#### Abstract

A theorem of Dubickas, affirming a conjecture of Kuba, states that a nonzero algebraic number is a root of a polynomial with positive rational coefficients if and only if none of its conjugates is a positive real number. A certain quantitative version of this result, yielding a growth factor for the coefficients of similar to the condition of the classical Eneström-Kakeya theorem of such polynomial, is derived. The bound for the growth factor so obtained is shown to be sharp for some particular classes of algebraic numbers.

#### 1. Introduction

A nonzero complex number is called* positively algebraic* if it is a root of a polynomial all of whose coefficients are positive rational numbers. In 2005, Kuba [1] conjectured that a necessary condition for an algebraic number to be positively algebraic is that none of its conjugates is a positive real number. This conjecture was confirmed affirmatively by Dubickas [2], in 2007, through the following result.

Theorem 1 (see [2]). *A nonzero complex number is a root of a polynomial with positive rational coefficients if and only if is an algebraic number such that none of its conjugates is a positive real number.*

In 2009, Brunotte [3] gave an elementary proof of Dubickas-Kuba theorem based on the following lemma ([3, Lemma 2]), which is originated from [4] (see also [5]):* if ** is a polynomial having no nonnegative roots, then there exists ** such that ** has only positive coefficients*. See also [6], where a bound for the degree of the polynomial with rational positive coefficients was given.

We fix the following notation and terminology throughout. Denote by the set of positive real numbers. For , let Some basic properties, which are needed in our work here, modified with the same proofs for from [7], are in the following lemma.

Lemma 2. *Let and . Then *(1)*, ;* *, ;*(2)*, where ;*(3)*;*(4)* and for some has a unique positive zero ; this zero is simple and all other roots have absolute values ;*(5)*let . Then for all ;*(6)*([7, Lemma 2]) are positive real numbers, and , where ;*(7)*([7, Lemma 3]) is a complex number which is not real positive is a root of a polynomial in for any .*

Our first main result is a certain quantitative improvement of Dubickas-Kuba theorem (Theorem 1 above).

Theorem 3. *Let be a nonzero algebraic number of degree (over ), let be all its conjugates, and let
**
Then all conjugates of are in if and only if there exists such that .*

The nontrivial half of Theorem 3 is its necessity part, and its main difficulty is to show the existence of a polynomial in all of whose coefficients are rational numbers. Should we need only a polynomial with nonnegative real coefficients, this necessity part follows from Roitman-Rubinstein’s result [7, Lemma 4] but with a bound for which depends not only on the conjugates of but also on other roots. Theorem 3 is proved in the next section.

Combining Theorem 3 with part (2) of Lemma 2, it is natural to ask for the least possible value of for which Theorem 3 holds. In the last section, we show that the bound is optimal for a class of algebraic numbers without nonnegative conjugates. The investigation in this second part leads us to some interesting connections with the classical result of Eneström-Kakeya [8], which gives upper and lower bounds for the absolute values of the roots of polynomials with positive coefficients.

#### 2. Proof of Theorem 3

The proof of Theorem 3 is done by analyzing the nature of the roots which are complex numbers, and we simplify by treating them separately in the next two lemmas.

Lemma 4. *Let with . Then, for any real , there exists such that .*

*Proof. *Let . Since , by Kronecker’s theorem ([9, Theorem 439, page 376]), the set is dense in the unit circle and so there exists such that
This yields
Let
Thus, (4) shows that all the coefficients of are . Again, using (4) with Lemma 2 part (2), we get
Thus, and so, by Lemma 2 part (5), . Let
so that . Consequently,
Putting
we see that
and and are roots of .

Lemma 5. *Let with . Then there exists such that .*

*Proof. *Since , there is an such that . Putting
we clearly see that , and so
The result follows by taking
noting that and are two roots of .

Observe that the proof of this last lemma remains true when is a negative real number (with ). The followimg result indicates how to obtain a multiple in of a given element of with a factor in .

Lemma 6. *Let , . If , for some , then there exists a polynomial such that .*

*Proof. *Writing
we have , and, by Lemma 2 part (1),
For each , since the th coefficient depends on the first coefficients of and and since any real number is a limit of a sequence of rational numbers from both sides, we can successively choose rational numbers so close to the corresponding coefficients of in such a way that when the polynomial
is multiplied to yielding
these new coefficients still satisfy
Lemma 2 part (1) shows then that .

The next lemma proves Theorem 3 for the case when the algebraic number and all its conjugates lie inside the unit circle.

Lemma 7. *Let be a nonzero algebraic number and let be all its conjugates. Assume that
**
where . Then all conjugates of are in if and only if there exists such that .*

*Proof. *Assume that the nonzero algebraic number and all its conjugates are in . Arrange these numbers in such a way that
with . (In the above arrangement, it is tacitly assumed that , , and are positive integers; indeed, should any one of the three sets be empty, the corresponding value(s) of , , or could be zero. Yet the remaining arguments below still work with only slight adjustments.) Since , for each , there exists an odd positive integer such that , for each , and , for each . Thus, for , we have
For , by Lemma 5, there exits a polynomial
For , by Lemma 4 and , there exits a polynomial
Let
and note that are its roots. From Lemma 2 part (6) with , we have . Thus,
Consequently,
and so
Since over , where is the minimal polynomial of , there exists such that . Thus,
From Lemma 6, there exists such that
Putting
we see that and .

Conversely, assume that there exists such that . Since , all conjugates of are zeros of and, since all coefficients of are positive, no real conjugate of can be positive.

We are now ready to prove Theorem 3.

*Proof of Theorem 3. *Assume that all conjugates of . Since and is dense in , there exists such that . Let
Then and , for all . Let
where is the minimal polynomial of . Then is the minimal polynomial of , and are conjugates of . By Lemma 7, there exists such that . Then ; that is, . Since and , we have . Let
Then and . Since , by Lemma 2 part (2), we have .

On the other hand, assume that there exists such that . Since , all conjugates of are zeros of and, since all coefficients of are positive, no real conjugate of can be positive.

It is worth noting that in [7, Lemma 4], the bound for reduces to if the polynomial mentioned in Theorem 3 is the minimal polynomial of . The next corollary contains the Dubickas-Kuba theorem and some equivalent assertions.

Corollary 8. *Let be a nonzero algebraic number, let be all its conjugates, and let . Then the following statements are equivalent: *(i)*all conjugates of are in ;*(ii)*there exists such that ;*(iii)*there exists such that ;*(iv)*there exist and such that .*

*Proof. *Assertions (i) and (ii) are equivalent by Theorem 3. Clearly, (ii) implies (iii). To show (iii) implies (iv), assume that there exists
such that , and so . Since , for all , none of can be real and positive. If for some , choose
Since is a factor of , we have . From and Lemma 2 part (4), we know that has a unique positive zero, say , which must then be distinct from all , as desired.

That assertion (iv) implies that assertion (i) is again an immediate consequence of Lemma 2 part (4).

#### 3. Eneström-Kakeya Theorem

To proceed further, we need a new notion.

*Definition 9. *For
its lower and upper Eneström-Kakeya quotients are defined, respectively, by

Part (1) of Lemma 2 is closely related to the classical Eneström-Kakeya theorem, [8, Theorem 1], which states the following.

Theorem 10. *Let all of whose coefficients are positive. Then all the zeros of are contained in the annulus , where and are, respectively, the lower and upper Eneström-Kakeya quotients of .*

In this section, we derive a proposition yielding conditions which are necessary for a product of two polynomials to be in .

Proposition 11. *Let
**
where , and let . If
**
then for all .*

*Proof. *Let
If , then Lemma 2 part (1) gives
where we adopt the convention that , for all , and , for all . From , and (42), we get . From (43) and (40), we have
From (44) and (40), we get
which together with previous results yield . Continuing in the same manner up to (46), we get , . Thus,
Since , the left-hand expression in (50) can be only when ; that is, , which is not possible.

The next proposition indicates the significance of the upper and lower Eneström-Kakeya quotients.

Proposition 12. *Let .*(i)*The upper Eneström-Kakeya quotient is the smallest such that .*(ii)*The lower Eneström-Kakeya quotient has the property that if with , then over .*

*Proof. *Part (i) follows from Lemma 2 part (1). Part (ii) is immediate from Proposition 11.

In passing, from the definition of Eneström-Kakeya quotients, it seems natural to ask whether one quotient can be a reciprocal of the other. This is indeed the case when the polynomial is self-reciprocal. Let be a polynomial. The reciprocal polynomial of is defined as
and we say that is* self-reciprocal* if .

Proposition 13. *Let . Then , and . Moreover, if is self-reciprocal, then and .*

*Proof. *Writing , we have
and so
From Proposition 12 part (i), we have
If is self-reciprocal, then clearly and .

Corollary 14. *If is self-reciprocal, then *(i)*, and*(ii)*, if and only if .*

*Proof. *(i) If , then . Since is self-reciprocal, by Proposition 13, , which contradicts Proposition 12 part (ii). Part (ii) follows readily from Proposition 13.

#### 4. Minimal Polynomials with Positive Coefficients

In this section, we aim to show that the bound , in Theorem 3, is best possible by showing that it cannot be improved for a subclass of the class of algebraic numbers whose minimal polynomials have positive coefficients. Let be a nonzero algebraic number, and its minimal polynomial. We say that is* positively minimal* if all its coefficients are positive. From Proposition 12, we have the following.

Proposition 15. *Let be an algebraic number and let
**
be its minimal polynomial. If is positively minimal, then, for any and any , we have .*

Proposition 15 leads naturally to the following definition.

*Definition 16. *Let be a nonzero algebraic number, whose minimal polynomial is positively minimal. Define the* growth factor* of an algebraic number with respect to , denoted by , to be the infimum of the set of nonnegative real numbers with the following property: for any , there exists such that .

*Remarks.* The growth factor enjoys the following basic properties:(i) is unique;(ii) (Theorem 3);(iii) (Propositions 12 and 15).

For a nonzero algebraic number whose minimal polynomial is positively minimal and whose growth factor is , let A class of algebraic numbers for which the bound in Theorem 3 cannot be improved is given in the next theorem.

Theorem 17. *Let be an algebraic number and assume that its minimal polynomial
**
is positively minimal.*(i)*If , then and .*(ii)*We have , if and only if
**for some .*

*Proof. *Assertion (i) follows at once from the preceding remarks.

(ii) If , then
which gives
The converse is trivial.

Theorem 17 poses a natural question whether a positively minimal polynomial must necessarily belong to . A negative answer is provided by the next proposition.

Proposition 18. *Let be an algebraic number of degree over . Assume that its minimal polynomial is positively minimal. If and all the conjugates of are negative real numbers, then
*

*and*.*Proof. *Since all conjugates of are negative real numbers, we have
Lemma 2 part (6) shows that
and so Proposition 12 (i) implies that . Since and , we have . Consequently,
By Proposition 12 (i) and the definition of the upper Eneström-Kakeya quotient, we conclude that .

It may be of interest to look at the growth factors and the Eneström-Kakeya quotients for positively minimal polynomials of small degrees.

Proposition 19. *Let be an algebraic number. Assume that its minimal polynomial is positively minimal and that all its conjugates are in . *(1)*If , then .*(2)*If , let and ; then,(2.1) for , we have ,(2.2)for and , we have ,(2.3)for and , we have .*(3)

*If , then .*

*Proof. *Part (1) is trivial. We now consider part (2), that is, . If , then , and so
If , without loss of generality, assume that with . Then and
Since is positively minimal, we have and . From
we deduce that, if , then and that, if , then .

Finally consider part (3), that is, . If and all its conjugates , , and are negative real numbers, the conclusion follows at once from Proposition 18. Assume then that
Since the minimal polynomial,
is positively minimal, we have , and , which gives
Since , is positive rational, and is positive irrational, we must have . By the remarks after Proposition 15, there remains only the verification that .

We split our consideration into three cases depending on the maximum absolute value of the conjugates. (i)Case . If , or if , then . Otherwise, we have and , which gives , a contradiction.(ii)Case . If , since is positive rational, we deduce that must be positive rational. Since is positive rational, we conclude that is positive rational, which is a contradiction. If , then . If , then
If , by (71), we get
yielding ; that is, . Consider now . If , since is positive rational, we deduce that is positive rational. Since is positive rational, we conclude that is positive rational, which is a contradiction. If , then . If , then , yielding .

We end this paper with another class of algebraic numbers for which the bound in Theorem 3 is optimal.

Proposition 20. *Let be an algebraic number, and let
**
be its minimal polynomial. For , if , then .*

*Proof. *Assume on the contrary that there is such that . Thus, , so that
If , by Lemma 2 part (1) we have , a contradiction.

If , let for some and the relation (74) becomes
Invoking upon Lemma 2 part (1), we get the following chain of inequalities:
Proceeding in the same manner, we have
Since , we have and
Thus, , which is a contradiction. The proofs for the cases and are similar but simpler.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.