On the Degree of Product of Two Algebraic Numbers

Abstract

A triplet $(a,b,c)$ of positive integers is said to be product-feasible if there exist algebraic numbers $\alpha$, $\beta$ and $\gamma$ of degrees (over $\mathbb{Q}$) a, b and c, respectively, such that $\alpha \beta \gamma =1$. This work extends the investigation of product-feasible triplets started by Drungilas, Dubickas and Smyth. More precisely, for all but five positive integer triplets $(a,b,c)$ with $a\le b\le c$ and $b\le 7$, we decide whether it is product-feasible. Moreover, in the Appendix we give an infinite family or irreducible compositum-feasible triplets and propose a problem to find all such triplets.

1. Introduction

Following [1], we say that a triplet $(a,b,c)\in {\mathbb{N}}^3$ is sum-feasible (resp., product-feasible) if there exist algebraic numbers $\alpha$, $\beta$, $\gamma$ of degrees $a,b,c$ (over $\mathbb{Q}$), respectively, such that $\alpha +\beta +\gamma =0$ (resp., $\alpha \beta \gamma =1$). In [1], the problem of finding all sum-feasible triplets was proposed. In the same paper and in its continuations [2,3,4], an analogous problem for number fields was considered. Namely, we say that a triplet $(a,b,c)\in {\mathbb{N}}^3$ is compositum-feasible if there exist number fields K and L of degrees a and b (over $\mathbb{Q}$), respectively, such that the degree of their compositum $KL$ is c. All sum-feasible triplets $(a,b,c)\in {\mathbb{N}}^3$, satisfying $a\le b\le c$, $b\le 7$, and all possible compositum-feasible triplets $(a,b,c)$, satisfying $a\le b\le c$, $b\le 9$, were determined in [1,2,4]. Moreover, it was proved in [1,4] that the three feasibility problems are related in the following way: if $\mathcal{C}$, $\mathcal{S}$ and $\mathcal{P}$ denote sets of all possible compositum-feasible, sum-feasible and product-feasible triplets, respectively, then

$$\mathcal{C}⊊\mathcal{S}⊊\mathcal{P}.$$

(1)

Therefore all sum-feasible triplets that were found in the preceding papers are also product-feasible, but they do not exhaust all possible product-feasible triplets $(a,b,c)$ for which $a\le b\le c$ and $b\le 7$. There comes a natural motivation to investigate the case of the product more closely.

In this paper, we consider product-feasible triplets $(a,b,c)$ under the same restrictions $a\le b\le c$, $b\le 7$. More precisely, we prove the following:

Theorem 1. 

All the triplets $(a,b,c)\in {\mathbb{N}}^3$ with $a\le b\le c$, $b\le 7$ that are product-feasible are given in Table 1, with five possible exceptions that are circled.

Table 1. Triplets $(a,b,c)$, $a\le b\le c$, and $b\le 7$, which are product-feasible with 5 possible exceptions.

$\mathit{b}\setminus \mathit{a}$	1	2	3	4	5	6	7
1	1
2	2	2, 4
3	3	3, 6	3, 6, 9
4	4	4, 8	6, 12	4, 6, 8,
				12, 16
5	5	10	15	5, 10, 20	5, 10,
					20, 25
6	6	6, 12	6, 9,	6, ➇	➉, ⑮	6, 8, 9,
			12, 18	12, 24	30	12, 15, 18,
						24, 30, 36
7	7	14	21	➆, ⑭	35	7, 14,	7, 14, 21,
				28		21, 42	28, 42, 49

Moreover, we obtain several results related to triplets that include prime components.

Theorem 2. 

The triplet $(n-1,n,n)$, $n\ge 2$, is product-feasible if and only if n is a prime number.

In [1] (Theorem 8), it was proved that the triplet $(2,t,t)\in {\mathbb{N}}^3$ is product-feasible if and only if $2|t$ or $3|t$. We obtain an analogous result for triplets $(p,t,t)\in {\mathbb{N}}^3$, where $p>2$ is a prime number.

Theorem 3. 

Suppose a prime number p and a positive integer t satisfy $t\ge p>2$. Then, the triplet $(p,t,t)$ is product-feasible if and only if $p|t$.

The following theorem, taking $d=1$, implies the sufficiency part of Theorem 2.

Theorem 4. 

For any prime number p and each divisor d of $p-1$, the triplet $(p-1,p,pd)$ is product-feasible.

It was conjectured in [1] that the set $\mathcal{C}$ of compositum-feasible triplets is a multiplicative semigroup, i.e., if $(a,b,c),(a^{\prime },b^{\prime },c^{\prime })\in \mathcal{C}$, then $(a{a}^{\prime },b{b}^{\prime },c{c}^{\prime })\in \mathcal{C}$. This conjecture was proved in [3] (Theorem 1.3) assuming the answer to the inverse Galois problem is positive, i.e., that every finite group occurs as a Galois group of some normal field extension of $\mathbb{Q}$. Therefore, it is natural to consider irreducible elements of $\mathcal{C}$. In Appendix A, we give an infinite family of irreducible elements of $\mathcal{C}$ (see Proposition A1). Finally, at the end of Appendix A, we propose a problem of finding all irreducible compositum-feasible triplets.

The paper is organized as follows. The proof of Theorem 1 is given in Section 3 and is based on Theorems 2–4. In Section 2, we state some auxiliary results. Appendix A is devoted to irreducible elements of $\mathcal{C}$.

2. Auxiliary Results

Lemma 1 

(Lemma 14, [1]). Suppose that a triplet $(a,b,c)$ is product-feasible. Then, $c|lcm(a,b)·t$ for some positive $t\le gcd(a,b)$.

Lemma 2 

(Proposition 19, [1]). For any positive integers a and b, the triplet $(a,b,ab)$ is compositum-feasible and hence both sum-feasible and product-feasible.

Lemma 3 

(Lemma 7, [4]). Suppose that positive integers $a\le b\le c$ satisfy $ab<2c$. Then, if the triplet $(a,b,c)\in {\mathbb{N}}^3$ is not compositum-feasible, then it is neither sum-feasible nor product-feasible.

Lemma 4 

(Theorem 8, [1]). The triplet $(2,t,t)\in {\mathbb{N}}^3$ is product-feasible if and only if $2|t$ or $3|t$.

Let p be a prime number and $n\in \mathbb{N}$. Denote by ${ord}_p\left(n\right)$ the exponent to which p appears in the prime factorization of n (if $p\nmid n$ set ${ord}_p\left(n\right)=0$). We say that a triplet $(a,b,c)$ satisfies the exponent triangle inequality with respect to a prime p if

$$\begin{array}{c}{ord}_p\left(a\right)+{ord}_p\left(b\right)\ge {ord}_p\left(c\right),{ord}_p\left(a\right)+{ord}_p\left(c\right)\ge {ord}_p\left(b\right)\mathrm{and}\\ {ord}_p\left(b\right)+{ord}_p\left(c\right)\ge {ord}_p\left(a\right).\end{array}$$

Lemma 5 

(Proposition 28, [1]). Suppose that the triplet $(a,b,c)\in {\mathbb{N}}^3$ satisfies the exponent triangle inequality with respect to any prime number. Then, for any product-feasible triplet $(a^{\prime },b^{\prime },c^{\prime })\in {\mathbb{N}}^3$, the triplet $(a{a}^{\prime },b{b}^{\prime },c{c}^{\prime })$ is also product-feasible.

Lemma 6 

(Proposition 21, [1]). Suppose that α and β are algebraic numbers of degrees m and n over $\mathbb{Q}$, respectively. Let ${\alpha }_1=\alpha ,{\alpha }_2,\cdots ,{\alpha }_m$ be the distinct conjugates of α, and let ${\beta }_1=\beta ,{\beta }_2,\cdots ,{\beta }_n$ be the distinct conjugates of β. If β is of degree n over $\mathbb{Q}\left(\alpha \right)$, then all the numbers ${\alpha }_i{\beta }_j$, $1\le i\le m$, and $1\le j\le n$ are conjugate over $\mathbb{Q}$ (although not necessarily distinct).

Let ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$ be the roots of a nonzero separable polynomial $f\left(x\right)\in \mathbb{Q}\left[x\right]$ of degree $n\ge 2$. A multiplicative relation between ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$ is a relation of the kind

$$\prod _{i=1}^n{\alpha }_i^{k_i}\in \mathbb{Q},$$

where all the $k_j\in \mathbb{Q}$. We call this multiplicative relation trivial if $k_1=k_2=\cdots =k_n$.

Lemma 7 

(Theorem 1, [5]). Let $p>2$ be a prime number and $f\left(x\right)\in \mathbb{Q}\left[x\right]$ an irreducible monic polynomial $\ne x^p+a_0$ of degree p over $\mathbb{Q}$. Then, there are no nontrivial multiplicative relations between the roots ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_p$ of $f\left(x\right)$.

Lemma 8 

(Problem 6523, [6]). Suppose $f\left(x\right)$ is an irreducible polynomial of degree d over the field of rational numbers, and suppose $f\left(x\right)$ has two roots α, β with $\frac{\alpha }{\beta }$ a primitive nth root of unity. Then, $\phi \left(n\right)\le d$.

Let G be a group acting transitively on a set S. If the cardinality of S equals $n\in \mathbb{N}$, we say G is a group of degreen. A nonempty subset $\Delta \subseteq S$ is called a block for G if for each $x\in G$ either ${\Delta }^x=\Delta$ or ${\Delta }^x\cap \Delta =\varnothing$, here ${\Delta }^x=\{{\delta }^x:\delta \in \Delta \}$. Every group acting transitively on S has S and the singletons $\left\{\alpha \right\}$, $\alpha \in S$, as blocks. These are called the trivial blocks. Any other block is called nontrivial. For example, the cyclic group $G=⟨(1,2,3,4,5,6)⟩$ acting on $S=\{1,2,3,4,5,6\}$ has nontrivial blocks $\{1,4\}$, $\{2,5\}$, $\{3,6\}$, $\{1,3,5\}$, $\{2,4,6\}$ and in fact these are the only nontrivial blocks for G (see Exercise 1.5.2, [7]). We say that a group G acting transitively on a set S is primitive if G has no nontrivial blocks on S. For instance, the symmetric group $S_n$ and the alternating group $A_n$ acting on $S=\{1,2,\dots ,n\}$ are primitive for any $n\in \mathbb{N}$. One more example—the cyclic group $G=⟨(1,2,3,4,5)⟩$ acting on $S=\{1,2,3,4,5\}$ is primitive (see Lemma 9).

Lemma 9 

(Theorem 8.3, [8]). A transitive group of prime degree is primitive.

Lemma 10 

(Proposition 1, [4]). Suppose that $n>4$ is a positive integer and $p>2$ is a prime number that is not a divisor of $n-1$. Moreover, assume that p does not divide the order of any transitive subgroup of the symmetric group $S_n$, except possibly for $A_n$ and $S_n$. Then, for any positive integer $k>n$ divisible by p, the triplet $(n,n,k)$ is not product-feasible.

Lemma 11 

(Theorem 3.3, [7]). Let G be a subgroup of the symmetric group $S_n$ acting on the set $\{1,2,\dots ,n\}$. Suppose that G is primitive and contains a cycle of length p, where p is a prime number. Then, either G contains the alternating group $A_n$ as a subgroup, or $n\le p+2$.

Lemma 12 

([8] (Theorem 3.7) Special case of [8] (Theorem 3.7) taking any Sylow subgroup U of G and any $\alpha \in fixU$.). In a transitive group G, the normalizer of every Sylow subgroup Q of G is transitive on the points left fixed by Q.

Lemma 13 

(N/C theorem, see, e.g., Example 2.2.2, [7]). Let H be a subgroup of a group G. Then, $C_G\left(H\right)⊲N_G\left(H\right)$ and the qoutient $N_G\left(H\right)/C_G\left(H\right)$ is isomorphic to some subgroup of $AutH$, here

$$N_G\left(H\right)=\{g\in G:gH=Hg\}\mathrm{and}{C}_G\left(H\right)=\{g\in G:gh=hg\forall h\in H\}$$

are the normalizer and the centralizer of H in G, respectively.

3. Proofs

Proof of Theorem 2.

Necessity. Suppose that the triplet $(n-1,n,n)$ is product-feasible. Then, there exist algebraic numbers $\alpha$ and $\beta$, such that $\left[\mathbb{Q}\right(\alpha ):\mathbb{Q}]=n-1$ and $\left[\mathbb{Q}\right(\beta ):\mathbb{Q}]=\left[\mathbb{Q}\right(\alpha \beta ):\mathbb{Q}]=n$. Since $\mathbb{Q}\left(\alpha \right)$ and $\mathbb{Q}\left(\beta \right)$ are subfileds of $\mathbb{Q}(\alpha ,\beta )$, we find that $\left[\mathbb{Q}\right(\alpha ,\beta ):\mathbb{Q}]$ is divisible both by $n-1$ and n. Then, $gcd(n-1,n)=1$ implies that $\left[\mathbb{Q}\right(\alpha ,\beta ):\mathbb{Q}]$ is divisible by $(n-1)n$. On the other hand, $\left[\mathbb{Q}\right(\alpha ,\beta ):\mathbb{Q}]\le \left[\mathbb{Q}\right(\alpha ):\mathbb{Q}]\left[\mathbb{Q}\right(\beta ):\mathbb{Q}]=(n-1)n$. Hence, $\left[\mathbb{Q}\right(\alpha ,\beta ):\mathbb{Q}]=(n-1)n$ and we have the following diagram (see Figure 1):

Figure 1. Diagram for $(n-1,n,n)$.

Let ${\beta }_1:=\beta ,{\beta }_2,\cdots ,{\beta }_n$ be the distinct conjugates of $\beta$ over $\mathbb{Q}$. All the numbers

$$\alpha {\beta }_1,\alpha {\beta }_2,\cdots ,\alpha {\beta }_n$$

are pairwise distinct and, by Lemma 6, they all are conjugate over $\mathbb{Q}$. Hence, these are all the algebraic conjugates of $\alpha \beta$. Consequently the product

$$\left(\alpha {\beta }_1\right)\cdots \left(\alpha {\beta }_n\right)={\alpha }^n{\beta }_1{\beta }_2\cdots {\beta }_n$$

is a nonzero rational number. On the other hand, ${\beta }_1{\beta }_2\cdots {\beta }_n\in \mathbb{Q}\backslash \left\{0\right\}$ too. So, ${\alpha }^n$ is a non-zero rational number, say r. Therefore, $\alpha$ is a root of the polynomial $x^n-r$. The minimal polynomial of $\alpha$ is of degree $n-1$ and divides the polynomial $x^n-r$. Hence, $x^n-r$ has a root that is a rational number, say $r_0$. Then, $r=r_0^n$. Assume that n is not a prime number. Then, there exist integers $a>1$ and $b>1$ such that $n=ab$. Note that

$$x^n-r=x^{ab}-r_0^{ab}=(x^a-r_0^a)(x^{a(b-1)}+\cdots +r_0^{a(b-1)}).$$

So, the minimal polynomial of $\alpha$ divides either $x^a-r_0^a$ or the polynomial $x^{a(b-1)}+\cdots +r_0^{a(b-1)}$. However, this is impossible since the degree of either of these polynomials is strictly less than $n-1$. Therefore, n is a prime number.

Sufficiency. Assume n is a prime number. Let ${\zeta }_n$ be the primitive nth root of unity. Then, the degree of $\alpha =\frac{1}{{\zeta }_n}$ equals $n-1$. The numbers $\beta =\sqrt[n]{2}{\zeta }_n$ and $\gamma =\frac{1}{\sqrt[n]{2}}$ are of degree n and $\alpha \beta \gamma =1$. Hence, the triplet $(n-1,n,n)$ is product-feasible. □

Proof of Theorem 3.

Necessity. Assume that the triplet $(p,t,t)$ is product-feasible. Suppose for the contrary, that $p\nmid t$. We have that there exist algebraic numbers $\alpha$ and $\beta$ such that $\left[\mathbb{Q}\right(\alpha ):\mathbb{Q}]=p$ and $\left[\mathbb{Q}\right(\beta ):\mathbb{Q}]=\left[\mathbb{Q}\right(\alpha \beta ):\mathbb{Q}]=t$. Since $gcd(p,t)=1$, we obtain similarly as in the proof of Theorem 2 the following diagram (see Figure 2):

Figure 2. Diagram for $(p,t,t)$.

Using Lemma 6 analogously as in the proof of Theorem 2, we find that ${\alpha }^t\in \mathbb{Q}$. Hence, $\alpha$ is a root of a binomial equation $x^t-a=0$, $a\in \mathbb{Q}\setminus \left\{0\right\}$. On the other hand, $deg\alpha =p>2$ is a prime number. Therefore, Lemma 7 implies that the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is of the form $x^p-b$, $b\in \mathbb{Q}\setminus \left\{0\right\}$. We find that $x^t-a$ is divisible by $x^p-b$. Let $t=pq+r$, where q and r are non-negative integers and $r<p$. Since $t\ge p$ and $p\nmid t$, we find that $r>0$ and $q>0$. Note that

$$x^t-a=x^{pq+r}-a={(x^p-b+b)}^q{x}^r-a\equiv b^q{x}^r-a(mod{x}^p-b).$$

The remainder polynomial $b^q{x}^r-a$ is of degree $r>0$, which is strictly less than p. Hence, $x^p-b$ does not divide the polynomial $x^t-a$. A contradiction. Therefore, t is divisible by p.

Sufficiency. Let $t\ge p>2$ and $t=pk$ for some positive integer k. The triplet $(1,k,k)$ is obviously product-feasible, whereas the triplet $(p,p,p)$ satisfies the exponent triangle inequality. By Lemma 5, the triplet $(p,t,t)=(p·1,p·k,p·k)$ is product-feasible. □

Proof of Theorem 4.

If $p=2$, the assertion is obvious. If $d=p-1$, our triplet is product-feasible by Lemma 2. Suppose that $d<p-1$. Consider a field extension $\mathbb{Q}\left({\zeta }_{2p}\right):\mathbb{Q}$, here ${\zeta }_{2p}=e^{\frac{2\pi i}{2p}}$. As a cyclotomic extension, it is normal for degree $\phi \left(2p\right)=p-1$ and its Galois group G is isomorphic to the multiplicative group of the ring of residues modulo $2p$ (see, e.g., [9]), which means G is cyclic (Recall a well-known fact that the multiplicative group of the ring of residues modulo $n>1$ is cyclic if and only if $n=2$, 4, $p^{\alpha }$ or $2p^{\alpha }$ where $p>2$ is prime and $\alpha \in \mathbb{N}$ (see, e.g., [10]).). Therefore, for every divisor d of $\left|G\right|=p-1$, the group G has a unique subgroup of order $(p-1)/d$, say H. Let K be an intermediate field that corresponds to the subgroup H in the Galois correspondence, i.e., K consists of all elements of the field $\mathbb{Q}\left({\zeta }_{2p}\right)$, which are left invariant by every automorphism in H. Then, the degree of K over $\mathbb{Q}$ equals $\left|G\right|/\left|H\right|=d$. By the primitive element theorem $K=\mathbb{Q}\left(\theta \right)$ for some $\theta \in \mathbb{Q}\left({\zeta }_{2p}\right)$. Let g be a primitive root modulo $2p$. Then, the automorphism $\sigma \in G$ defined by

$$\sigma :{\zeta }_{2p}\mapsto {\zeta }_{2p}^g$$

generates G. We claim that $deg\left(\theta {\zeta }_{2p}\right)=p-1$. It suffices to show that all the numbers

$${\sigma }^k\left(\theta {\zeta }_{2p}\right),k=1,2,\dots ,p-1,$$

are distinct. Indeed, assume that ${\sigma }^k\left(\theta {\zeta }_{2p}\right)={\sigma }^l\left(\theta {\zeta }_{2p}\right)$ for some $1\le k<l\le p-1$. So that ${\sigma }^k\left(\theta \right){\zeta }_{2p}^{g^k}={\sigma }^l\left(\theta \right){\zeta }_{2p}^{g^l}$ and

$$\frac{{\sigma }^k\left(\theta \right)}{{\sigma }^l\left(\theta \right)}=e^{\frac{(g^l-g^k)\pi i}{p}}.$$

Note that $g^l-g^k=2m$, where $p\nmid m$. Therefore, ${\sigma }^k\left(\theta \right)/{\sigma }^l\left(\theta \right)$ is a primitive pth root of unity, which contradicts Lemma 8 since $d<p-1$. Hence, $deg\left(\theta {\zeta }_{2p}\right)=p-1$.

Finally, take

$$\alpha =\theta {\zeta }_{2p},\beta =\sqrt[p]{2},\gamma ={\left(\sqrt[p]{2}{e}^{\frac{\pi i}{p}}\theta \right)}^{-1}.$$

We have $\alpha \beta \gamma =1$. It remains to show that $deg\gamma =pd$. Let $\theta ={\theta }^{\left(1\right)},{\theta }^{\left(2\right)},\cdots ,{\theta }^{\left(d\right)}$ be all the conjugates of $\theta$. Since the numbers $deg\left(\sqrt[p]{2}{e}^{\frac{\pi i}{p}}\right)=p$ and $deg\theta =d$ are coprime Lemma 6, it implies that all the numbers

$${\gamma }_k^{\left(l\right)}:={\left(\sqrt[p]{2}{e}^{\frac{\pi i}{p}}{e}^{\frac{2\pi ik}{p}}{\theta }^{\left(l\right)}\right)}^{-1},k=0,1,\dots ,p-1,l=1,2,\dots ,d,$$

(2)

are conjugate to $\gamma$. It suffices to show that all these numbers are distinct. Indeed, assume that ${\gamma }_{k_1}^{\left(l_1\right)}={\gamma }_{k_2}^{\left(l_2\right)}$, where $k_1,k_2\in \{0,1,\cdots ,p-1\}$, $l_1,l_2\in \{1,2,\cdots ,d\}$ and either $k_1\ne k_2$ or $l_1\ne l_2$. Note that if $k_1=k_2$, then $l_1=l_2$. Therefore, $k_1\ne k_2$ and the equality ${\gamma }_{k_1}^{\left(l_1\right)}={\gamma }_{k_2}^{\left(l_2\right)}$ implies

$$e^{\frac{2\pi i(k_1-k_2)}{p}}=\frac{{\theta }^{\left(l_2\right)}}{{\theta }^{\left(l_1\right)}}.$$

Since $e^{2\pi i(k_1-k_2)/p}$ is a primitive pth root of unity, by Lemma 8, we find that $p-1=\phi \left(p\right)\le deg\theta =d$. This is a contradiction. Hence, all the numbers in (2) are distinct, and therefore $deg\gamma =pd$. This completes the proof of the theorem. □

Proposition 1. 

The triplet $(6,6,10)$ is not product-feasible.

Proof. 

The proof of [1] (Theorem 38) can be modified easily to the multiplicative case. Using same notations, we finally obtain ${\beta }_6^6\in \mathbb{Q}$, hence the minimal polynomial of $\beta$ is of the form $x^6-r_2$, $r_2\in \mathbb{Q}$. Interchanging $\alpha$ and $\beta$ in the proof of [1] (Theorem 38), we find that the minimal polynomial of $\alpha$ is also of the form $x^6-r_1$, $r_1\in \mathbb{Q}$. Hence, $\alpha =\sqrt[6]{r_1}{\epsilon }_6$ and $\beta =\sqrt[6]{r_2}{\epsilon }_6^{\prime }$, here ${\epsilon }_6$ and ${\epsilon }_6^{\prime }$ are some 6th roots of unity. This yields $\alpha \beta =\sqrt[6]{r_1{r}_2}{\epsilon }_6{\epsilon }_6^{\prime }$ as a root of $x^6-r_1{r}_2$, thus $deg\left(\alpha \beta \right)\le 6$, a contradiction. □

Proof of Theorem 1.

Using Lemma 1, we determine all possible candidates to product-feasible triplets $(a,b,c)$ with $a\le b\le c$, $b\le 7$. They are listed in Table 2.

Table 2. Candidates to product-feasible triplets.

$\mathit{b}\setminus \mathit{a}$	1	2	3	4	5	6	7
1	1
2	2	2, 4
3	3	3, 6	3, 6, 9
4	4	4, 8	4, 6, 12	4, 6, 8,
				12, 16
5	5	5, 10	5, 15	5, 10, 20	5, 10, 15
					20, 25
6	6	6, 12	6, 9,	6, ➇	6, ➉,	6, 8, 9, 10
			12, 18	12, 24	⑮, 30	12, 15, 18,
						24, 30, 36
7	7	7, 14	7, 21	➆, ⑭	7, 35	7, 14,	7, 14, 21,
				28		21, 42	28, 35, 42,
							49

Blue-colored triplets are sum-feasible, as is proved in [1,2]. Therefore, all these triplets are also product-feasible by (1).

Green-colored triplets are product-feasible too: $(2,3,3)$ is product-feasible by Lemma 4, the triplets $(3,6,9)$, $(3,4,6)$ and $(6,6,8)$ by Lemma 5, $(4,5,5)$ and $(6,7,7)$ by Theorem 2, whereas $(4,5,10)$, $(6,7,14)$, $(6,7,21)$ are product-feasible by Theorem 4 taking $(p,d)=(5,2)$, $(7,2)$ and $(7,3)$, respectively.

Red-colored triplets are not product-feasible: the triplets $(3,4,4)$, $(3,5,5)$, $(3,7,7)$, $(5,6,6)$ and $(5,7,7)$ are not product-feasible by Theorem 3, $(2,5,5)$, $(2,7,7)$ by Lemma 4, $(6,6,10)$ by Proposition 1, whereas $(5,5,15)$ and $(7,7,35)$ are not product-feasible by Lemma 3 and [2] (Corollary 1.5).

The circled triplets have not been examined yet. □

Let p and n be a prime number and a positive integer, respectively. Suppose that the triplet $(p,p,n)$ is product-feasible. If $p\nmid n$, then, by Lemma 1, we find that $n<p$. Hence, if $n>p$, then $p\mid n$. Finally, we give another result related to product-feasible triplets containing prime components.

Proposition 2. 

Suppose p, q and w are prime numbers such that $2<w<q<p$, $p=2q+w$ and $w\neg \left|\right(q-1)$. Then, both triplets $(p,p,pq)$ and $(p,p,2pq)$ are not product-feasible.

For instance, none of the triplets $(19,19,19·7k)$, $(29,29,29·11k)$ and $(31,31,31·13k)$, $k=1,2$, are product-feasible. Moreover, suppose that p, q and w satisfy the conditions of Proposition 2. Then, for any positive integer $t\ge 3$, the triplet $(p,p,pqt)$ is not product-feasible, by Lemma 1.

Proof 

(Proof of the Proposition). Let G be a transitive subgroup of the symmetric group $S_p$ such that $G\ne A_p$ and $G\ne S_p$. We will show that q cannot divide the order of G. Then, Lemma 10 will imply that the triplets $(p,p,pq)$ and $(p,p,2pq)$ both are not product-feasible. (Note that from $p=2q+w$, $2<w<q<p$, it follows that $q\neg \left|\right(p-1)$.).

Suppose for the contrary that the order of G is divisible by a prime q. Denote by Q a Sylow q-subgroup of G. The order of Q equals q or $q^2$ since Q is a subgroup of $S_p$ and ${ord}_q\left|S_p\right|=ord(p!)=q^2$. We claim that $\left|Q\right|=q$. Indeed, assume that $\left|Q\right|=q^2$. Then, Q is a Sylow q-subgroup of $S_p$, too. Take any cycle $\tau \in S_p$ of length q. Then, a cyclic subgroup $⟨\tau ⟩$ is contained in some Sylow q-subgroup of $S_p$. Since any two Sylow q-subgroups are conjugated and conjugate elements in $S_p$ are of the same cyclic structure, we find that the subgroup Q of G also contains a cycle of length q. However, Lemma 9 implies G is primitive, therefore we obtain a contradiction by Lemma 11. Hence, $\left|Q\right|=q$, which means Q is a cyclic subgroup generated by an element $\sigma \in G$ of order q. If $\sigma$ were a cycle of length q, we would obtain a contradiction by Lemma 11. Since $p=2q+w<3q$, it follows that $\sigma$ must be a product of two disjoint cycles of length q, say, $\pi$ and $\rho \in G$. Therefore, $|fixQ|=p-2q=w$, here $fixQ:=\{n\in \{1,2,\dots ,p\}:n^{\tau }=n\forall \tau \in Q\}$.

Note that Lemma 12 implies the order of the normalizer $N_G\left(Q\right)$ is divisible by $|fixQ|=w$, which is prime. Hence, there exists an element $\tau \in N_G\left(Q\right)$ of order w. We claim that in fact $\tau \in C_G\left(Q\right)\subseteq N_G\left(Q\right)$. Indeed, if $\tau \notin C_G\left(Q\right)$, then the order of $\tau C_G\left(Q\right)$ in the qoutient group $N_G\left(Q\right)/C_G\left(Q\right)$ equals w. Therefore, by Lemma 13, we find that $\omega$ divides the order of $AutQ$. However, $|AutQ|=\phi \left(q\right)=q-1$ and $\omega \nmid (q-1)$ by our assumption (here $\phi$ denotes the Euler’s totient function—a contradiction).

We have proved $Q=⟨\pi ·\rho ⟩$, where $\pi ,\rho \in S_p$ are two disjoint q-cycles. Let us denote $\pi =(i_1,i_2,\dots ,i_q)$ and $\rho =(j_1,j_2,\dots ,j_q)$. Since

$$\tau \in C_G\left(Q\right)=\{\sigma \in G:\sigma \eta {\sigma }^{-1}=\eta \forall \eta \in Q\},$$

we obtain $\tau ·\left(\pi \rho \right){\tau }^{-1}=\pi \rho$, i.e.,

$$(i_1^{\tau },i_2^{\tau },\dots ,i_q^{\tau })(j_1^{\tau },j_2^{\tau },\dots ,j_q^{\tau })=(i_1,i_2,\dots ,i_q)(j_1,j_2,\dots ,j_q).$$

By the uniqueness of the cycle decomposition, there are two possible cases: either

$$(i_1^{\tau },i_2^{\tau },\dots ,i_q^{\tau })=(i_1,i_2,\dots ,i_q)\mathrm{and}(j_1^{\tau },j_2^{\tau },\dots ,j_q^{\tau })=(j_1,j_2,\dots ,j_q)$$

or

$$(i_1^{\tau },i_2^{\tau },\dots ,i_q^{\tau })=(j_1,j_2,\dots ,j_q)\mathrm{and}(j_1^{\tau },j_2^{\tau },\dots ,j_q^{\tau })=(i_1,i_2,\dots ,i_q).$$

In both cases, we find that

$$(i_1^{{\tau }^2},i_2^{{\tau }^2},\dots ,i_q^{{\tau }^2})=(i_1,i_2,\dots ,i_q)\mathrm{and}(j_1^{{\tau }^2},j_2^{{\tau }^2},\dots ,j_q^{{\tau }^2})=(j_1,j_2,\dots ,j_q).$$

Denote $\eta ={\tau }^2$. We will show that $\eta$ fixes every element of the set

$$\{i_1,i_2,\dots ,i_q,j_1,j_2,\dots ,j_q\}.$$

Firstly, note that $\eta \left(i_1\right)=i_1$. Indeed, suppose for the contrary that $\eta \left(i_1\right)=i_{1+k}$ for some $k\in \{1,\dots ,q-1\}$. Then,

$${\eta }^l\left(i_1\right)=i_{1+lk\left(\mathrm{mod}q\right)}=i_1\iff 1+lk\equiv 1\left(\mathrm{mod}q\right)\iff l\equiv 0\left(\mathrm{mod}q\right),$$

which implies that $\eta$ has a cycle of length q in its cycle decomposition, but this is impossible since the order of $\eta$ equals w and $\gcd (w,q)=1$. Hence, $\eta \left(i_1\right)=i_1$, and therefore $\eta \left(i_k\right)=i_k$ for every $k=1,\dots ,q$. Analogously, $\eta \left(j_k\right)=j_k$ for every $k=1,\dots ,q$.

Hence, there are at most $p-2q=w$ elements in the set $\{1,2,\dots ,p\}$ that are not fixed under $\eta$. Since the order of $\eta$ equals $\omega$, it follows that $\eta$ is a cycle of length w, which leads to a contradiction by Lemma 11. This completes the proof of the proposition. □