3 with a ≤ b ≤ c is a sum-feasible, a product-feasible or a compositum-feasible triplet, then (in any of the three cases) we must have c ≤ a b .

In the study by Drungilas et al. [1], it was shown that

Later, in the study by Drungilas and Dubickas [2], it was proved that

Consequently, if a triplet is not product-feasible, then it is neither sum-feasible nor compositum-feasible.

The simple example ( 2 , 3 , 3 ) shows that a triplet can be product-feasible but not sum-feasible, so these two questions are not equivalent. Indeed, the product γ of the quadratic number α = e 2 π i 3 and the cubic number β = 2 3 is cubic, because γ = α β = e 2 π i 3 2 3 is conjugate to β . On the other hand, since the integers 2 and 3 are coprime, the sum γ of any quadratic number α and any cubic number β must be of degree 6 (and so γ = α + β cannot be cubic); see, e.g., [3].

In the study by Drungilas et al. [1], all sum-feasible and also all compositum-feasible triplets ( a , b , c ) ∈ N 3 satisfying a ≤ b ≤ c , with b ≤ 6 , have been described except for one special case ( 6 , 6 , 8 ) . the study by Drungilas et al. [4], the missing case ( 6 , 6 , 8 ) from that classification has been treated by showing that the triplet ( 6 , 6 , 8 ) is not sum-feasible, and the classification has been extended to b ≤ 7 . Then, in the study by Drungilas and Maciulevičius [5], all possible compositum-feasible triplets ( a , b , c ) satisfying a ≤ b ≤ c , with b ≤ 9 , have been determined. In the special case when a = b = p is a prime number, a complete characterization of all compositum-feasible triplets ( p , p , p s ) has been given in [6].

Recently, in the study by Maciulevičius [7], the second named author described all possible product-feasible triplets ( a , b , c ) satisfying a ≤ b ≤ c , with b ≤ 7 , except for the following five cases:

By a different approach, Virbalas [8] showed that the triplets ( 4 , 7 , 7 ) , ( 4 , 7 , 14 ) , ( 5 , 6 , 10 ) , and ( 5 , 6 , 15 ) are not product-feasible. More generally, he proved the following:

This leaves us with the only undecided case ( 4 , 6 , 8 ) . In this note, we complete the picture of [7] by showing that

This implies that

Combining the aforementioned results with [7, Theorem 1], we obtain the following table that describes all possible product-feasible triplets ( a , b , c ) , where a ≤ b ≤ c and b ≤ 7 (Table 1).

In Table 1, there are nine triplets (with c written in italics) that are not sum-feasible by the results in [4] and [1], namely,

We also prove the following more general theorem:

In the proof of Theorem 4, assuming that there are α and β of degree 4 and 6 over Q , respectively, whose product α β is of degree 8, we will first show that the conjugates of α must all be of the same modulus. In 1969, Robinson [9] described algebraic integers whose conjugates including α itself are all of the same moduli; see also [10] for the description of algebraic numbers with conjugates of two distinct moduli. Here, we need a more specific result for quartic algebraic numbers α whose all four conjugates have equal moduli. The proposition below is a new ingredient that turns out to be a useful tool in the proof of Theorem 4.

This approach apparently cannot be used in the proof of a more general Theorem 6, although some parts of the proof of Theorem 4 will be used in the more general setting of Theorem 6.

Of course, Theorem 6 immediately implies Theorem 4 by selecting ( n , k ) = ( 4 , 2 ) . However, we present both proofs of the case ( a , b , c ) = ( 4 , 6 , 8 ) , since a separate proof of Theorem 4 contains some ideas that may be useful in treating similar problems for algebraic numbers of small degrees, but not only. As observed by one of the referees, in the first proof, the case ( 4 , 6 , 8 ) is obtained via descent, using the fact that the triplet ( 2 , 4 , 6 ) is not product-feasible. The main auxiliary result in the proof of Theorem 6 is Lemma 11.

In the next section, we will prove Proposition 7 and also give several auxiliary lemmas, which will be useful in the proofs of Theorems 4 and 6. Then, in Section 3, we will prove Theorem 4. The proof of Theorem 6 is a bit more involved. First, in Section 4, we will give some preparation under the assumption that the triplet ( n , ( n − 1 ) k , n k ) is product feasible. Second, in Section 5, using some previous results, we will show that the triplet ( a , b , c ) = ( n , ( n − 1 ) k , n k ) , where n is a prime number and k ≥ 1 , is product-feasible. Finally, we will complete the proof of Theorem 6 by getting a contradiction for composite n .

2 Proof of Proposition 7

Write

Assume first that p has a real root. By the assumptions of the proposition, not all four roots of p can be real. Thus, p must have a pair of complex conjugate roots and another real root. So, without loss of generality, we may assume that the four roots of p in equation (1) are { α 1 , α 2 } = { α , − α } and { α 3 , α 4 } = { α e i ϕ , α e − i ϕ } for some α > 0 and ϕ ∈ ( 0 , π ) . It follows that a 0 = α 1 α 2 α 3 α 4 = − α 4 is a negative rational number, say, − r , where r ∈ Q > 0 . Furthermore, since p has four roots, at least one real, on the circle ∣ z ∣ = α , by Ferguson’s result [11] (or even by an earlier result of Boyd [12]), we must have p ( x ) = g ( x 4 ) with some g ∈ Z [ x ] . The polynomials p and g have the same constant coefficient, so g is a monic linear polynomial x − r , which implies that p is as in ( i ) .

Next, we consider the case when p has no real roots. Then, there are ϱ > 0 and 0 < ϕ < ξ < π such that the roots of p are { α 1 , α 2 } = { ϱ e i ϕ , ϱ e − i ϕ } and { α 3 , α 4 } = { ϱ e i ξ , ϱ e − i ξ } . This time, by equation (1), we obtain a 0 = α 1 α 2 α 3 α 4 = ϱ 4 ∈ Q > 0 . Furthermore, by Vieta’s formulas,

and

Likewise, by Vieta’s formula, we deduce

If a 3 = 0 , then a 1 = ϱ 2 a 3 = 0 and cos ( ϕ ) = − cos ( ξ ) by equation (2). Thus, equation (3) yields a 2 = 2 ϱ 2 ( 1 − 2 cos 2 ( ξ ) ) = − 2 ϱ 2 cos ( 2 ξ ) ∈ Q . Therefore, setting r ≔ a 0 = ϱ 4 ∈ Q > 0 and s ≔ a 2 = − 2 r cos ( 2 ξ ) ∈ Q , we obtain the polynomial p ( x ) = x 4 + s x 2 + r ∈ Q [ x ] , which is the case ( i i ) , because s 2 = 4 r cos 2 ( 2 ξ ) ≤ 4 r . Here, the inequality must be strict, i.e., s 2 < 4 r , since otherwise p ( x ) has a real root.

It remains to consider the case a 3 ≠ 0 . Then, a 1 = ϱ 2 a 3 ≠ 0 and ϱ 2 = a 1 ∕ a 3 ∈ Q . Set r ≔ ϱ 2 ∈ Q > 0 . Since α 1 α 2 = α 3 α 4 = r , setting

and

by equation (1), we obtain

Here, a 3 = u + u ′ ≠ 0 , a 2 = u u ′ + 2 r , a 1 = ( u + u ′ ) r , and a 0 = r 2 . Since r ∈ Q > 0 , these four numbers are all rational if and only if u + u ′ and u u ′ are both rational. Furthermore, u and u ′ are both real by equations (4) and (5). They are both irrational, since otherwise p ( x ) would be reducible. Since u and u ′ are the roots of the quadratic polynomial

which is irreducible over Q due to u , u ′ ∉ Q , the numbers u and u ′ must be real quadratic conjugates. Moreover, since the roots of x 2 + u x + r and x 2 + u ′ x + r are all nonreal, we must have u 2 − 4 r < 0 and u ′ 2 − 4 r < 0 . Consequently, p is a polynomial of the form described in case ( i i i ) . This completes the proof of the proposition.

In the proofs of Theorems 4 and 6, we will also use the following two lemmas:

Of course, the numbers α i β j , where 1 ≤ i ≤ a and 1 ≤ j ≤ b , cover all possible conjugates of α β . Lemma 8 describes the situation when they are all conjugate over Q .

Let p be a prime number and n ∈ N . Denote by ord p ( n ) the exponent to which p appears in the prime factorization of n (for p ∤ n , we set ord p ( n ) = 0 ). We say that a triplet ( a , b , c ) satisfies the exponent triangle inequality with respect to a prime number p if

The next lemma will be used in the proof of Theorem 6.

3 Proof of Theorem 4

Suppose there exist algebraic numbers α and β satisfying

Since Q ( β ) and Q ( α β ) are subfields of Q ( α , β ) , we find that [ Q ( α , β ) : Q ] is divisible by both 6 and 8 and therefore divisible by lcm ( 6 , 8 ) = 24 . On the other hand,

This implies [ Q ( α , β ) : Q ] = 24 , and we have the picture as in Figure 1.

Figure 1

Diagram for the product-feasible triplet ( 4 , 6 , 8 ) .

Let α 1 ≔ α , α 2 , α 3 , α 4 be the four distinct conjugates of α over Q , and let β 1 ≔ β , β 2 , … , β 6 be the 6 distinct conjugates of β over Q . For i = 1 , 2 , 3 , 4 , we denote

The diagram and Lemma 8 with ( a , b ) = ( 4 , 6 ) implies that the numbers in the union Γ 1 ∪ ⋯ ∪ Γ 4 are all conjugate over Q . Since this set contains the number α β , it must have exactly 8 distinct elements. Therefore,

which implies ∣ Γ 1 ∩ Γ 2 ∣ ≥ 4 . Consequently, there exist distinct indices i 1 , i 2 , i 3 , and i 4 and distinct indices j 1 , j 2 , j 3 , and j 4 in { 1 , 2 , … , 6 } such that

Clearly, { i 1 , i 2 , i 3 , i 4 } ∩ { j 1 , j 2 , j 3 , j 4 } ≠ ∅ . Also, i 1 ≠ j 1 , since α 1 ≠ α 2 . Without loss of generality, we may assume that i 1 = j 2 . Dividing equation (6) by equation (7) we obtain β i 1 2 = β j 1 β i 2 . If j 1 ≠ i 2 , then Lemma 9 implies β j 1 m 1 = β i 1 m 1 for some m 1 ∈ N . Hence, by (6),

If j 1 = i 2 , then multiplying (6) and (7) we obtain α 1 2 = α 2 2 , and so equation (8) holds for m 1 = 2 .

Repeating the analogous argument for the pairs ( Γ 1 , Γ 3 ) and ( Γ 1 , Γ 4 ) , we deduce

for some m 2 , m 3 ∈ N . Now, by equations (8) and (9), we conclude that all four conjugates of α 1 = α have equal moduli.

By Lemma 8 with ( a , b ) = ( 4 , 6 ) , the full list of conjugates of α β of degree 8 is, for instance,

for some i , j ∈ { 2 , 3 , 4 } and some t , l ∈ { 1 , 2 , 3 , 4 , 5 , 6 } . In equation (10) we can choose any distinct products α i β t and α j β l , which do not belong to Γ 1 . In particular, for indices i and j , we must have

Adding and multiplying all eight conjugates in equation (10), we obtain

where r 1 ≔ β 1 + β 2 + ⋯ + β 6 ∈ Q and r 3 ≔ β 1 β 2 ⋯ β 6 ∈ Q \ { 0 } . This yields

and hence,

Squaring equation (12) and multiplying it by α i 2 β t 2 , we find that

Thus, β t is a root of a degree 4 polynomial over the field Q ( α , α i 2 ) . As we have shown earlier, all the conjugates of α have equal moduli. Hence, there are three possible cases for the minimal polynomials of α that are listed in Proposition 7.

In case ( i ) , we have α i = α ε for some ε ∈ { − 1 , ± i } . Then, α i 2 = α 2 or α i 2 = − α 2 , which implies Q ( α , α i 2 ) = Q ( α ) . But then the degree of β t over Q ( α ) is at most 4. Since the degree of β over Q ( α ) equals 6 (see the diagram), β and β t are conjugate over Q ( α ) . So the degree of β t over Q ( α ) equals 6, a contradiction.

In case ( i i ) , since both α 2 and α i 2 are the roots of the polynomial x 2 + s x + r , we must have either α i 2 = α 2 or α i 2 = − s − α 2 . In both cases, Q ( α , α i 2 ) = Q ( α ) , and we obtain the same contradiction.

It remains to consider case ( i i i ) . Then, α is nonreal. Assume, without loss of generality, that α 2 is the complex conjugate of α 1 = α . Note that in case ( i i i ) , both α 1 and α 2 = α 1 ¯ are the roots of the same quadratic factor x 2 + u x + r or x 2 + u ′ x + r , since u , u ′ , r ∈ R . Hence, α 2 α = r .

If Γ 1 ≠ Γ 2 , then we can take in equation (10) i = 2 , and so α i = α 2 . From α i = α 2 = r ∕ α , we obtain Q ( α , α i 2 ) = Q ( α ) , so that equation (13) leads to the same contradiction again.

In the alternative case, when Γ 1 = Γ 2 , we obtain

where { τ ( 1 ) , τ ( 2 ) , … , τ ( 6 ) } = { 1 , 2 , … , 6 } . This time, by equation (11), we cannot take i = 2 , so that i ∈ { 3 , 4 } . Multiplying all equalities in equation (14) we obtain α 6 = α 2 6 . Since α 2 = α ¯ and α 2 α = r , this yields r 6 = α 6 α 2 6 = α 12 . Consequently, α 6 is a rational number r 5 ∈ { − r 3 , r 3 } . Adding all equalities in equation (14), we derive that r 1 = β 1 + β 2 + … + β 6 = 0 , since α ≠ α 2 . Thus, by equation (12), we must have

This means that δ = α i β t is a rational number or a quadratic number. Evidently, δ cannot be rational, since α i and β t are of distinct degrees 4 and 6, respectively. On the other hand, if δ were quadratic, then the product of the quadratic number δ and the quartic number 1 ∕ α i would be the sextic number β t . However, the triplet ( 2 , 4 , 6 ) is not product-feasible by [7, Theorem 1]. This completes the proof of Theorem 4.

4 The case ( a , b , c ) = ( n , ( n ‒ 1 ) k , n k )

Set a = n , b = ( n − 1 ) k , and c = n k , where n , k > 1 are integers. Throughout this section, we assume that there exist algebraic numbers α and β satisfying

The beginning of the argument is essentially the same as that in Section 3. Since Q ( β ) and Q ( α β ) are subfields of Q ( α , β ) , we find that [ Q ( α , β ) : Q ] is divisible by both b and c and therefore divisible by

On the other hand, [ Q ( α , β ) : Q ] ≤ [ Q ( α ) : Q ] [ Q ( β ) : Q ] = a b . This implies [ Q ( α , β ) : Q ] = a b and

Thus, we have the following diagram:

Let α 1 ≔ α , α 2 , …, α a be the distinct conjugates of α over Q , and let β 1 ≔ β , β 2 , …, β b be the distinct conjugates of β over Q . Denote

Figure 2 and Lemma 8 imply that the numbers in the set

are all conjugate over Q . Since the set Γ contains the number α β , it must have exactly c distinct elements. This means that [ Q ( α i β j ) : Q ] = c for any indices i ∈ A and j ∈ B .

Figure 2

Degree diagram for α , β and α β .

Therefore, the same degree diagram as that in Figure 2 holds for any pair of indices i ∈ A , j ∈ B (Figure 3).

Figure 3

Degree diagram for α i , β j and α i β j .

We conclude this section with the following lemma and its corollary: