β=a+b√m, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | 4+2√3+3√1−m4≤a<2+√1−m and b=0 |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥a | none |
Subcase 2.2: |b|<a | 3≤a<√5+4√1−m and b=±1 |
In the present work, the authors developed the scheme for time Fractional Partial Diffusion Differential Equation (FPDDE). The considered class of FPDDE describes the flow of fluid from the higher density region to the region of lower density, macroscopically it is associated with the gradient of concentration. FPDDE is used in different branches of science for the modeling and better description of those processes that involve flow of substances. The authors introduced the novel concept of fractional derivatives in term of both time and space independent variables in the proposed FPDDE. We provided the approximate solution for the underlying generalized non-linear time PFDDE in the sense of Caputo differential operator via Laplace transform combined with Adomian decomposition method known as Laplace Adomian Decomposition Method (LADM). Furthermore, we established the general scheme for the considered model in the form of infinite series by aforementioned techniques. The consequent results obtained by the proposed technique ensure that LADM is an effective and accurate technique to handle nonlinear partial differential equations as compared to the other available numerical techniques. At the end of this paper, the obtained numerical solution is visualized graphically by Matlab to describe the dynamics of desired solution.
Citation: Amjad Ali, Iyad Suwan, Thabet Abdeljawad, Abdullah. Numerical simulation of time partial fractional diffusion model by Laplace transform[J]. AIMS Mathematics, 2022, 7(2): 2878-2890. doi: 10.3934/math.2022159
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In the present work, the authors developed the scheme for time Fractional Partial Diffusion Differential Equation (FPDDE). The considered class of FPDDE describes the flow of fluid from the higher density region to the region of lower density, macroscopically it is associated with the gradient of concentration. FPDDE is used in different branches of science for the modeling and better description of those processes that involve flow of substances. The authors introduced the novel concept of fractional derivatives in term of both time and space independent variables in the proposed FPDDE. We provided the approximate solution for the underlying generalized non-linear time PFDDE in the sense of Caputo differential operator via Laplace transform combined with Adomian decomposition method known as Laplace Adomian Decomposition Method (LADM). Furthermore, we established the general scheme for the considered model in the form of infinite series by aforementioned techniques. The consequent results obtained by the proposed technique ensure that LADM is an effective and accurate technique to handle nonlinear partial differential equations as compared to the other available numerical techniques. At the end of this paper, the obtained numerical solution is visualized graphically by Matlab to describe the dynamics of desired solution.
Determining the irreducibility of a polynomial has been one of the most intensively studied problems in mathematics. Among many irreducibility criteria for polynomials in Z[x], a classical result of A. Cohn [1] states that if we express a prime p in the decimal representation as
p=an10n+an−110n−1+⋯+a110+a0, |
then the polynomial f(x)=anxn+an−1xn−1+⋯+a1x+a0 is irreducible in Z[x]. This result was subsequently generalized to any base b by Brillhart et al. [2]. In 2002, Murty [3] gave another proof of this fact that was conceptually simpler than the one in [2].
In the present work, we are interested in studying the result of A. Cohn in any imaginary quadratic field. Let K=Q(√m) with a unique squarefree integer m≠1, be a quadratic field. We have seen that the quadratic field K is said to be real if m>0 and imaginary if m<0. The set of algebraic integers that lie in K is denoted by OK. Indeed,
OK={a+bσm∣a,b∈Z}, |
where
σm:={√mif m≢1 (mod 4),1+√m2if m≡1 (mod 4) |
[4]. Clearly, OQ(i)=Z[i], the ring of Gaussian integers, where i=√−1. It is well known that OK is an integral domain and K is its quotient field. Then the set of units in OK[x] is U(OK), the group of units in OK.
In general, we know that a prime element in OK is an irreducible element and the converse holds if OK is a unique factorization domain. A nonzero polynomial p(x)∈OK[x] is said to be irreducible in OK[x] if p(x) is not a unit and if p(x)=f(x)g(x) in OK[x], then either f(x) or g(x) is a unit in OK. Polynomials that are not irreducible are called reducible. For β=a+bσm∈OK, we denote the norm of β by
N(β)={a2−mb2 if m≢1 (mod 4),a2+ab+b2(1−m4) if m≡1 (mod 4). |
Clearly, N(β)∈Z for all β∈OK. To determine whether α∈OK is an irreducible element, we often use the fact that if N(α)=±p, where p is a rational prime, then α is an irreducible element [4].
For α,β∈OK with α≠0, we say that α divides β, denoted by α∣β, if there exists δ∈OK such that β=αδ. For α,β,γ∈OK with γ≠0, we say that α is congruent to β modulo γ and we write α≡β (mod γ), if γ∣(α−β). By a complete residue system modulo β in OK, abbreviated by CRS(β) [5], we mean a set of |N(β)| elements C={α1,α2,…,α|N(β)|} in OK, which satisfies the following.
(i) For each α∈OK, there exists αi∈C such that α≡αi (mod β).
(ii) For all i,j∈{1,2,…,|N(β)|} with i≠j, we have αi≢αj (mod β).
We have seen from [6] that
C={x+yi∣x=0,1,…,a2+b2d−1 and y=0,1,…,d−1} | (1.1) |
is a CRS(β), where β=a+bi∈Z[i] and d=gcd(a,b). It is clear that
C′:={x+yi∣x=0,1,…,max{|a|,|b|}−1 and y=0,1,…,d−1}⊆C. |
In 2017, Singthongla et al. [7] established the result of A. Cohn in OK[x], where K is an imaginary quadratic field such that OK is a Euclidean domain, namely m=−1,−2,−3,−7, and −11 [4]. Regarding the complete residue system (1.1), they established irreducibility criteria for polynomials in Z[i][x] as the following results.
Theorem A. [7] Let β∈{2±2i,1±3i,3±i} or β=a+bi∈Z[i] be such that |β|≥2+√2 and a≥1. For a Gaussian prime π, if
π=αnβn+αn−1βn−1+⋯+α1β+α0=:f(β), |
with n≥1, Re(αn)≥1, and α0,α1,…,αn−1∈C′ satisfying Re(αn−1)Im(αn)≥Re(αn)Im(αn−1), then f(x) is irreducible in Z[i][x].
In the proof of Theorem A in [7], the inequality
|β|≥3+√1+4M2, | (1.2) |
where M=√(max{a,|b|}−1)2+(d−1)2 is necessary. It can be verified that for β=a+bi∈Z[i], if |β|<2+√2 and a≥1, then β∈{3±i,2±2i,2±i,1±3i,1±2i,1±i,3,2,1}. It is clear that the Gaussian integers 2±2i, 1±3i, and 3±i satisfy (1.2), while 2±i,1±2i,1±i,3,2,1 do not. Consequently, we cannot apply Theorem A for these numbers. However, there is an irreducibility criterion for polynomials in Z[i][x] using β=3 in [7].
Theorem B. [7] If π is a Gaussian prime such that
π=αn3n+αn−13n−1+⋯+α13+α0, |
where n≥3, Re(αn)≥1, and α0,α1,…,αn−1∈C′ satisfying the conditions
Re(αn−1)Im(αn)≥Re(αn)Im(αn−1), |
Re(αn−2)Im(αn)≥Re(αn)Im(αn−2), |
Re(αn−2)Im(αn−1)≥Re(αn−1)Im(αn−2), |
then the polynomial f(x)=αnxn+αn−1xn−1+⋯+α1x+α0 is irreducible in Z[i][x].
In 2017, Tadee et al. [8] derived three explicit representations for a complete residue system in a general quadratic field K=Q(√m). We are interested in the first one and only the case m≢1 (mod 4) because the complete residue system in another case, m≡1 (mod 4) is inapplicable for our work. The CRS(β) for m≢1 (mod 4) in [8] is the set
C:={x+y√m∣x=0,1,…,|N(β)|d−1 and y=0,1,…,d−1}, | (1.3) |
where β=a+b√m and d=gcd(a,b).
Recently, Phetnun et al. [9] constructed a complete residue system in a general quadratic field K=Q(√m) for the case m≡1 (mod 4), which is similar to that in (1.3). They then determined the so-called base-β(C) representation in OK and generalized Theorem A for any imaginary quadratic field by using such representation. These results are as the following.
Theorem C. [9] Let K=Q(√m) be a quadratic field with m≡1 (mod 4). If β=a+bσm∈OK∖{0} with d=gcd(a,b), then the set
C={x+yσm∣x=0,1,…,|N(β)|d−1 and y=0,1,…,d−1} | (1.4) |
is a CRS(β).
From (1.3) and (1.4), we have shown in [9] for any m<0, that the set
C′:={x+yσm∣x=0,1,…,max{|a|,|b|}−1 and y=0,1,…,d−1}⊆C. | (1.5) |
Moreover, if d=1, then C′={0,1,…,max{|a|,|b|}−1}, while b=0 implies C′={x+yσm∣x,y=0,1,…,|a|−1}=C.
Definition A. [9] Let K=Q(√m) be an imaginary quadratic field. Let β∈OK∖{0} and C be a CRS(β). We say that η∈OK∖{0} has a base-β(C) representation if
η=αnβn+αn−1βn−1+⋯+α1β+α0, | (1.6) |
where n≥1, αn∈OK∖{0}, and αi∈C (i=0,1,…,n−1). If αi∈C′ (i=0,1,…,n−1), then (1.6) is called a base-β(C′) representation of η.
Theorem D. [9] Let K=Q(√m) be an imaginary quadratic field with m≢1 (mod 4). Let β=a+b√m∈OK be such that |β|≥2+√1−m and a≥1+√1−m. For an irreducible element π in OK with |π|≥|β|, if
π=αnβn+αn−1βn−1+⋯+α1β+α0=:f(β) |
is a base-β(C′) representation with Re(αn)≥1 satisfying Re(αn−1)Im(αn)≥Re(αn)Im(αn−1), then f(x) is irreducible in OK[x].
Theorem E. [9] Let K=Q(√m) be an imaginary quadratic field with m≡1 (mod 4). Let β=a+bσm∈OK be such that |β|≥2+√(9−m)/4, a≥1, and a+(b/2)≥1. For an irreducible element π in OK with |π|>√(9−m)/4(|β|−1), if
π=αnβn+αn−1βn−1+⋯+α1β+α0=:f(β) |
is a base-β(C′) representation with Re(αn)≥1 satisfying Re(αn−1)Im(αn)≥Re(αn)Im(αn−1), then f(x) is irreducible in OK[x].
In this work, we first establish further irreducibility criteria for polynomials in OK[x], where K=Q(√m) is an imaginary quadratic field, which extend Theorem D and Theorem E. We observe that the result for the case m≢1 (mod 4) is a generalization of Theorem B. Furthermore, we provide elements of β that can be applied to the new criteria but not to the previous ones.
In this section, we establish irreducibility criteria for polynomials in OK[x], where K is an imaginary quadratic field. To prove this, we first recall the essential lemmas in [7,10] as the following.
Lemma 1. [10] Let K=Q(√m) be an imaginary quadratic field. Then |β|≥1 for all β∈OK∖{0}.
We note for an imaginary quadratic field K that |α|=1 for all α∈U(OK).
Lemma 2. [7] Let f(x)=αnxn+αn−1xn−1+⋯+α1x+α0∈C[x] be such that n≥3 and |αi|≤M (0≤i≤n−2) for some real number M≥1. If f(x) satisfies the following:
(i) Re(αn)≥1, Re(αn−1)≥0, Im(αn−1)≥0, Re(αn−2)≥0, and Im(αn−2)≥0,
(ii) Re(αn−1)Im(αn)≥Re(αn)Im(αn−1),
(iii) Re(αn−2)Im(αn)≥Re(αn)Im(αn−2), and
(iv) Re(αn−2)Im(αn−1)≥Re(αn−1)Im(αn−2),
then any complex zero α of f(x) satisfies |α|<M1/3+0.465572 if |argα|≤π/6; otherwise
Re(α)<√32(1+√1+4M2). |
We note that the inequality |α|<M1/3+0.465572 appears in Lemma 2 follows from the proof of the lemma in [7] as follows: It was shown in [7] that
0=|f(α)αn|>|α|3−|α|2−M|α|2(|α|−1)=:h(|α|)|α|2(|α|−1), | (2.1) |
where h(x)=x3−x2−M. To obtain such inequality, the authors suppose to the contrary that |α|≥M1/3+0.465572. One can show that h(x) is increasing on (−∞,0)∪(2/3,∞). Since M1/3+0.465572>2/3, it follows that
h(|α|)≥h(M1/3+0.465572)=0.396716M2/3−0.280872138448M1/3−0.115841163475170752>0.396716M2/3−0.280873M1/3−0.115842=M1/3(0.396716M1/3−0.280873)−0.115842≥0.000001, since M≥1>0, |
which contradicts to (2.1).
Now, we proceed to our first main results. To obtain an irreducibility criterion for the case m≢1 (mod 4), we begin with the following lemma.
Lemma 3. Let K=Q(√m) be an imaginary quadratic field with m≢1 (mod 4). Let β=a+b√m∈OK be such that a>1 and
M:=√(max{a,|b|}−1)2−m(d−1)2, | (2.2) |
where d=gcd(a,b). Then M≥1.
Proof. If b=0, then M=√(a−1)2−m(a−1)2=√1−m(a−1)>1. Now, assume that b≠0 and we treat two separate cases.
Case 1: |b|≥a. Then M=√(|b|−1)2−m(d−1)2≥√(|b|−1)2=|b|−1≥1.
Case 2: |b|<a. Then M=√(a−1)2−m(d−1)2≥√(a−1)2=a−1≥1.
From every case, we conclude that M≥1.
By applying Lemmas 1–3, we have the following.
Theorem 1. Let K=Q(√m) be an imaginary quadratic field with m≢1 (mod 4). Let β=a+b√m∈OK be such that |β|≥M1/3+1.465572 and a≥1+(√3/2)((1+√1+4M)/2), where M is defined as in (2.2). For an irreducible element π in OK, if
π=αnβn+αn−1βn−1+⋯+α1β+α0=:f(β) |
is a base-β(C′) representation with n≥3 and Re(αn)≥1 satisfying conditions (ii)–(iv) of Lemma 2, then f(x) is irreducible in OK[x].
Proof. Suppose to the contrary that f(x) is reducible in OK[x]. Then f(x)=g(x)h(x) with g(x) and h(x) in OK[x]∖U(OK). We first show that either degg(x)≥1 and |g(β)|=1 or degh(x)≥1 and |h(β)|=1. It follows from degf(x)≥3 that g(x) or h(x) is a positive degree polynomial. If either degg(x)=0 or degh(x)=0, we may assume that h(x)=α∈OK. Then degg(x)=degf(x) and f(x)=αg(x) so that π=αg(β). Since π is an irreducible element and α∉U(OK), we obtain g(β)∈U(OK) and thus, |g(β)|=1. Otherwise, both degg(x)≥1 and degh(x)≥1, we have that π=g(β)h(β). Using the irreducibility of π again, we deduce that either g(β) or h(β) is a unit and hence, either |g(β)|=1 or |h(β)|=1, as desired.
We now assume without loss of generality that degg(x)≥1 and |g(β)|=1. We will show that this cannot happen. Note that M≥1 by Lemma 3. Moreover, since αi∈C′ for all i∈{0,1,…,n−1}, where C′ is defined as in (1.5), we have
|αi|≤|(max{a,|b|}−1)+(d−1)√m|=√(max{a,|b|}−1)2−m(d−1)2=M |
for all i∈{0,1,…,n−1}. Since degg(x)≥1, g(x) can be expressed in the form
g(x)=ε∏i(x−γi), |
where ε∈OK is the leading coefficient of g(x) and the product is over the set of complex zeros of g(x). It follows from Lemma 2 that any complex zero γ of g(x) satisfies either
|γ|<M1/3+0.465572 or Re(γ)<√32(1+√1+4M2). |
In the first case, it follows from |β|≥M1/3+1.465572 that
|β−γ|≥|β|−|γ|>|β|−(M1/3+0.465572)≥1. |
In the latter case, it follows from a≥1+(√3/2)((1+√1+4M)/2) that
|β−γ|≥Re(β−γ)=Re(β)−Re(γ)=a−Re(γ)>a−√32(1+√1+4M2)≥1. |
From both cases, by using Lemma 1, we obtain
1=|g(β)|=|ε|∏i|β−γi|≥∏i|β−γi|>1, |
which is a contradiction. This completes the proof.
By taking β=3 together with m=−1 in Theorem 1, we obtain Theorem B. This shows that Theorem 1 is a generalization of Theorem B. We will show in the next section that if β=a+bi∈Z[i]∖{0} with b=0, then β=3 is the only element that can be applied to Theorem 1.
Next, we illustrate the use of Theorem 1 by the following example.
Example 1. Let K=Q(√−5), β=3+√−5∈OK, and π=−9069−5968√−5. Then d=1 and so C′={0,1,2}. Note that M=√(3−1)2+5(1−1)2=2, |β|=√14>M1/3+1.465572, a=3>1+(√3/2)((1+√1+4M)/2), and π is an irreducible element because N(π)=(−9069)2+5(−5968)2=260331881 is a rational prime. Now, we have
π=(13+8√−5)β5+2β4+2β3+β2+2β+1 |
is its base-β(C′) representation with n=5 and Re(αn)=13 satisfying conditions (ii)–(iv) of Lemma 2.
By using Theorem 1, we obtain that
f(x)=(13+8√−5)x5+2x4+2x3+x2+2x+1 |
is irreducible in OK[x].
Note from Example 1 that we cannot apply Theorem D to conclude the irreducibility of the polynomial f(x) because |β|=|3+√−5|<2+√6=2+√1−m. Moreover, we see that a=3<1+√6=1+√1−m.
For the case m≡1 (mod 4), we start with the following lemma.
Lemma 4. Let K=Q(√m) be an imaginary quadratic field with m≡1 (mod 4). Let β=a+bσm∈OK be such that a+(b/2)≥1+(√3/2)((1+√1+4M)/2) and
M:=√(max{|a|,|b|}−1)2+(max{|a|,|b|}−1)(d−1)+(d−1)2(1−m4), | (2.3) |
where d=gcd(a,b). Then M≥1.
Proof. If b=0, then a≥1+(√3/2)((1+√1+4M)/2)>1. It follows that
M=√(a−1)2+(a−1)2+(a−1)2(1−m4)>√(a−1)2=a−1≥1. |
If a=0, then b/2≥1+(√3/2)((1+√1+4M)/2)>1. Thus, b>2 and so
M=√(b−1)2+(b−1)2+(b−1)2(1−m4)>√(b−1)2=b−1>1. |
Now, assume that |a|≥1 and |b|≥1. If |a|=1 and |b|=1, then M=0, yielding a contradiction because a+(b/2)≥1+(√3/2)((1+√1+4M)/2). Then |a|>1 or |b|>1. It follows from d≥1 that
M≥√(2−1)2+(2−1)(d−1)+(d−1)2(1−m4)≥√(2−1)2=1. |
By applying Lemmas 1, 2 and 4, we obtain an irreducibility criterion for the case m≡1 (mod 4) as the following theorem.
Theorem 2. Let K=Q(√m) be an imaginary quadratic field with m≡1 (mod 4). Let β=a+bσm∈OK be such that |β|≥M1/3+1.465572 and a+(b/2)≥1+(√3/2)((1+√1+4M)/2), where M is defined as in (2.3). For an irreducible element π in OK, if
π=αnβn+αn−1βn−1+⋯+α1β+α0=:f(β) |
is a base-β(C′) representation with n≥3 and Re(αn)≥1 satisfying conditions (ii)–(iv) of Lemma 2, then f(x) is irreducible in OK[x].
Proof. Suppose to the contrary that f(x) is reducible in OK[x]. Then f(x)=g(x)h(x) with g(x) and h(x) in OK[x]∖U(OK). It can be proved similarly to the proof of Theorem 1 that either degg(x)≥1 and |g(β)|=1 or degh(x)≥1 and |h(β)|=1. We may assume without loss of generality that degg(x)≥1 and |g(β)|=1. We will show that this cannot happen. By Lemma 4, we have M≥1. For i∈{0,1,…,n−1}, since αi∈C′, it follows from the definition of C′ in (1.5) that
|αi|≤|(max{|a|,|b|}−1)+(d−1)(1+√m2)|=|((max{|a|,|b|}−1)+d−12)+(d−12)√m|=√(max{|a|,|b|}−1)2+(max{|a|,|b|}−1)(d−1)+(d−1)2(1−m4)=M. |
The remaining proof is again similar to that of Theorem 1 by using Lemmas 1, 2 and Re(β)=a+(b/2).
We illustrate the use of Theorem 2 by the following example.
Example 2. Let K=Q(√−3), β=4−σ−3, and π=359−278σ−3. Then d=1 and so C′={0,1,2,3}. Note that M=√(4−1)2+(4−1)(1−1)+(1−1)2=3, |β|=√13>M1/3+1.465572, a+(b/2)=3.5>1+(√3/2)((1+√1+4M)/2), and π is an irreducible element because N(π)=3592−359⋅278+(−278)2=106363 is a rational prime. Now, we have
π=β4+3β3+β2+2β+1 |
is its base-β(C′) representation with n=4 and Re(αn)=1 satisfying conditions (ii)–(iv) of Lemma 2.
By using Theorem 2, we obtain that
f(x)=x4+3x3+x2+2x+1 |
is irreducible in OK[x].
From Example 2, we emphasize that we cannot apply Theorem E to conclude the irreducibility of the polynomial f(x) because |β|=|4−σ−3|<2+√3=2+√(9−m)/4, although a=4>1 and a+(b/2)=4−(1/2)>1.
Let K=Q(√m) be an imaginary quadratic field. In this section, we will try to find elements of β=a+bσm∈OK∖{0} that can be applied to Theorem 1, respectively, Theorem 2 but not to Theorem D, respectively, Theorem E. We are only interested in two cases, namely b=0 and b≠0 with d=gcd(a,b)=1 because the remaining case, b≠0 with d>1 requires us to solve a multi-variable system of inequalities, which is more complicated. To proceed with this objective, we begin with the following remarks.
Remark 1. Let a and m be integers with m<0. Then the following statements hold.
(i) a≥1+√32(1+√1+4(a−1)2) if and only if a≥3.
(ii) a≥1+√32(1+√1+4√1−m(a−1)2) if and only if a≥4+2√3+3√1−m4.
(iii) a≥1+√32(1+√1+4√(9−m)/4(a−1)2) if and only if a≥4+2√3+3√(9−m)/44.
Proof. For convenience, we let A=a−1. We have for any real number x>0 that
a≥1+√32(1+√1+4x(a−1)2) ifandonlyif A≥√34(1+√1+4xA), ifandonlyif (4√3A3−1)2≥1+4xA, ifandonlyif 16A23−(8√3+12x)A3≥0, ifandonlyif A[4A−(2√3+3x)]≥0, ifandonlyif 4A−(2√3+3x)≥0, ifandonlyif A≥2√3+3x4, ifandonlyif a≥4+2√3+3x4. | (3.1) |
Substituting x=1, x=√1−m, and x=√(9−m)/4 in (3.1) lead to (i)–(iii), respectively, as desired.
To compare Theorem 1 with Theorem D and to compare Theorem 2 with Theorem E, we require the following remark.
Remark 2. For any real number x, the following statements hold.
(i) 4+2√3+3√x4≥(x+√x)1/3+1.465572 for all x∈[3,∞).
(ii) √x2+5≥(x−1)1/3+1.465572 for all x∈[1,∞).
(iii) √3x+1≥(x−1)1/3+1.465572 for all x∈[1,∞).
(iv) √x22+1≥(x−1)1/3+1.465572 for all x∈[4,∞).
(v) x≥(√2(x−1))1/3+1.465572 for all x∈[2.85,∞).
(vi) √x2+1≥(x−1)1/3+1.465572 for all x∈[3,∞).
(vii) √−73−121x>4+√9−x for all x∈(−∞,−2].
(viii) √29−9x>4+√9−x for all x∈(−∞,−3].
Proof of Remark 2. By using the WolframAlpha computational intelligence (www.wolframalpha.com), it can be verified by considering the graphs of both left and right functions of each inequality.
Let K=Q(√m) be an imaginary quadratic field with m≢1 (mod 4). In this subsection, we will find elements of β∈OK∖{0} that can be applied to Theorem 1 but not to Theorem D. Now, let β=a+b√m be a nonzero element in OK that can be applied to Theorem 1 but not to Theorem D. Then |β|≥M1/3+1.465572 and a≥1+(√3/2)((1+√1+4M)/2), where M is defined as in (2.2). Since β cannot be applied to Theorem D, one can consider two possible cases, namely, |β|<2+√1−m or |β|≥2+√1−m as follows:
Case A: |β|<2+√1−m. Then, we now try to find elements of β that satisfy the following inequality system:
|β|<2+√1−m|β|≥M1/3+1.465572a≥1+√32(1+√1+4M2). | (3.2) |
We consider two cases as follows:
Case 1: b=0. Then β=a and M=√(a−1)2−m(a−1)2=√1−m(a−1). Thus, the system (3.2) becomes
a<2+√1−m | (3.3) |
a≥(√1−m(a−1))1/3+1.465572 | (3.4) |
a≥1+√32(1+√1+4√1−m(a−1)2). | (3.5) |
By (3.5) and Remark 1(ii), we have a≥(4+2√3+3√1−m)/4, which together with (3.3) yield
4+2√3+3√1−m4≤a<2+√1−m. | (3.6) |
To show that the integers β=a satisfying (3.6) are solutions of the system above, we must show that they also satisfy (3.4). If m=−1, then a≥(4+2√3+3√2)/4≈2.93. It follows from Remark 2(v) with x=a that a≥(√2(a−1))1/3+1.465572=(√1−m(a−1))1/3+1.465572. Assume that m≤−2. By taking x=1−m in Remark 2(i), we obtain that
4+2√3+3√1−m4≥(1−m+√1−m)1/3+1.465572=(√1−m(2+√1−m−1))1/3+1.465572>(√1−m(a−1))1/3+1.465572, by (3.3), |
implying (3.4).
We note for m=−1 that the inequality (3.6) implies a=3. Hence, β=3∈Z[i] is the only element that can be applied to Theorem 1 but not to Theorem D.
Case 2: b≠0 and d=1. There are two further subcases:
Subcase 2.1: |b|≥a. Then |β|=√a2−mb2 and M=√(|b|−1)2=|b|−1. Thus, the system (3.2) becomes
√a2−mb2<2+√1−m | (3.7) |
√a2−mb2≥(|b|−1)1/3+1.465572a≥1+√32(1+√1+4(|b|−1)2). | (3.8) |
Since |b|≥a, we obtain from (3.8) that a≥1+(√3/2)((1+√1+4(a−1))/2). Using Remark 1(i), we have that a≥3. It follows from |b|≥a, a≥3, and m≤−1 that
√a2−mb2≥√a2−ma2=√a2(1−m)≥√9(1−m)=3√1−m>2+√1−m, |
which is contrary to (3.7). Thus, the system above has no integer solution (a,b). This means that the assumptions in the system generate no pairs (a,b) that are solutions to Theorem 1 and that are also not solutions to Theorem D.
Subcase 2.2: |b|<a. Then |β|=√a2−mb2 and M=√(a−1)2=a−1. Thus, the system (3.2) becomes
√a2−mb2<2+√1−m | (3.9) |
√a2−mb2≥(a−1)1/3+1.465572 | (3.10) |
a≥1+√32(1+√1+4(a−1)2). | (3.11) |
Using Remark 1(i) and (3.11), we have a≥3. Since m≤−1, we obtain (6−5m)2=25m2−60m+36>16m2−52m+36=4(9−4m)(1−m), yielding 6−5m>2√(9−4m)(1−m). It follows that
(√9−4m−√1−m)2=10−5m−2√(9−4m)(1−m)>4 |
and so √9−4m−√1−m>2. If |b|≥2, then √a2−mb2≥√9−4m>2+√1−m, which is contrary to (3.9). Thus, |b|=1. Using (3.9) and a≥3, we have √9−m≤√a2−m<2+√1−m and so 9≤a2<5+4√1−m, i.e., 3≤a<√5+4√1−m. We next show that the pairs (a,b) with
3≤a<√5+4√1−m and b=±1 | (3.12) |
also satisfy (3.10). Since |b|=1, a≥3, and Remark 2(vi) with x=a, we have
√a2−mb2=√a2−m≥√a2+1≥(a−1)1/3+1.465572, |
yielding (3.10). Thus, we conclude that the pairs (a,b) satisfying (3.12) are solutions of the system above.
Case B: |β|≥2+√1−m. Since we cannot apply the element β to Theorem D, we have a<1+√1−m. Now, we try again to find elements of β that satisfy the following inequality system:
|β|≥2+√1−ma<1+√1−m|β|≥M1/3+1.465572a≥1+√32(1+√1+4M2). | (3.13) |
We consider two cases as follows:
Case 1: b=0. Then a<1+√1−m<2+√1−m≤|β|=a, which is a contradiction. Hence, the system (3.13) has no integer solution β=a. In other words, the assumptions in the system generate no pairs (a,b) that are solutions to Theorem 1 and that are also not solutions to Theorem D.
Case 2: b≠0 and d=1. There are two further subcases:
Subcase 2.1: |b|≥a. Then |β|=√a2−mb2 and M=√(|b|−1)2=|b|−1. Thus, the system (3.13) becomes
√a2−mb2≥2+√1−m | (3.14) |
a<1+√1−m | (3.15) |
√a2−mb2≥(|b|−1)1/3+1.465572 | (3.16) |
a≥1+√32(1+√1+4(|b|−1)2). | (3.17) |
Since |b|≥a, we obtain from (3.17) that a≥1+(√3/2)((1+√1+4(a−1))/2). It follows from Remark 1(i) that a≥3. Since d=1, we have |b|>a. By using (3.15) together with a≥3, we have 3≤a<1+√1−m, implying m≤−5. It can be verified by using (3.17) that |b|≤((4√3(a−1)−3)2+27)/36. Now, we have that
3≤a<1+√1−m and a<|b|≤(4√3(a−1)−3)2+2736. | (3.18) |
To show that the pairs (a,b) satisfying (3.18) are solutions of the system, it remains to show that they also satisfy (3.14) and (3.16). Since |b|>a≥3 and m<0, we obtain
√a2−mb2>√a2−ma2=a√1−m≥3√1−m>2+√1−m, |
yielding (3.14). From Remark 2(ii) with x=|b|, we have
√a2−mb2>√5+b2≥(|b|−1)1/3+1.465572, |
showing (3.16).
Subcase 2.2: |b|<a. Then |β|=√a2−mb2 and M=√(a−1)2=a−1. Thus, the system (3.13) becomes
√a2−mb2≥2+√1−m | (3.19) |
a<1+√1−m | (3.20) |
√a2−mb2≥(a−1)1/3+1.465572 | (3.21) |
a≥1+√32(1+√1+4(a−1)2). | (3.22) |
Again, using Remark 1(i) and (3.22), we obtain a≥3. By using (3.20) together with a≥3, we have 3≤a<1+√1−m, implying m≤−5. Using (3.19), we can verify that |b|≥√(5−m+4√1−m−a2)/(−m). Now, we have that
3≤a<1+√1−m and √5−m+4√1−m−a2−m≤|b|<a. | (3.23) |
To show that the pairs (a,b) satisfying (3.23) are solutions of the system, it remains to show that they also satisfy (3.21). It follows from b2≥1, m≤−5, and Remark 2(ii) with x=a that
√a2−mb2≥√a2+5≥(a−1)1/3+1.465572, |
yielding (3.21).
From every case, we conclude that elements of β=a+b√m∈OK∖{0} with m≢1 (mod 4) that can be applied to Theorem 1 but not to Theorem D are shown in the following tables.
We note from Subcase 2.2 in Table 1 that the number of a roughly grows as 24√1−m. To see this, since 84√1−m>1, we have
5+4√1−m<4√1−m+84√1−m+4=(24√1−m+2)2 |
and so 3≤a<√5+4√1−m<24√1−m+2. This means that the number of such a is approximately 24√1−m.
β=a+b√m, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | 4+2√3+3√1−m4≤a<2+√1−m and b=0 |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥a | none |
Subcase 2.2: |b|<a | 3≤a<√5+4√1−m and b=±1 |
We note from Table 2 that the complicated lower bound in Subcase 2.2 is actually very close to 1. Indeed, we show that
√5−m+4√1−m−a2−m<2. |
Since m≤−1, it follows that
(4−3m)2−16(1−m)=(9m2−24m+16)−16+16m=9m2−8m=m(9m−8)>0, |
showing (4−3m)2>16(1−m) and so 4−3m>4√1−m. Using 3≤a<1+√1−m, we have that −2+m−2√1−m<−a2≤−9. It follows that
0<3+2√1−m−m=(5−m+4√1−m)+(−2+m−2√1−m)−m<(5−m+4√1−m)−a2−m≤(5−m+4√1−m)−9−m=−4−m+4√1−m−m<−4−m+(4−3m)−m=4. |
This shows that √(5−m+4√1−m−a2)/(−m)<√4=2, as desired.
β=a+b√m, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | none |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥a | 3≤a<1+√1−m and a<|b|≤(4√3(a−1)−3)2+2736 |
Subcase 2.2: |b|<a | 3≤a<1+√1−m and √5−m+4√1−m−a2−m≤|b|<a |
Let K=Q(√m) be an imaginary quadratic field with m≡1 (mod 4). In this subsection, we find elements of β∈OK∖{0} that can be applied to Theorem 2 but not to Theorem E. Now, let β=a+bσm be a nonzero element in OK that can be applied to Theorem 2 but not to Theorem E. Then |β|≥M1/3+1.465572 and a+(b/2)≥1+(√3/2)((1+√1+4M)/2), where M is defined as in (2.3). Since β cannot be applied to Theorem E, one can consider two possible cases, namely, |β|<2+√(9−m)/4 or |β|≥2+√(9−m)/4 as follows:
Case A: |β|<2+√(9−m)/4. Then we will find elements of β that satisfy the inequality system:
|β|<2+√9−m4|β|≥M1/3+1.465572a+b2≥1+√32(1+√1+4M2). | (3.24) |
We consider two cases as follows:
Case 1: b=0. Then β=a and M=√(a−1)2+(a−1)(a−1)+(a−1)2(1−m)/4=√(9−m)/4(a−1). Thus, the system (3.24) becomes
a<2+√9−m4 | (3.25) |
a≥(√9−m4(a−1))1/3+1.465572 | (3.26) |
a≥1+√32(1+√1+4√(9−m)/4(a−1)2). | (3.27) |
By (3.27) and Remark 1(iii), we have that a≥(4+2√3+3√(9−m)/4)/4, which together with (3.25) yield
4+2√3+3√(9−m)/44≤a<2+√9−m4. | (3.28) |
To show that the integers β=a satisfying (3.28) are solutions of the system above, we must show that they also satisfy (3.26). By taking x=(9−m)/4 in Remark 2(i) and using (3.25), we obtain that
4+2√3+3√(9−m)/44≥(9−m4+√9−m4)1/3+1.465572=[√9−m4(2+√9−m4−1)]1/3+1.465572>(√9−m4(a−1))1/3+1.465572. | (3.29) |
It follows from (3.28) and (3.29) that a>(√(9−m)/4(a−1))1/3+1.465572, yielding (3.26).
Case 2: b≠0 and d=1. There are two further subcases:
Subcase 2.1: |b|≥|a|. Then |β|=√a2+ab+b2(1−m)/4 and M=√(|b|−1)2=|b|−1. Thus, the system (3.24) becomes
√a2+ab+b2(1−m4)<2+√9−m4 | (3.30) |
√a2+ab+b2(1−m4)≥(|b|−1)1/3+1.465572a+b2≥1+√32(1+√1+4(|b|−1)2). | (3.31) |
In this subcase, we now show that the system has no integer solution (a,b). If a<0, then it follows from (3.31) that b>0 and so (b/2)−1≥1+(√3/2)((1+√1+4(b−1))/2). Then b2−(11+√3)b+(19+4√3)≥0, implying b≥11. It follows from a2≥1, a>1−(b/2), b≥11, and Remark 2(vii) with x=m that
√a2+ab+b2(1−m4)>√1+(1−b2)b+b2(1−m4)=√b2(−1−m4)+b+1≥√−121−121m+484=12√−73−121m>12(4+√9−m)=2+√9−m4, | (3.32) |
which is contrary to (3.30). Thus, a≥0. If a=0, then |b|=1 because d=1. This contradicts to (3.31), so a≥1. If |b|=1, then a=1 and so (3.31) is false. Thus, |b|≥2 and so |b|>a because d=1. It follows from (3.31) and |b|≥2 that a+(b/2)>2.4 and so |b|+(b/2)>2.4. This implies that b≥2 or b≤−5. If b=2, then we obtain that 2=|b|>a≥(√3/2)((1+√1+4(2−1))/2)>1.4, which is a contradiction. If b=3, then we obtain that 3=|b|>a≥(√3/2)((1+√1+4(3−1))/2)−(1/2)>1.2, which implies that a=2. It follows that
√a2+ab+b2(1−m4)=√22+2⋅3+32(1−m4)=14(√49−9m+√49−9m)>14(8+√36−4m)=2+√9−m4, |
which is contrary to (3.30). If b≥4, then
√a2+ab+b2(1−m4)≥√1+4+16(1−m4)=12(√9−4m+√9−4m)>12(4+√9−m)=2+√9−m4, |
which is contrary to (3.30). If b≤−5, then
a−52≥a+b2≥1+√32(1+√1+4(|b|−1)2)≥1+√32(1+√1+4(5−1)2)>3.22, |
showing a≥6. Since b≤−5 and a≥6, it follows from −b=|b|>a that
√a2+ab+b2(1−m4)>√a2−b2+b2(1−m4)=√b2(1−m4−1)+a2≥√25(1−m4−1)+36=14(√69−25m+√69−25m)>14(8+√36−4m)=2+√9−m4, |
which is contrary to (3.30).
Thus, in this subcase, we conclude that the assumptions in the system generate no pairs (a,b) that are solutions to Theorem 2 and that are also not solutions to E.
Subcase 2.2: |b|<|a|. Then |β|=√a2+ab+b2(1−m)/4 and M=√(|a|−1)2=|a|−1. Thus, the system (3.24) becomes
√a2+ab+b2(1−m4)<2+√9−m4 | (3.33) |
√a2+ab+b2(1−m4)≥(|a|−1)1/3+1.465572 | (3.34) |
a+b2≥1+√32(1+√1+4(|a|−1)2). | (3.35) |
If a<0, then it follows from a+(b/2)>1 that b>0. Since |a|>|b|=b and (3.35), we obtain (b/2)−1≥a+(b/2)>1+(√3/2)((1+√1+4(b−1))/2), implying b≥11. Now, we have that a2>1, a>1−(b/2), and b≥11. It can be proved similarly to (3.32) that
√a2+ab+b2(1−m4)>2+√9−m4, |
which is contrary to (3.33). Thus, a≥0. If a=0 or a=1, then 0<|b|<|a|≤1, which is impossible so that a≥2. If b=−1, then it follows from (3.35) that a−(1/2)≥1+(√3/2)((1+√1+4(a−1))/2), implying a≥4. By taking x=a in Remark 2(iii), we have
√a2−a+1−m4=√a(a−1)+1−m4≥√3a+1≥(a−1)1/3+1.465572, |
yielding (3.34). It can be verified by (3.33) with b=−1 that a<(√8√9−m+25+1)/2. This shows that
4≤a<√8√9−m+25+12, when b=−1. | (3.36) |
If b=1, then it follows from (3.35) that a+(1/2)≥1+(√3/2)((1+√1+4(a−1))/2), implying a≥2. By taking x=a in Remark 2(iii), we have that
√a2+a+1−m4=√a(a+1)+1−m4≥√3a+1≥(a−1)1/3+1.465572, |
yielding (3.34). It can be verified by (3.33) with b=1 that a<(√8√9−m+25−1)/2 and thus
2≤a<√8√9−m+25−12, when b=1. | (3.37) |
We next show for b≥2 or b≤−2 that the system above has no integer solution (a,b). If b≥2, then a=|a|>|b|=b≥2 and so a≥3. It follows that
√a2+ab+b2(1−m4)≥√32+3⋅2+22(1−m4)=12(√16−m+√16−m)>12(4+√9−m)=2+√9−m4, |
which is contrary to (3.33). If b=−2, then we obtain from (3.35) that a−1≥1+(√3/2)((1+√1+4(a−1))/2), implying a≥4. Since d=1 and b=−2, we have that a≥5. Hence,
√a2−2a+1−m=√a(a−2)+1−m≥√5(3)+1−m=12(√16−m+√16−m)>12(4+√9−m)=2+√9−m4, |
which is contrary to (3.33). If b≤−3, then we have a−(3/2)≥a+(b/2)≥1+(√3/2)((1+√1+4(a−1))/2). This implies that a≥5. Since a>|b|=−b, we obtain that −b≤a−1 and so ab≥−a2+a. It follows from b≤−3, a≥5, ab≥−a2+a, and Remark 2(viii) with x=m that
√a2+ab+b2(1−m4)≥√a2−a2+a+b2(1−m4)≥√9(1−m4)+5=12√29−9m>12(4+√9−m)=2+√9−m4, |
which is contrary to (3.33).
Thus, in this subcase, we obtain that the pairs (a,b) with b≠0 and d=1 satisfying (3.36) or (3.37) are integer solutions of the system (3.24).
Case B: |β|≥2+√(9−m)/4. Since a+(b/2)>1 and we cannot apply β to Theorem E, it follows that a<1. Thus, we have to find elements of β that satisfy the following inequality system:
|β|≥2+√9−m4, a<1|β|≥M1/3+1.465572a+b2≥1+√32(1+√1+4M2). | (3.38) |
Note that M≥1 by Lemma 4. Then b/2≥1+(√3/2)((1+√5)/2)>2.4 and so b≥5. If b<|a|, then a≤−6 and so a+(b/2)<a+b<a+|a|=0, which is a contradiction. Thus, b≥|a|=−a and so M=√(b−1)2=b−1. Hence, the system (3.38) becomes
√a2+ab+b2(1−m4)≥2+√9−m4, a<1 | (3.39) |
√a2+ab+b2(1−m4)≥(b−1)1/3+1.465572 | (3.40) |
a+b2≥1+√32(1+√1+4(b−1)2). | (3.41) |
Since b≥5 and d=1, we have a≤−1. It follows by (3.41) that (b/2)−1≥1+(√3/2)((1+√1+4(b−1))/2), implying b≥11. Note that b≥−a, b≥11, and d=1 imply b>−a. That is, −b<a≤−1. Now, we have that
b≥11 and 1+√32(1+√1+4(b−1)2)−b2≤a≤−1 | (3.42) |
To show that the pairs (a,b) satisfying (3.42) are solutions of the system, it remains to show that they also satisfy (3.39) and (3.40). Since a2≥1, a>1−(b/2), and b≥11, we obtain by Remark 2(vii) with x=m that
√a2+ab+b2(1−m4)>√1+(1−b2)b+b2(1−m4)=√b2(−1−m4)+b+1≥√121(−1−m4)+12=12√−73−121m>12(4+√9−m)=2+√9−m4, |
showing (3.39). It follows from a2≥1, a>1−(b/2), m≤−3, and Remark 2(iv) with x=b that
√a2+ab+b2(1−m4)>√1+(1−b2)b+b2>√b22+1≥(b−1)1/3+1.465572, |
yielding (3.40), as desired.
From every case, we conclude that elements of β=a+bσm∈OK∖{0} with m≡1 (mod 4) that can be applied to Theorem 2 but not to Theorem E are shown in the following tables.
We note from Subcase 2.2 in Table 3 that when b=−1, the number of a roughly grows as 4√4(9−m). Otherwise, b=1 implies that the number of a roughly grows as 4√4(9−m)+1. To see these, one can see that
8√9−m+25<8√9−m+204√4(9−m)+25=(24√4(9−m)+5)2 |
and so √8√9−m+25<24√4(9−m)+5. If b=−1, then
4≤a<√8√9−m+25+12<24√4(9−m)+62=4√4(9−m)+3, |
showing that the number of such a is approximately 4√4(9−m). If b=1, we obtain
2≤a<√8√9−m+25−12<24√4(9−m)+42=4√4(9−m)+2, |
showing that the number of such a is approximately 4√4(9−m)+1.
β=a+bσm, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | 4+2√3+3√(9−m)/44≤a<2+√9−m4 and b=0 |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥|a| | none |
Subcase 2.2: |b|<|a| | 4≤a<√8√9−m+25+12, when b=−1, |
2≤a<√8√9−m+25−12, when b=1 |
From Table 4, one can verify that if b≥|a| and d=1, then b≥11 and
4.2−b2≈1+√32(1+√1+4(11−1)2)−b2≤1+√32(1+√1+4(b−1)2)−b2≤a≤−1. |
This implies that the number of possible values of a is at most ⌊(b/2)−4.2⌋, the greatest integer less than or equal to (b/2)−4.2.
β=a+bσm, d=gcd(a,b) | Integer solutions (a,b) |
b<|a| | none |
b≥|a| and d=1 | b≥11 and 1+√32(1+√1+4(b−1)2)−b2≤a≤−1 |
Let K=Q(√m) be an imaginary quadratic field with OK its ring of integers. In this paper, further irreducibility criteria for polynomials in OK[x] are established which extend the authors' earlier works (Theorems D and E). Moreover, elements of β∈OK that can be applied to the new criteria but not to the previous ones are also provided.
This work was supported by the Science Achievement Scholarship of Thailand (SAST) and Department of Mathematics, Faculty of Science, Khon Kaen University, Fiscal Year 2022.
All authors declare no conflicts of interest in this paper.
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β=a+b√m, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | 4+2√3+3√1−m4≤a<2+√1−m and b=0 |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥a | none |
Subcase 2.2: |b|<a | 3≤a<√5+4√1−m and b=±1 |
β=a+b√m, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | none |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥a | 3≤a<1+√1−m and a<|b|≤(4√3(a−1)−3)2+2736 |
Subcase 2.2: |b|<a | 3≤a<1+√1−m and √5−m+4√1−m−a2−m≤|b|<a |
β=a+bσm, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | 4+2√3+3√(9−m)/44≤a<2+√9−m4 and b=0 |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥|a| | none |
Subcase 2.2: |b|<|a| | 4≤a<√8√9−m+25+12, when b=−1, |
2≤a<√8√9−m+25−12, when b=1 |
β=a+bσm, d=gcd(a,b) | Integer solutions (a,b) |
b<|a| | none |
b≥|a| and d=1 | b≥11 and 1+√32(1+√1+4(b−1)2)−b2≤a≤−1 |
β=a+b√m, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | 4+2√3+3√1−m4≤a<2+√1−m and b=0 |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥a | none |
Subcase 2.2: |b|<a | 3≤a<√5+4√1−m and b=±1 |
β=a+b√m, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | none |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥a | 3≤a<1+√1−m and a<|b|≤(4√3(a−1)−3)2+2736 |
Subcase 2.2: |b|<a | 3≤a<1+√1−m and √5−m+4√1−m−a2−m≤|b|<a |
β=a+bσm, d=gcd(a,b) | Integer solutions (a,b) |
Case 1: b=0 | 4+2√3+3√(9−m)/44≤a<2+√9−m4 and b=0 |
Case 2: b≠0 and d=1 | |
Subcase 2.1: |b|≥|a| | none |
Subcase 2.2: |b|<|a| | 4≤a<√8√9−m+25+12, when b=−1, |
2≤a<√8√9−m+25−12, when b=1 |
β=a+bσm, d=gcd(a,b) | Integer solutions (a,b) |
b<|a| | none |
b≥|a| and d=1 | b≥11 and 1+√32(1+√1+4(b−1)2)−b2≤a≤−1 |