JN Journal of Numbers 2314-842X 2356-7511 Hindawi Publishing Corporation 10.1155/2014/296828 296828 Research Article Algebraic Numbers Satisfying Polynomials with Positive Rational Coefficients http://orcid.org/0000-0002-7357-1061 Laohakosol Vichian 1 Tadee Suton 2 Laghribi Ahmed 1 Department of Mathematics Faculty of Science Kasetsart University Bangkok 10900 Thailand ku.ac.th 2 Department of Mathematics and Statistics Faculty of Science and Technology Thepsatri Rajabhat University Lopburi 15000 Thailand tru.ac.th 2014 1672014 2014 02 05 2014 30 06 2014 16 7 2014 2014 Copyright © 2014 Vichian Laohakosol and Suton Tadee. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

Theorem 1 (see [<xref ref-type="bibr" rid="B4">2</xref>]).

A nonzero complex number γ is a root of a polynomial with positive rational coefficients if and only if γ is an algebraic number such that none of its conjugates is a positive real number.

In 2009, Brunotte  gave an elementary proof of Dubickas-Kuba theorem based on the following lemma ([3, Lemma 2]), which is originated from  (see also ): if f(x)R[x] is a polynomial having no nonnegative roots, then there exists mN such that (1+x)mf(x) has only positive coefficients. See also , 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 R+ the set of positive real numbers. For rR+, let (1)P={f(x)=j=0JcjxjR[x];aaaaaj=0JcJ>0,cj0(0jJ-1),JN{0}},Q={f(x)=j=0JcjxjR[x];aaaaaj=0JcJ>0,cj0(0jJ-1),JN{0}},Q(r)={f(x)=j=0JcjxjR[x];aaaaaj=0J(x-r)f(x)Q,JN{0}},Q(0)=Q,Q+(r)=Q(r)R+[x],Q+(>r)=s>rQ+(s). Some basic properties, which are needed in our work here, modified with the same proofs for Q+(r) from , are in the following lemma.

Lemma 2.

Let r>0 and f(x)=j=0JcjxjR[x]{0}. Then

f(x)Q(r)c00, cjrcj+1(0j<J);

f(x)Q+(r)c0>0, cjrcj+1(0j<J);

0<a<bX(a)X(b), where X{Q,Q+};

Q+(r)Q(r)P;

f(x)Q and cj0 for some j{0,,J-1}f(x) has a unique positive zero z0; this zero is simple and all other roots have absolute values z0;

let Y{P,Q,Q(r),Q+(r)}. Then f(x)Yf(xk)Y for all kN;

([7, Lemma 2]) a1,a2,,as are positive real numbers, and fk(x)X(ak)(1ks)k=1sfk(x)X(k=1sak), where X{Q,Q+};

([7, Lemma 3]) γ is a complex number which is not real positive γ is a root of a polynomial in Q(r) for any r>|γ|.

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

Theorem 3.

Let γ be a nonzero algebraic number of degree n (over Q), let γ1(=γ),,γn be all its conjugates, and let (2)r>|γ|¯=max{|γt|;1tn}. Then all conjugates of γ are in CR+ if and only if there exists p(x)Q+(r)Q[x] such that p(γ)=0.

The nontrivial half of Theorem 3 is its necessity part, and its main difficulty is to show the existence of a polynomial in Q+(r) 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 r which depends not only on the conjugates of α 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 r for which Theorem 3 holds. In the last section, we show that the bound |γ|¯ 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 , which gives upper and lower bounds for the absolute values of the roots of polynomials with positive coefficients.

2. Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref>

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 λ=|λ|e2πiθ0 with θ(0,1)Q. Then, for any real r>|λ|, there exists F(x)Q+(r) such that F(λ)=0.

Proof.

Let u=r/|λ|>1. Since θRQ, by Kronecker’s theorem ([9, Theorem 439, page 376]), the set {e2πinθ:nN} is dense in the unit circle and so there exists m(=mθ)N such that (3)-12<cos(2πmθ)<-12u. This yields (4)1<-2ucos(2πmθ)<u,1u<-2cos(2πmθ)<1<u. Let (5)f1(x)=(x-(λ|λ|)m)(x-(λ¯|λ¯|)m)=x2-2xcos(2πmθ)+1. Thus, (4) shows that all the coefficients of f1 are >1/u>0. Again, using (4) with Lemma 2 part (2), we get (6)f1(x)Q+(u)Q+(um). Thus, (x-um)f1(x)Q and so, by Lemma 2 part (5), (xm-um)f1(xm)Q. Let (7)f2(x)=(xm-umx-u)f1(xm), so that (x-u)f2(x)Q. Consequently, (8)(x|λ|-r|λ|)f2(x|λ|)Q. Putting (9)F(x)=1|λ|f2(x|λ|), we see that (10)F(x)=1|λ|3m(xm-rmx-r)(xm-λm)(xm-λ¯  m)Q+(r), and λ and λ¯ are roots of F(x).

Lemma 5.

Let λ=|λ|e2πiθ0 with θ(0,1)Q. Then there exists G(x)Q+(|λ|) such that G(λ)=0.

Proof.

Since θQ, there is an M(=Mθ)N such that (λ/|λ|)M=1. Putting (11)g(x)=xM-(λ/|λ|)Mx-1, we clearly see that (x-1)g(x)Q, and so (12)(x|λ|-1)g(x|λ|)Q. The result follows by taking (13)G(x)=1|λ|g(x|λ|)=1|λ|M(xM-λMx-|λ|)=1|λ|{(x|λ|)M-1+(x|λ|)M-2aaaaaaa++(x|λ|)2+x|λ|+1}Q+(|λ|), noting that λ and λ¯ are two roots of G(x).

Observe that the proof of this last lemma remains true when λ is a negative real number (with M=2). The followimg result indicates how to obtain a multiple in Q+(r) of a given element of R[x] with a factor in Q[x].

Lemma 6.

Let A(x), B(x)R[x]. If A(x)B(x)Q+(r), for some r>0, then there exists a polynomial B(x)Q[x] such that A(x)B(x)Q+(r).

Proof.

Writing (14)A(x)B(x)=ckxk+ck-1xk-1++c1x+c0 we have cj>0(0jk), and, by Lemma 2 part (1), (15)0<c0rc1r2c2r3c3rkck. For each j{0,1,,k}, since the jth coefficient cj depends on the first j coefficients of A(x) and B(x) and since any real number is a limit of a sequence of rational numbers from both sides, we can successively choose rational numbers b0,b1,,bm so close to the corresponding coefficients of B(x) in such a way that when the polynomial (16)B(x)=bmxm+bm-1xm-1++b1x+b0Q[x] is multiplied to A(x) yielding (17)A(x)B(x)=c~kxk+c~k-1xk-1++c~1x+c~0, these new coefficients still satisfy (18)0<c~0rc~1r2c~2r3c~3rkc~k. Lemma 2 part (1) shows then that A(x)B(x)Q+(r).

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

Lemma 7.

Let γ be a nonzero algebraic number and let γ1(=γ),,γn be all its conjugates. Assume that (19)|γ|¯=max1tn{|γt|}<r, where r(0,1]. Then all conjugates of γ are in CR+ if and only if there exists p(x)Q+(r)Q[x] such that p(γ)=0.

Proof.

Assume that the nonzero algebraic number γ and all its conjugates γ1(=γ),,γn are in CR+. Arrange these numbers in such a way that (20){γ1,γ2,,γn}={γjR-;1jJ}{γJ+k=|γJ+k|e2πiθJ+k;θJ+kQ,1kK}{γJ+K+l=|γJ+K+l|e2πiθJ+K+l;θJ+K+lRQ,1lL} with J+K+L=n. (In the above arrangement, it is tacitly assumed that J, K, and L are positive integers; indeed, should any one of the three sets be empty, the corresponding value(s) of J, K, or L could be zero. Yet the remaining arguments below still work with only slight adjustments.) Since |γt|<r, for each 1tn, there exists an odd positive integer N such that |γt|N<r/n, for each 1tn, and γjN<0, for each 1jJ. Thus, for 1jJ, we have (21)Hj(x)=(x-γjN)Q+(|γj|N)Q+(rn)aaaaaaaaaaaaaaaaawithHj(γjN)=0. For 1kK, by Lemma 5, there exits a polynomial (22)Gk(x)Q+(|γJ+k|N)Q+(rn)withGk(γJ+kN)=0. For 1lL, by Lemma 4 and |γJ+K+l|N<r/n, there exits a polynomial (23)Fl(x)Q+(rn)withFl(γJ+K+lN)=0. Let (24)E(x)=1jJHj(x)1kKGk(x)1lLFl(x), and note that γ1N,,γnN are its roots. From Lemma 2 part (6) with J+K+L=n, we have E(x)Q+(r). Thus, (25)(x-1)E(x)Qandso(xN-1)E(xN)Q. Consequently, (26)(x-1)(xN-1+xN-2++x+1)E(xN)Q and so (27)(xN-1+xN-2++x+1)E(xN)Q+(1). Since M(x)E(xN) over R[x], where M(x)Q[x] is the minimal polynomial of γ, there exists B(x)R[x] such that E(xN)=M(x)B(x). Thus, (28)(xN-1+xN-2++x+1)M(x)B(x)Q+(r). From Lemma 6, there exists B(x)Q[x] such that (29)(xN-1+xN-2++x+1)M(x)B(x)Q+(r)Q[x]. Putting (30)p(x)=(xN-1+xN-2++x+1)M(x)B(x), we see that p(x)Q+(r)Q[x] and p(γ)=0.

Conversely, assume that there exists p(x)Q+(r)Q[x] such that p(γ)=0. Since p(x)Q[x], all conjugates of γ are zeros of p(x) and, since all coefficients of p(x) are positive, no real conjugate of γ can be positive.

We are now ready to prove Theorem 3.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref>.

Assume that all conjugates of γCR+. Since r>|γ|¯ and Q is dense in R, there exists r1Q such that |γ|¯<r1<r. Let (31)λt=γtr1(1tn),(32)λ=λ1. Then |λt|<1 and λtCR+, for all 1tn. Let (33)F(x)=M(r1x), where M(x)Q[x] is the minimal polynomial of γ. Then F(x) is the minimal polynomial of λ, and λ1(=λ),,λn are conjugates of λ. By Lemma 7, there exists p1(x)Q+(1)Q[x] such that p1(λ)=0. Then F(x)p1(x); that is, M(x)p1(x/r1). Since p1(x)Q+(1)Q[x] and r1Q, we have p1(x/r1)Q+(r1)Q[x]. Let (34)p(x)=p1(xr1). Then p(x)Q+(r1)Q[x] and p(γ)=0. Since 0<r1<r, by Lemma 2 part (2), we have p(x)Q+(r)Q[x].

On the other hand, assume that there exists p(x)Q+(r)Q[x] such that p(γ)=0. Since p(x)Q[x], all conjugates of γ are zeros of p(x) and, since all coefficients of p(x) are positive, no real conjugate of γ can be positive.

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

Corollary 8.

Let γ be a nonzero algebraic number, let γ1(=γ),γ2,,γn be all its conjugates, and let r>|γ|¯. Then the following statements are equivalent:

all conjugates of γ are in CR+;

there exists p(x)Q+(r)Q[x] such that p(γ)=0;

there exists g(x)R+[x]Q[x] such that g(γ)=0;

there exist q(x)QQ[x] and wR+{γ1,,γn} such that q(γ)=q(w)=0.

Proof.

Assertions (i) and (ii) are equivalent by Theorem 3. Clearly, (ii) implies (iii). To show (iii) implies (iv), assume that there exists (35)g(x)=xm+g1xm-1++gm-1x+gmR+[x]Q[x] such that g(γ)=0, and so g(γ1)==g(γn)=0. Since gh>0, for all h{1,,m}, none of γ1,,γn can be real and positive. If γj0 for some j{1,2,,n}, choose (36)q(x)=(xm)2-(g1xm-1+g2xm-2++gm-1x+gm)2Q. Since g(x) is a factor of q(x), we have q(γ1)==q(γn)=0. From γj0 and Lemma 2 part (4), we know that q(x) has a unique positive zero, say w, which must then be distinct from all γ1,,γn, 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 (37)f(x)=j=0JcjxjR+[x]R, its lower and upper Eneström-Kakeya quotients are defined, respectively, by (38)α[f]=min{c0c1,c1c2,,cJ-2cJ-1,cJ-1cJ},β[f]=max{c0c1,c1c2,,cJ-2cJ-1,cJ-1cJ}.

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

Theorem 10.

Let p(x)R[x]R all of whose coefficients are positive. Then all the zeros of p(x) are contained in the annulus α[p]|x|β[p], where α[p] and β[p] are, respectively, the lower and upper Eneström-Kakeya quotients of p(x).

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

Proposition 11.

Let (39)f(x)=cJxJ+cJ-1xJ-1++c1x+c0R+[x], where J1, and let rR+. If (40)cj>rcj+1(0j<J), then f(x)d(x)Q(r) for all d(x)R[x]{0}.

Proof.

Let (41)d(x)=k=0KdkxkR[x]{0}. If f(x)d(x)Q(r), then Lemma 2 part (1) gives (42)0c0d0(43)c0d0r(c1d0+c0d1)(44)c1d0+c0d1r(c2d0+c1d1+c0d2)(45)(46)cK-1d0+cK-2d1++c0dK-1r(cKd0+cK-1d1++c0dK)(47)cKd0+cK-1d1++c0dKr(cK+1d0+cKd1++c1dK), where we adopt the convention that cj=0, for all j>J, and dk=0, for all k>K. From c0>0, and (42), we get d00. From (43) and (40), we have (48)(c0-rc1)d0rc0d1,d10. From (44) and (40), we get (49)(c1-rc2)d0+(c0-rc1)d1rc0d2, which together with previous results yield d20. Continuing in the same manner up to (46), we get d30, d40,,dK0. Thus, (50)(cK-rcK+1)d0+(cK-1-rcK)d1++(c1-rc2)dK-1+(c0-rc1)dK0. Since cj>rcj+1(j=0,1,2,,J-1), the left-hand expression in (50) can be 0 only when d0=d1==dK=0; that is, d(x)0, which is not possible.

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

Proposition 12.

Let f(x)R+[x].

The upper Eneström-Kakeya quotient β[f] is the smallest r>0 such that f(x)Q+(r).

The lower Eneström-Kakeya quotient α[f] has the property that if p(x)Q+(r) with 0<r<α[f], then f(x)p(x) over R[x].

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 g(x) be a polynomial. The reciprocal polynomial of g(x) is defined as (51)g*(x)=xdeggg(1x), and we say that g(x) is self-reciprocal if g(x)=g*(x).

Proposition 13.

Let f(x)R+[x]R. Then β[f*]=1/α[f], and f*(x)Q+(1/α[f]). Moreover, if f(x) is self-reciprocal, then β[f]=1/α[f] and f(x)Q+(1/α[f]).

Proof.

Writing f(x)=cJxJ+cJ-1xJ-1++c1x+c0R+[x]R, we have (52)f*(x)=xJf(1x)=cJ+cJ-1x++c1xJ-1+c0xJ and so (53)β[f*]=max{c1c0,c2c1,,cJ-1cJ-2,cJcJ-1}=1α[f]. From Proposition 12 part (i), we have (54)f*(x)Q+(β[f*])=Q+(1α[f]). If f(x) is self-reciprocal, then clearly β[f]=1/α[f] and f(x)Q+(1/α[f]).

Corollary 14.

If f(x)R+[x]R is self-reciprocal, then

α[f]1, and

α[f]=1, if and only if β[f]=1.

Proof.

(i) If α[f]>1, then 1/α[f]<1<α[f]. Since f is self-reciprocal, by Proposition 13, f(x)Q+(1/α[f]), 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 |γ|¯, 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 γ be a nonzero algebraic number, and M(x)Q[x] its minimal polynomial. We say that M(x) is positively minimal if all its coefficients are positive. From Proposition 12, we have the following.

Proposition 15.

Let γ be an algebraic number and let (55)M(x)=xn+an-1xn-1++a1x+a0 be its minimal polynomial. If M(x) is positively minimal, then, for any r(0,α[M]) and any p(x)Q+(r)Q[x], we have p(γ)0.

Proposition 15 leads naturally to the following definition.

Definition 16.

Let γ be a nonzero algebraic number, whose minimal polynomial M(x) is positively minimal. Define the growth factor of an algebraic number γ with respect to M(x), denoted by rγ, to be the infimum of the set of nonnegative real numbers with the following property: for any r>rγ, there exists p(x)Q+(r)Q[x] such that p(γ)=0.

Remarks. The growth factor enjoys the following basic properties:

rγ is unique;

rγ|γ|¯ (Theorem 3);

α[M]rγβ[M] (Propositions 12 and 15).

For a nonzero algebraic number γ whose minimal polynomial is positively minimal and whose growth factor is rγ, let (56)Pγ={p(x)Q+(rγ)Q[x];p(γ)=0}. A class of algebraic numbers for which the bound |γ|¯ in Theorem 3 cannot be improved is given in the next theorem.

Theorem 17.

Let γ be an algebraic number and assume that its minimal polynomial (57)M(x)=xn+an-1xn-1++a1x+a0 is positively minimal.

If α[M]=β[M], then α[M]=rγ=β[M] and M(x)Pγ.

We have α[M]=β[M], if and only if (58)M(x)=xn+axn-1+a2xn-2++an-1x+an

for some aQ+.

Proof.

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

(ii) If α[M]=β[M], then (59)a0a1=a1a2==an-2an-1=an-1, which gives (60)M(x)=xn+an-1xn-1+an-12xn-2++an-1n-1x+an-1nQ+[x]. The converse is trivial.

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

Proposition 18.

Let γ be an algebraic number of degree n over Q. Assume that its minimal polynomial M(x) is positively minimal. If n2 and all the conjugates of γ1=γ,,γn are negative real numbers, then (61)α[M]rγ|γ|¯<β[M]=t=1n|γt|and M(x)Pγ.

Proof.

Since all conjugates of γ are negative real numbers, we have (62)x-γtQ+(|γt|)(t=1,,n). Lemma 2 part (6) shows that (63)M(x)=(x-γ1)(x-γ2)(x-γn)=xn+an-1xn-1++a1x+a0Q+(t=1n|γt|)Q[x], and so Proposition 12 (i) implies that β[M]t=1n|γt|. Since n2 and an-1=t=1n|γt|, we have |γ|¯<t=1n|γt|=an-1β[M]. Consequently, (64)α[M]rγ|γ|¯<β[M]=t=1n|γt|. By Proposition 12 (i) and the definition of the upper Eneström-Kakeya quotient, we conclude that M(x)Pγ.

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 γ be an algebraic number. Assume that its minimal polynomial M(x) is positively minimal and that all its conjugates are in CR+.

If degγ=1, then α[M]=rγ=β[M]=|γ|¯=|γ|.

If degγ=2, let γ=a+bi(a,bR) and M(x)=x2+a1x+a0; then,

for b=0, we have a0/a1=α[M]<|γ|¯<β[M]=a1,rγ[α[M],|γ|¯],

for b0 and 3a2b2, we have α[M]<|γ|¯=|γ|<β[M],rγ[α[M],|γ|¯],

for b0 and 3a2=b2, we have α[M]=rγ=β[M]=|γ|¯=|γ|.

If degγ=3, then α[M]rγ|γ|¯<β[M].

Proof.

Part (1) is trivial. We now consider part (2), that is, degγ=2. If b=0, then a124a0>0, and so (65)α[M]=a0a1<|γ|¯=a12+a124-a0<a1=β[M]. If b0, without loss of generality, assume that γ=a+bi with bR+. Then |γ|¯=a2+b2 and (66)M(x)=(x-a-bi)(x-a+bi)=x2-2ax+a2+b2Q[x]. Since M(x) is positively minimal, we have a<0 and a2+b2>0. From (67)a2+b2-2aa2+b24a2b23a2, we deduce that, if 3a2b2, then α[M]<|γ|¯<β[M] and that, if 3a2=b2, then α[M]=rγ=β[M]=|γ|¯=|γ|.

Finally consider part (3), that is, degγ=3. If γ and all its conjugates γ1, γ2, and γ3 are negative real numbers, the conclusion follows at once from Proposition 18. Assume then that (68)γ1=-A,γ2=a+bi,γ3=a-biaawithAR+Q,aR,bR+. Since the minimal polynomial, (69)M(x)=(x+A)(x-a-bi)(x-a+bi)=x3+(A-2a)x2+(a2+b2-2Aa)x+A(a2+b2)Q[x], is positively minimal, we have A-2a>0, a2+b2-2Aa>0 and A(a2+b2)>0, which gives (70)β[M]=max{A(a2+b2)a2+b2-2Aa,a2+b2-2AaA-2a,A-2a}. Since degγ=3, A-2a is positive rational, and A is positive irrational, we must have a0. By the remarks after Proposition 15, there remains only the verification that |γ|¯<β[M].

We split our consideration into three cases depending on the maximum absolute value of the conjugates.

Case |γ|¯=A.

If A<A-2a, or if A<A(a2+b2)/(a2+b2-2Aa), then |γ|¯<β[M]. Otherwise, we have AA-2a and AA(a2+b2)/(a2+b2-2Aa), which gives a=0, a contradiction.

Case |γ|¯=a2+b2.

If A(a2+b2)/(a2+b2-2Aa)=a2+b2, since A(a2+b2)/(a2+b2-2Aa) is positive rational, we deduce that a2+b2 must be positive rational. Since A(a2+b2) is positive rational, we conclude that A is positive rational, which is a contradiction.

If A(a2+b2)/(a2+b2-2Aa)>a2+b2, then β[M]>|γ|¯.

If A(a2+b2)/(a2+b2-2Aa)<a2+b2, then (71)2Aa<a2+b2(a2+b2-A).

If a>0, by (71), we get (72)a2+b2(a2+b2-A)+2aa2+b2>2Aa

yielding (a2+b2-2Aa)/(A-2a)>a2+b2; that is, β[M]>|γ|¯.

Consider now a<0.

If A-2a=a2+b2, since A-2a is positive rational, we deduce that a2+b2 is positive rational. Since A(a2+b2) is positive rational, we conclude that A is positive rational, which is a contradiction.

If A-2a>a2+b2, then β[M]>|γ|¯.

If A-2a<a2+b2, then a2+b2(a2+b2-A+2a)>0>2Aa, yielding β[M]>|γ|¯.

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

Proposition 20.

Let γ be an algebraic number, and let (73)M(x)=xn+a(nN,aQ,a>0) be its minimal polynomial. For 0<r<|γ|¯=a1/n, if q(x)Q+(r)Q[x], then q(γ)0.

Proof.

Assume on the contrary that there is q(x)Q+(r)Q[x] such that q(γ)=0. Thus, M(x)q(x), so that (74)q(x)=(xn+a)×(xm+cm-1xm-1++c3x3+c2x2+c1x+c0)=xn+m+cm-1xn+m-1++c3xn+3+c2xn+2+c1xn+1+c0xn+axm+acm-1xm-1++ac3x3+ac2x2+ac1x+ac0. If m<n-1, by Lemma 2 part (1) we have 0<a0, a contradiction.

If m>n, let m=n+t for some tN and the relation (74) becomes (75)q(x)=x2n+t+cn+t-1x2n+t-1+cn+t-2x2n+t-2++ct+1xn+t+1+(ct+a)xn+t+(ct-1+acn+t-1)xn+t-1+(ct-2+acn+t-2)xn+t-2++(c4+acn+4)xn+4+(c3+acn+3)xn+3+(c2+acn+2)xn+2+(c1+acn+1)xn+1+(c0+acn)xn+acn-1xn-1+acn-2xn-2++ac3x3+ac2x2+ac1x+ac0. Invoking upon Lemma 2 part (1), we get the following chain of inequalities: (76)0<ac0c0>0ac0rac1c1>0,c0rc1ac1rac2c2>0,c1rc2,c0r2c2acn-2racn-1cn-1>0,cn-2rcn-1,cn-3r2cn-1,,c0rn-1cn-1acn-1r(c0+acn)rncn-1+racn<acn-1+racn0<racn,cn>0c0+acnr(c1+acn+1)r(c0+acn)r2c1+r2acn+1acn-1r2c1+r2acn+1rncn-1+r2acn+1<acn-1+r2acn+10<r2acn+1,cn+1>0c1+acn+1r(c2+acn+2)r(c1+acn+1)r2c2+r2acn+2c0+acnr2c2+r2acn+2r(c0+acn)r3c2+r3acn+2acn-1r3c2+r3acn+2rncn-1+r3acn+2<acn-1+r3acn+20<r3acn+2,cn+2>0. Proceeding in the same manner, we have (77)c2+acn+2r(c3+acn+3)cn+3>0c3+acn+3r(c4+acn+4)cn+4>0ct-2+acn+t-2r(ct-1+acn+t-1)cn+t-1>0. Since n+t-1t, we have ct>0 and (78)cn+t-1r,cn+t-2rcn+t-1r2,cn+t-3rcn+t-2r3,,ct+1rn-1. Thus, a<ct+arct+1rn<a, which is a contradiction. The proofs for the cases m=n-1 and n are similar but simpler.

Conflict of Interests

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

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