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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# Groups and monoids of Pythagorean triples connected to conics

/ Marco Abrate
/ Stefano Barbero
/ Umberto Cerruti
Published Online: 2017-11-10 | DOI: https://doi.org/10.1515/math-2017-0111

## Abstract

We define operations that give the set of all Pythagorean triples a structure of commutative monoid. In particular, we define these operations by using injections between integer triples and 3 × 3 matrices. Firstly, we completely characterize these injections that yield commutative monoids of integer triples. Secondly, we determine commutative monoids of Pythagorean triples characterizing some Pythagorean triple preserving matrices. Moreover, this study offers unexpectedly an original connection with groups over conics. Using this connection, we determine groups composed by Pythagorean triples with the studied operations.

MSC 2010: 11A99

## 1 Introduction

Algebraic structures, like groups and monoids, are widely studied in mathematics for their importance. It is very interesting to give particular sets an algebraic structure, in order to deepen their study. The set of Pythagorean triples and their properties have been extensively studied for several years and aroused the interests of many mathematicians. In spite of these studies, it is a hard work to give Pythagorean triples algebraic structures. In [1] and [2], an operation over Pythagorean triples has been studied, showing a group of primitive Pythagorean triples. In [3] and [4], a monoid structure has been provided.

In this paper, we show an original approach that allows to characterize a family of operations which determine commutative monoids of Pythagorean triples. In section 2, we define injections, which we call natural, between integer triples and 3 × 3 matrices. We use these injections to define products between integer triples, focusing on Pythagorean ones. In section 3, we characterize matrices generating natural injections that determine commutative monoids over ℤ3. Similarly, in section 4, we characterize a subset of Pythagorean triple preserving matrices, such that the set of all Pythagorean triples has a commutative monoid structure. Finally, in section 5, we find a surprising connection between Pythagorean triples and conics. In this way, we determine a group composed by Pythagorean triples by means of a morphism between points over conics and Pythagorean triples. Section 6 is devoted to conclusions.

## 2 Commutative monoids of Pythagorean triples

Let 𝓟 be the set of Pythagorean triples $P={(x,y,z)∈Z3:x2+y2=z2}.$

Several operations can be defined over this set. In particular, Taussky [1] and Eckert [2] studied the following operation $(x1,y1,z1)⋅(x2,y2,z2)=(x1x2−y1y2,x1y2+y1x2,z1z2)$

which defines a structure of free abelian group over primitive Pythagorean triples. In [5], this result has been generalized from ℤ to any ring of integers of an algebraic number field. In [3] and [4], the following operation has been studied $(x1,y1,z1)⋅(x2,y2,z2)=(x1x2,y1z2+y2z1,y1y2+z1z2),$

determining a commutative monoid (i.e., a commutative semigroup with identity) over 𝓟.

Here, we study new operations on Pythagorean triples starting from an injection between triples and matrices. These operations provide new commutative monoid structures over 𝓟. Let ν be an injective function $ν:Z3→M3,$

where M3 is the set of 3 × 3 matrices whose entries belong to ℤ. Clearly, ν can be used to induce products over ℤ3. Let us denote with ”.” the usual matrix product. One natural way to induce a product over ℤ3 is given by $a∙1b=ν−1(ν(a).ν(b)),∀a,b∈Z3,$

whenever ν is invertible. A second natural way is given by $a∙2b=ν(a).b,∀a,b∈Z3.$

Let us consider the following definition.

#### Definition 2.1

We say that injection ν is natural when

1. ν(ν(a).b) = ν(a).ν(b), ∀ a, b ∈ ℤ3

2. ν(a).b = ν(b).a, ∀ a, b ∈ ℤ3

3. ν(1,0,1) = I3 (identity matrix)

#### Remark 2.2

We consider an element a ∈ ℤ3 like a column vector or a row vector in order that products between a and matrices are consistent.

In the following, we are interested in natural ν. Thus, we will consider the induced product ∙ = ∙1 = ∙2, i.e., $a∙b=ν(a).b=ν−1(ν(a).ν(b)),∀a,b∈Z3.$

If ν is natural, then (ℤ3, ∙) is a commutative monoid, whose identity is (1, 0, 1).

We also want to determine when the product ∙ preserves Pythagorean triples, i.e., $∀a,b∈P,ν(a).b∈P.$

This study is clearly related to research Pythagorean triple preserving matrices.

#### Definition 2.3

A matrix A is called a Pythagorean triple preserving matrix (PTPM) if given a Pythagorean triple a, then A. a is still a Pythagorean triple.

In [6] and [7], the authors characterized PTPMs. Precisely, they are of the form $12(r2−t2−s2+u2)rs−tu12(r2−t2+s2−u2)rt−suru+strt+su12(r2+t2−s2−u2)rs+tu12(r2+t2+s2+u2)$

for some values of r, s, t, u. Further results can be found in [8]. Moreover, Tikoo [9] studied special cases of PTPMs provided by matrices $B1=x000zy0yzB2=y000zx0xzB3=z0x0y0x0zB4=z0y0x0y0zB5=−xy0yx000z$

for (x, y, z) a given Pythagorean triple. All these matrices induce products that preserve Pythagorean triples. In other words, we can consider, e.g., the injection $νB1:Z3→M3,νB1(x,y,z)=B1.$

Thus, the induced product ∙B1 preserves Pythagorean triples, i.e., $∀a,b∈P,a∙B1b∈P.$

Moreover, with a bit of calculation, it is possible to check that (𝓟, ∙B1) is a commutative monoid or equivalently νB1 is natural. Similarly, we can consider the injection νB2 and the product ∙B2. Clearly, ∙B2 preserves Pythagorean triples. However, ∙B2 is not commutative nor associative and there is not the identity. In this case (𝓟, ∙B2) is not a commutative monoid or equivalently νB2 is not natural.

Thus, it is natural to ask when PTPMs determine commutative and associative products with identity. In the following we answer to this question finding a characterization for these matrices.

## 3 Matrices yielding commutative monoids of triples

In this section we determine all matrices that yield natural injections ν. In other words, we determine all matrices such that (ℤ3, ∙) is a commutative monoid whose identity is (1,0,1). Using previous notation, we have that ν(x, y, z)∈ M3 has entries depending on triple (x, y, z).

We consider generic 3 × 3 matrices whose entries are functions of (x, y, z): $ν(x,y,z)=a(x,y,z)b(x,y,z)c(x,y,z)d(x,y,z)e(x,y,z)f(x,y,z)g(x,y,z)h(x,y,z)i(x,y,z).$

Triple (1,0,1) is the identity with respect to ∙ if and only if $(x,y,z)∙(1,0,1)=(1,0,1)∙(x,y,z)=(x,y,z),∀(x,y,z)∈Z3$

i.e., if and only if $ν(x,y,z).(1,0,1)=ν(1,0,1).(x,y,z)=(x,y,z).$

Direct calculations show that these equalities are satisfied if and only if $ν(x,y,z)=a(x,y,z)b(x,y,z)x−a(x,y,z)d(x,y,z)e(x,y,z)y−d(x,y,z)g(x,y,z)h(x,y,z)z−g(x,y,z).$

Moreover, ∙ is commutative if and only if $(x,y,z)∙(r,s,t)=(r,s,t)∙(x,y,z),∀(x,y,z),(r,s,t)∈Z3$

From this condition, we obtain the following relations: $a(x,y,z)r+b(x,y,z)s+(x−a(x,y,z))t=a(r,s,t)x+b(r,s,t)y+(r−a(r,s,t))zd(x,y,z)r+e(x,y,z)s+(y−d(x,y,z))t=d(r,s,t)x+e(r,s,t)y+(s−d(r,s,t))zg(x,y,z)r+h(x,y,z)s+(z−g(x,y,z))t=g(r,s,t)x+h(r,s,t)y+(t−g(r,s,t))z$

In particular, if (r, s, t) = (1, 0, 0) we obtain $a(x,y,z)=a(1,0,0)x+b(1,0,0)y+(1−a(1,0,0))zd(x,y,z)=d(1,0,0)x+e(1,0,0)y−d(1,0,0)zg(x,y,z)=g(1,0,0)x+h(1,0,0)y−g(1,0,0)z$(1)

On the other hand, when (r, s, t) = (0,1,0) we have $b(x,y,z)=a(0,1,0)x+b(0,1,0)y−a(0,1,0)ze(x,y,z)=d(0,1,0)x+e(0,1,0)y+(1−d(0,1,0))zh(x,y,z)=g(0,1,0)x+h(0,1,0)y−g(0,1,0)z$(2)

From (1), when (x, y, z) = (0,1,0), we get $a(0,1,0)=b(1,0,0)d(0,1,0)=e(1,0,0)g(0,1,0)=h(1,0,0)$

These relations allow to rewrite equations (1) and (2) introducing the parameters $α=a(1,0,0)β=a(0,1,0)=b(1,0,0)ϕ=b(0,1,0)δ=d(1,0,0)γ=d(0,1,0)=e(1,0,0)ρ=e(0,1,0)σ=g(1,0,0)θ=g(0,1,0)=h(1,0,0)λ=h(0,1,0)$

and to finally find $α(x−z)+βy+zβ(x−z)+ϕy(1−α)(x−z)−βyδ(x−z)+γyγ(x−z)+ρy+z−δ(x−z)+(1−γ)yσ(x−z)+θyθ(x−z)+λy−σ(x−z)−θy+z$

Elements of ν(x, y, z) are all linear functions of (x, y, z) depending on the introduced parameters. This fact is very useful in order to verify also condition 1 in Definition 2.1. Indeed, we may only consider such a condition for triples (1, 0, 0), (0, 1, 0), and (0, 0, 1). A (not so) little bit of calculation shows that parameters must satisfy the following system $αγ+(ρ+θ−β)δ+(1−γ)σ=γ2−(ϕ−λ)α+ϕγ−λ=β(ρ+θ−β)−σ(ϕ−λ)+λγ=θ(ρ+θ−β)αθ+δλ−βσ=θγδ(ϕ−λ)=βγ−θγ+θ$

We consider two cases: ϕ = λ and ϕ≠λ. When ϕ = λ, we have $αγ+(ρ+θ−β)δ+(1−γ)σ=γ2ϕγ−λ=β(ρ+θ−β)ϕγ=θ(ρ+θ−β)αθ+δϕ−βσ=θγβγ=θ(γ−1)$

In this case, when γ ≠ 0 we find the solution $\begin{array}{}\alpha =\frac{{\gamma }^{3}+{\gamma }^{2}\sigma -\gamma \sigma -\delta \left(\rho \gamma +\theta \right)}{{\gamma }^{2}},\beta =\frac{\theta \left(\gamma -1\right)}{\gamma },\varphi =\frac{\theta \left(\rho \gamma +\theta \right)}{{\gamma }^{2}},\end{array}$ corresponding to matrices $A=γ3+σγ2−σγ−δ(ργ+θ)γ2(x−z)+θ(γ−1)γy+zθ(γ−1)γ(x−z)+θ(ργ+θ)γ2y(γ2−γ3−σγ2+σγ+δ(ργ+θ)γ2)(x−z)−θ(γ−1)γyδ(x−z)+γyγ(x−z)+ρy+z−δ(x−z)+(1−γ)yσ(x−z)+θyθ(x−z)+θ(ργ+θ)γ2−σ(x−z)−θy+z$

On the other hand, if γ = 0 we find solution θ = γ = 0, ϕ = λ = −β(ρβ), σ = −δ(ρβ) and corresponding matrices are $B=α(x−z)+βy+zβ(x−z)−β(ρ−β)y(1−α)(x−z)−βyδ(x−z)ρy+z−δ(x−z)+y−δ(ρ−β)(x−z)−β(ρ−β)yδ(ρ−β)(x−z)+z$

When ϕ ≠ λ solution is $\begin{array}{}\alpha =\frac{\varphi \gamma -\lambda -\beta \left(0+\theta -\beta \right)}{\varphi -\lambda },\delta =\frac{\beta \gamma -\theta \gamma +\theta }{\varphi -\lambda },\sigma =\frac{\lambda \gamma -\theta \left(0+\theta -\beta \right)}{\varphi -\lambda },\end{array}$ giving matrices $C=ϕγ−λ−β(ρ+θ−β)ϕ−λ(x−z)+βy+zβ(x−z)+ϕy(ϕ(1−γ)+β(ρ+θ−β)ϕ−λ)(x−z)−βyβγ−θγ+θϕ−λ(x−z)+γyγ(x−z)+ρy+z−βγ−θγ+θθ−γ(x−z)+(1−γ)yλγ−θ(ρ+θ−β)ϕ−γ(x−z)+θyθ(x−z)+γy−λγ−θ(ρ+θ−β)ϕ−λ(x−z)−θy+z$

## 4 Matrices yielding commutative monoids of Pythagorean triples

In the previous section, we have found all matrices ν(x,y,z) such that injection ν is natural and consequently (ℤ3, ∙) is a commutative monoid. Now we want to show that a particular subset of PTPMs lies in one of families 𝓐, 𝓑 and 𝓒.

Let us consider the set 𝓟 = {(x, y, z) ∈ ℤ3 : x2 + y2 = z2} and matrices ν(x, y, z), where (x, y, z)∈ 𝓟. We recall that generic form of ν(x, y, z), with (x, y, z)∈ 𝓟, belonging to one of families 𝓐, 𝓑, 𝓒, is $ν(x,y,z)=A1(x−z)+B1y+zA2(x−z)+B2yA3(x−z)+B3yA4(x−z)+B4yA5(x−z)+B5y+zA6(x−z)+B6yA7(x−z)+B7yA8(x−z)+B8yA9(x−z)+B9y+z.$(3)

#### Lemma 4.1

Necessary conditions for (3) to be a PTPM are $B1=B5=B9,B7=B3,B8=B6,B4=−B2.$(4)

#### Proof

Using the well–known parametrization x = m2n2, y = 2mn and z = m2 + n2, matrix (3) becomes $−2A1n2+2B1mn+m2+n2−2A2n2+2B2mn−2A3n2+2B3mn−2A4n2+2B4mn−2A5n2+2B5mn+m2+n2−2A6n2+2B6mn−2A7n2+2B7mn−2A8n2+2B8mn−2A9n2+2B9mn+m2+n2.$

We find, with a little bit of calculation, that condition $ν(m2−n2,2mn,m2+n2).(u2−v2,2uv,u2+v2)∈P$

is valid for all Pythagorean triples (m2n2, 2mn, m2 + n2) and (u2ν2, 2uv, u2 + ν2) if the following equality holds $[M12+M22−M32−M1(u2−v2)−2M2uv+M3(u2+v2)]n4++[N1(u2−v2)+2N2uv−N3(u2+v2)−2M1N1−2M2N2+2M3N3]n3m++[N12+N22−N32−M1(u2−v2)−2M2uv+M3(u2+v2)]m2n2+[N1(u2−v2)+2N2uv−N3(u2+v2)]nm3=0$(5)

where, for j = 0, 1, 2, we have $Mj+1=A3j+1(u2−v2)+2A3j+2uv+A3j+3(u2+v2)Nj+1=B3j+1(u2−v2)+2B3j+2uv+B3j+3(u2+v2).$(6)

Equality (5) is independent by the choice of involved triples. Thus, all coefficients within square brackets must be equal to 0. In particular, if we consider the equality $N1(u2−v2)+2N2uv−N3(u2+v2)=0$(7)

and we substitute the second relations of (6) into (7) we find $(B1+B3−B7−B9)u4+2(B2+B4+2B6−2B8)u3v−2(B1−2B5+B9)u2v2+−2(B2+B4−B6+B8)uv3+(B1−B3+B7−B9)v4=0$

Since this equality must hold for all u and v, coefficients within round brackets must be 0, leading to the system of equations $B1+B3−B7−B9=0B2+B4+2B6−2B8=0B1−2B5+B9=0B2+B4−B6+B8=0B1−B3+B7−B9=0.$

which easily gives necessary conditions (4).□

Now, using previous lemma, we prove that PTPMs belong to family 𝓒.

#### Theorem 4.2

The PTPMs v(x, y, z) = Mβ,γ(x, y, z), with (x, y, z)∈ 𝓟, are subset of the family 𝓒 and they have the following form $(γ2−γ+1−β2)(x−z)+βy+zβ(x−z)−γy(β2−γ2+γ)(x−z)−βy−β(2γ−1)(x−z)+γyγ(x−z)+βy+zβ(2γ−1)(x−z)+(1−γ)y(β2+γ2−γ)(x−z)−βy−β(x−z)+(1−γ)y−(β2+γ2−γ)(x−z)+βy+z$(8)

#### Proof

First we show that conditions (4) are satisfied only by a subset of 𝓒. Let us consider a matrix of the form 𝓐. Conditions (4) correspond to relations $θ(γ−1)γ=ρ=−θ,θ=−θ(γ−1)γ2,θ(ργ+θ)γ2=1−γ,γ=θ(ργ+θ)γ2,$

which are clearly inconsistent.

If we examine a matrix belonging to family 𝓑, we have that conditions (4) lead to equalities $β=ρ=0,β=0,−β(ρ−β)=1,−β(ρ−β)=0,$

and also in this case they are clearly inconsistent.

Finally, for matrices belonging to family C, conditions (4) are equivalent to $β=ρ=−θ,θ=−β,λ=1−γ,ϕ=−γ.$

They are compatible, providing matrices of the form (8). We can prove that these matrices are PTPMs considering two triples generated by m, n and u, v.

We have $(m2−n2,2mn,m2+n2)∙(u2−v2,2uv,u2+v2)=(A,B,C)$

where $A=(mu+(1−2γ)nv)2−(mv+n(u+2βv))2,B=2(mu+(1−2γ)nv)(mv+n(u+2βv)),C=(mu+(1−2γ)nv)2+(mv+n(u+2βv))2,$

which is clearly a Pythagorean triple.

On the other hand $(u2−v2,2uv,u2+v2).ν(m2−n2,2mn,m2+n2)=(A′,B′,C′)$

where $A′=(mu+nv)2−(mv+n(u−2γu+2βv))2,B′=2(mu+nv)(mv+n(u−2γu+2βvC′=(mu+nv)2+(mv+n(u−2γu+2βv))2,$

which is clearly another Pythagorean triple. □

The previous theorem characterizes matrices Mβ,γ, depending on two parameters β and γ, such that $ν:P→M3,ν(x,y,z)=Mβ,γ(x,y,z)$

is natural, i.e., said ∙β,γ the induced product, (𝓟, ∙β,γ) is a commutative monoid.

## 5 Products over conics and groups of Pythagorean triples

Since we have the classical parametrization for the Pythagorean triples (x,y,z)∈ 𝓟 given by $ϕ:Z2→P,ϕ(u,v)=(u2−v2,2uv,u2+v2),$

matrices Mβ,γ(x,y,z) can be rewritten as $Mβ,γ(u,v)=u2+2βuv+(2β2−2γ2+2γ−1)v2−2v(γu−βv)2(−β2v−βu+γ(γ−1))v−2(βv−γ(u+2βv))vu2+2βuv+(−2γ+1)v22(−γu+u+βv−2βγv)v−2(β2v+βu+γ(γ−1)v)v2(−γu+u+βv)vu2+2βuv+(2β2+2γ2−2γ+1)v2.$

Now, we define a product between points of ℝ2 such that it is well–defined with respect to multiplication of matrices Mβ,γ(u,v).

#### Definition 5.1

For any (u,v),(s,t)∈ ℝ2, we define $(u,v)∗β,γ(s,t)=(su+tv(1−2γ),tu+sv+2βtv).$(9)

The reader can easily prove that $Mβ,γ(u,v).Mβ,γ(s,t)=Mβ,γ((u,v)∗β,γ(s,t)).$

The determinant of Mβ,γ(u,v) is surprisingly (u2+2β uv−(1−2γ)v2)3. Thus, function $δ(Mβ,γ(u,v))=u2+2βuv−(1−2γ)v2$

is multiplicative, i.e., $δ(Mβ,γ(u,v).Mβ,γ(u′,v′))=δ(Mβ,γ(u,v))⋅δ(Mβ,γ(u′,v′)),$

since it is the cubic root of the determinant.

In this way, study of commutative monoids of Pythagorean triples offers an unexpected connection with conics. Indeed, it is now natural to consider conics $αβ,γ(z)={(x,y)∈R2:x2+2βxy−(1−2γ)y2=z}.$

From previous notation and remarks, it follows that $(u,v)∗β,γ(s,t)∈αβ,γ(wz),∀(u,v)∈αβ,γ(w),(s,t)∈αβ,γ(z).$(10)

#### Proposition 5.2

Let (u,v)n*β,γ be the nth power of (u,v) with respect to*β,γ, then $(Mβ,γ(u,v))n=Mβ,γ((u,v)n∗β,γ).$

Consequently, power of matrices of the form (8) are still of this form.

#### Proof

The proof directly follows from previous observations. □

Conic αβ,γ(1) is especially interesting, due to equation (10). In [10], the authors studied the conic $x2+hxy−dy2=1,$(11)

which is a group with the operation $(u,v)⋅(s,t)=(tu+svd,tu+sv+tvh).$(12)

The identity is the point (1,0) and the inverse of a generic point (x,y) is (x+hy, −y). This conic and this product clearly coincide with αβ,γ(1) and *β,γ for h=2β and d = 1−2γ. Thus, (αβ,γ(1),*β,γ) is a group.

#### Theorem 5.3

If β2−2γ+1 is a positive square–free integer then we have (u1β v1,v1) ∈ αβ,γ(1), where (u1, v1) is the minimal solution of the Pell equation $x2−(β2−2γ+1)y2=1$

#### Proof

The proof directly follows from the equality $x2+2βxy−(1−2γ)y2=(x+βy)2−(β2−2γ+1)y2.$

As a consequence, the set $Nβ,γ={(u,v)∈αβ,γ(1):u,v∈Z}$

includes infinite points and (Nβ,γ, *β,γ) is a group.

Let us consider ϕ = ϕ|αβ,γ(1), then ϕ is a morphism, i.e., $ϕ¯((u,v)∗β,γ(s,t))=ϕ¯(u,v)∙β,γϕ¯(s,t),∀(u,v),(s,t)∈αβ,γ(1).$

Moreover, kerϕ = {(± 1,0)} and $(Im(ϕ¯),∙β,γ)≅(Nβ,γ)/{(±1,0)}.$

Thus we have the following theorem.

#### Theorem 5.4

(Im(ϕ),∙β,γ) is a group of Pythagorean triples and the inverse of the Pythagorean triple ϕ(u,v) is $ϕ¯(u+2βv,−v)=(−v2+(u+2βv)2,−2v(u+2βv),v2(u+2βv)2).$

#### Example

Let us consider the Pythagorean triple (3,4,5). Since we have ϕ(2,1) = (3,4,5), we are interested in conics αβ,γ(1) containing the point (2,1). It is easy to see that (2,1) ∈ αβ,−2β−1(1), ∀β∈ ℝ. However, we only consider β∈ ℤ. By Theorem (5.4), inverse of (2,1) is (2(β+1),−1) and inverse of (3,4,5) is $ϕ¯(2(β+1),−1)=(4β2+8β+3,−4(β+1),4(β+1)2+1).$

Let us set, e.g., β = 1, the conic α1,−3(1) is $x2+2xy−7y2=1.$

The inverse of (2,1) is (4,−1) and inverse of (3,4,5), with respect to1,−3, is (15,−8,17). Indeed, $(3,4,5)∙1,−3(15,−8,17)=M1,−3(3,4,5).(15,−8,17)=−151018−261530−30183515−817=101.$

Moreover, we can evaluate powers of Pythagorean triples: $(3,4,5)2∙1.−3=−151018−261530−301835345=85132157.$

The Pythagorean triple (85,132,157) corresponds to the point (2,1)2*1,−3 = (11,6) ∈ α1,−3(1). Thus, the inverse of (11,6) is (23,−6) and the inverse of (85,132,157) is ϕ(23,−6) = (493, −276,565 ).

## 6 Conclusion

The paper provides new structures of commutative monoids and commutative groups over sets of Pythagorean triples, starting from natural injections between triples and 3 × 3 matrices. Moreover, this study has connection with groups over conics.

As future developments, we find interesting to study operations *β,γ and ∙β,γ over finite fields ℤp for p prime. Moreover, this study offers the possibility to use Pythagorean triples in the approximation of quadratic irrationalities. Indeed, in [10], conic (11) and product (12) have been profitably used in Diophantine approximation for quadratic irrationalities. Thus, the above connections with Pythagorean triples show that they can be used in this context.

## References

• [1]

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## About the article

Accepted: 2017-09-22

Published Online: 2017-11-10

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1323–1331, ISSN (Online) 2391-5455,

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