On the Unsolvable Quintic Equation x5 + x2 + x + 1/π = 0

T he theorem credited to Niels Henrik Abel (1802-1829)and Paolo Ruffini (1765-1822), the Abel–Ruffini theo-

rem, briefly states that the quintic equation

a5x5 + a4x

4 + a3x3 + a2x

2 + a1x+ a0 = 0, (1)

with constant coefficients a5, a4, a3, a2, a1, a0, cannot besolved in radicals.

There are, however, some of these equations that can besolved, subject to some properties. Spearman and Williams[1] proposed that quintic equations of the form

x5 + a1x+ a0 = 0, (2)

are solvable for rational values of a0 and a1.Similarly, in [2] Nikos Bagis proposed Gamma functions

and nested radical functions. The property Kulkarni [3] intro-duced, required that the sum of its four roots be equal to thesum of its remaining three roots. A similar property was usedby Faggal and Lazard [4]. They considered polynomials whoseroots are the sum of two different roots of the input (for thequintics) or the sum of three different roots of the input (forthe septics).

Here, on the other hand, use the property that the quinticequation has a real root. A contra-posed view of this of thisproperty, is that one solution of the quintic equation is alreadyavailable; available without without any effort. It is given. Thisfollows from the fact that roots of polynomials, starting withthe quadratic equation, assume the general form

x = Re[x] + iIm[x]. (3)

Therefore, if one says the quintic equation has a real root, thenone is effectively saying it one solution exists, and is

Im[x] = 0. (4)

This is the given solution. It is the property we use tospecifically investigate

a5x5 + a2x

2 + a1x+ a0 = 0 (5)

for solutions. A numerical test is conducted on

x5 + x2 + x+1

π= 0. (6)

First we solve a particular second-order differential equa-tion. It plays a primary role in our approach to the quinticequation. We first demonstrate why we cannot rely on the con-temporary solution, then also Sophus Lie’s symmetry grouptheoretical methods too fails. After which, we introduce ourmodified Lie symmetries, a concept we discuss extensively in[5], [6] and [7]. This we do in Section II.

We also need quaternions, discussed briefly in Section III.The quintic equation is transformed into a fifth-order linearordinary differential equation in Section IV.

In Section V, we first demonstrate our procedure on thequadratic equation. This being because of its simplicity. Sec-tion VI is on the solution of equation (1), simplified to (4),following the steps we demonstrated on the quadratic equation

The common approach to the solutions of the equation

d2q

ds2− αq = 0, (7)

is to let

q = eλs, (8)

which reduces the problem to that of solving the polynomial

λ2 − α = 0. (9)

Subsequently, it is determined that

qα>0 = Ae−√αs +Be

√αs, (10)

for α > 0. Here A and B are treated as integration constants.The truth is that {exp(−

√α); exp(

√α)} forms the basis of

the solution sought, as such these constants are simply linearcombination parameters.

For α = we have q′′ = 0, so that

qα=0 = Cs+D. (11)

1. Introduction

2. Solving q′′ = αq

On the Unsolvable Quintic Equation x5 + x2 + x + 1/π = 0

JACOB M. MANALE Department of Mathematical Sciences, University of South Africa, Pretoria,

SOUTH AFRICA

Abstract: According to Galois’ theory, a solution x for the quintic equation cannot be solved for in radicals. Here we do not solve for x. We construct it. We note that radical numbers a and b can be found, such that a line that connects the points (a; f(a)) and (b; f(b)) passes through the interior point (x; 0), a root of the quintic function f(x). We then test this on f(x) = x5 +x2 + x+1/π. The values found for a and b, converted to decimal form, are a = −0.6012227544458956 and b = −0.6012348965947528, from which it is determined that x = −0.6012235335340363. The corresponding value, obtained through the software Mathematica numerical routines, is xn = −0.6012235335340362. Keywords: Quintic Equations, Differential Equations, Symmetry Analysis, Quaternions.

Received: May 9, 2021. Revised: September 10, 2021. Accepted: September 16, 2021. Published: October 8, 2021.

WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2021.20.52 Jacob M. Manale

E-ISSN: 2224-2880 496 Volume 20, 2021

Here C and D are genuinely integration constants.For α < 0 we have

qα<0 = E cos(√αs) + F sin(

√αs). (12)

Here < cos(√αs) + F sin(

√αs) > is the solution basis.

We are not particularly thrilled by this approach, because thesolutions are inconsistent. Note that when α = 0 we requirethat

qα<0 = qα=0 = qα>0, (13)

which suggests

A+B = Cs+D = E. (14)

For consistency, we would be required to let C = 0, whichcontradicts basic calculus principles, in that it suggests q′′ = 0should lead to q = constant.

In the next subsection, we give (15) a Lie symmetry grouptreatment.

Before we begin, we let s = t/√α, to simplify, because the

method involves too much details, so that

d2q

dt2− q = 0. (15)

In addressing (15) through Lie symmetry group methods, weseek the transformation of the point (t; q) to (t̄; q̄), involvinga parameter a, so that

t̄ = ϕ(t, q, a), q̄ = ψ(t, q, a), (16)

where ϕ = ϕ(t, q, a) and ψ = ψ(t, q, a) are functions.The transformation is expected to be invertible. That is,

there should exist ϕ−1 = ϕ−1(t̄, q̄, a) and ψ−1 = ψ−1(t̄, q̄, a),such that

t = ϕ−1(t̄, q̄, a), q = ψ−1(t̄, q̄, a). (17)

This property, therefore, requires that∣∣∣∣ ϕt ϕqψt ψq

∣∣∣∣ ̸= 0. (18)

The generators of symmetry groups requires are determinedfor small values of a, involving the Taylor-Maclaurin expan-sion. Subsequently

X = ξ(t, q)∂

∂t+ η(t, q)

∂

∂q, (19)

where

ξ(t, q) =∂ϕ(t, q, a)

∂a

∣∣∣∣a=0

, η(t, q) =∂ψ(t, q, a)

∂a

∣∣∣∣a=0

. (20)

For (15), we get the monomials

1 : ηtt + (ηy − 2ξx)y = 0, (21)q̇ : 2ηxy − ξxx − 3yξy = 0, (22)q̇2 : ηyy − 2ξxy = 0, (23)q̇3 : ξyy = 0. (24)

They lead to the infinitesimals

ξ = A1q +A2

2t+

A3

2(25)

and

η = A2q +B1. (26)

The resulting symmetries are

X1 = q∂

∂t, (27)

X2 =∂

∂q, (28)

X3 =t

2

∂

∂t+ q

∂

∂q, (29)

X4 =1

2

∂

∂t. (30)

(31)

TABLE ITHE COMMUTATOR TABLE

X1 X2 X3 X4

X1 0 2X212X2 0

X2 -2X2 0 X2 0

X3 - 12X2 −X2 0 1

2X4

X4 0 0 − 12X4 0

From the commutator table, Table I, we deduce that the set{X1, X2, X3, X4} is a Lie algebra. It has the three dimen-sional sub-algebras {X1, X2} ⊂ {X1, X2, X3}. {X2, X4} ⊂{X2, X2, X4} and {X3, X4} ⊂ {X2, X3, X4}, and the twodimensional sub-algebras {X2} ⊂ {X1, X2} and {X2} ⊂{X2, X3}. The Abelian algebras are {X1, X4} and {X2, X4}.

TABLE IITHE PSEUDO-SCALAR PRODUCT TABLE

X1 X2 X3 X4

X1 0 y y2 y2

X2 −y 0 0 0

X3 −y2 0 0 x4

X4 − y2

0 −x2

0

Equation (15), being a second-order equation, we focus onlyon two-dimensional algebras. We note from the tables I andII that the two-dimensional algebras suggests the followingapproaches for obtaining solutions.

1) Type I procedures: Type I procedures follow from

[Xi, Xj ] = 0, Xi ∨Xj ̸= 0. (32)

The algebras satisfying this are from the Abelian groups. Thereis only one: {X1, X4}.

2.1 A Lie Symmetry Group Theoretical Approach


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The expected form of the solution is

q′′ = f(q′). (33)

The symmetries are

X1 = q∂

∂t, X4 =

1

2

∂

∂t. (34)

That is,

X1(t) = 1, X4(t) = 0, X1(q) = 0, X4(q) = 1. (35)

To determine t̄, we use

q∂t̄

∂t= 1,

1

4

∂t̄

∂t= 0. (36)

The first equation suggests t̄ = j(q). That is, t̄ depends onq. From the second part we have

X̄1(q) = 0, X̄4(q) = 1. (37)

That is,

q∂q̄

∂t= 0,

1

2

∂q̄

∂t= 1. (38)

It follows from the first part that q̄ = k(q). The second partsuggests that q̄ = 2t+ k(q). A comparison leads to 2t = 0. Acontradiction.

2) Type II procedures: Type II procedures follow from

[Xi, Xj ] = 0, Xi ∨Xj = 0. (39)

The algebras satisfying this are also from the Abelian groups.There is only one: {X2, X4}.


q′′ = f(t). (40)

The symmetries are

X2 =∂

∂q, X4 =

1

2

∂

∂t. (41)

That is,

X̄2(t) = 0, X̄4(t) = 0, X̄1(q) = 1, X̄4(q) = t. (42)


∂t̄

∂q= 0,

1

4

∂t̄

∂t= 0, (43)

from (42). They give

t̄ = 0, (44)

leading nowhere.

3) Type III procedures: Type IV procedures follow from

[Xi, Xj ] = Xi, Xi ∨Xj = 0. (45)

The algebras satisfying this are also from the non-Abeliangroups. There is only one of them. Namely, {X2} ⊂{X2, X3}.


q′′ =1

tf(q′). (46)

The symmetries are

X1 =∂

∂t, X3 =

x

2

∂

∂t+ q

∂

∂q. (47)

That is,

X̄2(t) = 0, X̄3(t) = t, X̄2(q) = 1, X̄3(q) = q. (48)


∂t̄

∂q= 0,

t

2

∂t̄

∂t+ q

∂t̄

∂q= t. (49)

They give

t̄ = 2t+ f( qt2

)and t̄ = g(t), (50)

a contradiction.From the second part, we have

X̄2(q) = 1, X̄3(q) = q. (51)

∂q̄

∂q= 1,

t

2

∂q̄

∂t+ q

∂q̄

∂q= q. (52)

They give

q̄ = q + f( qt2

)and q̄ = q + g(t), (53)

another contradiction.4) Type IV procedures: Type IV procedures follow from

[Xi, Xj ] = Xi, Xi ∨Xj ̸= 0. (54)

The algebras satisfying this are also from the non-Abeliangroups. There is only one. Namely, {X2} ⊂ {X1, X2}.


q′′ = f(t)q′. (55)

The symmetries are

X1 = q∂

∂t, X3 =

x

2

∂

∂t+ q

∂

∂q. (56)

That is,

X̄1(t) = 0, X̄3(t) = 0, X̄1(q) = 1, X̄3(q) = q. (57)


q∂t̄

∂t= 0,

t

2

∂t̄

∂t+ q

∂t̄

∂q= 0, (58)

which gives t̄ = C, a constant, leading nowhere.


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From the second part, we have

X̄2(q) = 1, X̄3(q) = q. (59)

∂q̄

∂q= 1,

t

2

∂q̄

∂t+ q

∂q̄

∂q= q. (60)

They give

q̄ = q + f (t̄) and q̄ = g(t), (61)

a contradiction.We clear all these contradiction through what we call

modified Lie symmetries. An infinitesimal parameter ω isintroduced to address them.

Modified Lie Symmetry is a concept we developed. A briefdescription, in the current coordinates, would be they resultfrom a two-parameter, a and ϵ, transformation, from point(t; q) to (t̃; q̃). Alternatively, it is a one parameter, ω from thepoint (t̄; q̄), as described in (16), of one point to another, whatIn addressing (15) through Lie symmetry group methods, weseek the transformation of the point (t; q) to (t̄; q̄), involvinga parameter a, to (t̃; q̃). The only requirement is that for aninfinitesimal ω, point (t̄; q̄) and (t̄; q̄) are indistinguishablefrom one another.

A simple procedure, for realizing such symmetries, wouldbe to note that sin(ω[at+ b])/ω transforms into at+ b whenω = 0. Therefore, (26) transforms into

ξ̃ = A1

sin(ωq + ω A3

2A1

)ω

+A2

2

sin(ωt+ ωA3

A2

)ω

(62)

and

η̃ = A2tqω sin

(ωt+ ω

A3

A2

)+A2q cos

(ωt+ ω

A3

A2

)+D0. (63)

The modified symmetries that result from these are

X̃1 =sin (ωq)

ω

∂

∂t, (64)

X̃2 =∂

∂q, (65)

X̃3 =1

2

sin (ωt)

ω

∂

∂t

+

{tqω sin (ωt) + q cos (ωt)

}∂

∂q, (66)

X̃4 =

{A1

sin(ωq + ω

2A1

)ω

∣∣∣∣∣∣A1=0

+A2

2

sin(ωt+ ω

A2

)ω

∣∣∣∣∣∣A2=0

}∂

∂t

+

{A2tqω sin

(ωt+

ω

A2

)∣∣∣∣A2=0

+A2q cos

(ωt+

ω

A2

)∣∣∣∣A2=0

}∂

∂q. (67)

The two dimensional modified symmetry algebra <X̃1, X̃4 > then lead to

q = Asin(i[ωt+ ϕ])

iω, (68)

where A and ϕ are constants of integration.

According to the Abel-Ruffini theorem, there is absolutelyno way of obtaining the solution to the quintic equation. This,we think, is a bit too broad. We interpret it here to mean thereis absolutely no way of obtaining the solution, as a complexnumber, to the quintic equation. It is however possible as aquaternion, as supported by the following.

Theorem 1: A number xj exists, such that x = xr + ixi +jxj + kxk, a quaternion, is a solution of the equation

x5 + x2 + a4x+ a5 = 0.

To transform (1) into a differential equation, we first intro-duce the factor exp(xξ) to the equation. That is,

(a5x5 + a4x

4 + a3x3 + a2x

2 + a1x+ a0)exξ = 0, (69)

or

a5x5exξ + a4x

4exξ + a3x3exξ + a2x

2exξ

+a1xexξ + a0e

xξ = 0, (70)

so that

a5d5

dξ5exξ + a4

d4

dξ4exξ + a3

d3

dξ3exξ + a2

d2

dξ2exξ

+a1d

dξexξ + a0e

xξ = 0 (71)

Letting

y = exξ (72)

with y = y(ξ), suggests

y(5) =d5exξ

dξ5, y(4) =

d4exξ

dξ4, y(3) =

d3exξ

dξ3,

y′′ =d2exξ

dξ2and y′ =

dexξ

dξ, (73)

leading to the ordinary differential equation

a5y(5) + a4y

(4) + a3y(3) + a2y

′′ + a1y′ + a0y = 0. (74)

This is the equation we solve, a linear ordinary differentialequation. We intend to use the solution of (74) to solve(1). This goes against popular practice, because it is usuallythe solutions of polynomials that are used to solve lineardifferential equations.

2.2 Modified Lie Symmetries

3. The Role of Quaternions

4. The Transformation of Equation (1) Into a Differential Equation


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The quadratic equation

ax2 + bx+ c = 0, (75)

with the parameters a, b and c, has the solutions

x =−b±

√b2 − 4ac

2a. (76)

What we wish to achieve here is demonstrate that themethod we are proposing, to be used on the quintic, lead tothe same result.

First we introduce the factor (72) as before, so that

ay′′ + by′ + cy = 0. (77)

There is a popular belief that this differential equation istrivial, and that it can easily be solved. Euler introduced aformula that it is supposed to solve it centuries ago, see [8],[9], [10], [11] and [12].

It is still used today and taught to all students, in all fields ofmathematics, engineering and the physical sciences. We arguein [7] and [13] that there is an error in Euler’s formula, andthat the correct solution is

y = Ae−b/(2a) sin(iωξ + ϕ)

iω, (78)

where A and ϕ are constants of integration, and

ω =±√b2 − 4ac

2a. (79)

Equating (72) to (78) by removing the common y gives

exξ = Ae−ξb/(2a) sin(iωξ + ϕ)

iω, (80)

so that

xξ = ln

[A

iω

]− b

2aξ + ln [sin(iωξ + ϕ)] , (81)

or

x =ln[Aiω

]ξ

− b

2a+

ln [sin(iωξ + ϕ)]

ξ. (82)

This can be written in the form

x =ln[Aiω

]ξ

− b

2a+

ln[− eωξ−iϕ−e−ωξ+iϕ

2i

]ξ

. (83)

The limit as ξ → ±∞ gives

x = − b

2a− ω

eωξ−iϕ + e−ωξ+iϕ

eωξ−iϕ − e−ωξ+iϕ. (84)

That is,

x = − b

2a− ω

eωξ−iϕ

eωξ−iϕ, (85)

and subsequently

x = − b

2a− ω, (86)

or

x = − b

2a− ±

√b2 − 4ac

2a. (87)

This result is the same as the one in (76).

An equivalent space arises from (69) when

y = y′′ = 0. (88)

This can be expressed in the form

y′′

y=y(3)

y′, (89)

which integrates into

y = asin(iωξ + ϕ)

iω, (90)

where A, ϕ and ω are constants of integration. Substitutingthis result into (74) gives an equation that determines ω. Thatis

−Aa5 (iω)5cos(iω[ξ + ϕ])

iω+Aa1 cos(iω[ξ + ϕ]) = 0. (91)

This gives

ω1 = (−1)1/4 4

√a1a5, (92)

ω2 = −(−1)1/4 4

√a1a5, (93)

ω3 = (−1)3/4 4

√a1a5, (94)

ω4 = −(−1)3/4 4

√a1a5. (95)

Note that ω1, ω2, ω3 and ω4 appeared to be just fourquantities. But each can be interpreted in four possible ways,bring the quantities to 16. It is best to express them in theform

ω1 = (p1 + jq1) 4

√a1a5, (96)

ω2 = (p2 + jq2) 4

√a1a5, (97)

ω3 = (p3 + jq3) 4

√a1a5, (98)

ω4 = (p4 + jq4) 4

√a1a5, (99)

where p1, p2, q1, q2 are solutions of z4 + 1 = 0, andp3, p4, q3, q4 solutions of z4+(−1)3 = 0, with the quaternionunit j. Our choice for j, instead of sticking with the regulari, will also help to keep of the imaginary solution, mentionedin the Introduction.

There are four possible ways that we can extract x. Firstwe have

−a5y(5) = a4y(4) + a3y

(3) + a2y′′ + a1y

′ + a0y. (100)

That is,

y(5) = −a4y(4) + a3y

(3) + a2y′′ + a1y

′ + a0y

a5, (101)

5. An Illustrative Example: the Quadratic Equation 6. Solving the Differential Equation

6.1 The Case for (96)


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or

d5exξ

dξ5=a4y

(4) + a3y(3) + a2y

′′ + a1y′ + a0y

a5= 0, (102)

so that

x5exξ = x5y = −a4y(4) + a3y

(3) + a2y′′ + a1y

′ + a0y

a5.(103)

Hence,

x5 =

− (a4y(4) + a3y

(3) + a2y′′ + a1y

′ + a0y)a0a5(a0y)

. (104)

Hence,

x5 =(a4y

(4) + a3y(3) + a2y

′′ + a1y′ + a0y)a0

a5(a5y(5) + a4y(4) + a3y(3) + a2y′′ + a1y′).(105)

With y given by (78), equation (105) reduces to

x5 = i

(a3a5ω3 − a1

a5ω3

)cot(iωξ + ϕ)

−a0a5

− a2a5ω2 − a4

a5ω4. (106)

The root can be presented in several different ways.

Substituting solution (90) into (105), then separating theimaginary component from the real, and subsequently equatingthe imaginary component to zero leads to

x5 =△1 +△2

△3 +△4, with

△1 = a0

(ia4ω

3 − ia2ω +ia0ω

)sin θ

△2 = a0(ia1 − ia3ω

2)cos θ

△3 = a5(ia4ω

3 − ia2ω)sin θ

△4 = a5(ia5ω

4 − ia3ω2 + ia1

)cos θ, (107)

with θ = iξω + ϕ. The expression can be presented in theform

x5 = ψr + iψi. (108)

Our interest is in the real root, this means the imaginarycomponent is zero, so that

x5 = ψr. (109)

That is,

x = 5√ψr. (110)

We get expressions for a and b by varying the radicalsigns. This is not different from the process followed whenextracting quadratic roots, with −√ assigned to one root, and+√ assigned to another. They are

a = Re

(j

(2a5

A13A15

(a25 − i)A16

)1/5)

(111)

and

b = Re

(−a5

B14B16j(a5 + (−1)3/4

)B17 (B8 −B11)

)1/5

. (112)

The parameters are

A1 = coth−1

1 +2j(a5+

4√−1)

2(−1)3/4−2ja5

−1 +2j(a5+

4√−1)

2(−1)3/4−2ja5

, (113)

A2 = log

((1 + j)

√a5 +

4√−1√

2(−1)3/4 − 2ja5

), (114)

A3 = −1 +2j(a5 +

4√−1)

2(−1)3/4 − 2ja5, (115)

A4 = 1 +2j(a5 +

4√−1)

2(−1)3/4 − 2ja5, (116)

A5 = sinh (A1 − (1− j)A2) , (117)

A6 = cosh (A1 − (1− j)A2) , (118)

A7 = cos (4 (−A2 + jA1)− 4jA2) , (119)

A8 = −a5A25 + a5A

26 +

4√−1A5√1− A2

3

A24

+(−1)3/4A3A5√

1− A23

A24A4

− (−1)3/4A6√1− A2

3

A24

−4√−1A3A6√1− A2

3

A24A4

, (120)

A9 = a25A7 + (−1)3/4a5 sin (4 (−A2 + jA1) + (1− 3i)A2) ,

(121)

A10 = a25A7 − 2(−1)3/4a5 sin ((1 + j)A2)

+2(−1)3/4a5 sin (4 (−A2jA1) + (1− 3j)A2)

+i cos (4 (−A2jA1) + (2− 2j)A2) , (122)

A11 =√T1 + T2 + T3, (123)

T1 =a5A

25

A8−

4√−1A5√

1− A23

A24A8

,

T2 =a5A

26

A8−

4√−1A3A6√

1− A23

A24A4A8

,

T3 =

√−a25 +A10 − 2i cos ((2 + 2i)A2)− i√

2A8

,

A12 = cosh (log (−A11)− (1 + j)A2) , (124)

6.2 The Radical Root


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A13 = jA5 cosh (log (−A11)− (1− i)A2)

−iA6 sinh (log (−A11)− (1− j)A2) , (125)

A14 = −A5 cosh (log (−A11)− 2A2)

+(1− j)a5

√1 + ja25 sinh (log (−A11)− (1 + i)A2) ,

(126)

A15 = 4√−1(1− i)

√1 + ja25A12

−a35A6 sinh (log (−A11)− 2A2)

−ia25A6 sinh (log (−A11)− 2A2)

+ja5A6 sinh (log (−A11)− 2A2)

+a35A5 (− cosh (log (−A11)− 2A2))

−ia25A5 cosh (log (−A11)− 2A2)

+ia5A5 cosh (log (−A11)− 2A2)

+A14 −A6 sinh (log (−A11)− 2A2) , (127)

A16 = cos (2 (iA1 − (1 + j)A2))

− cosh (2 (log (−A11)−A2) + 2jA2) , (128)

B1 = coth−1

1− 2j(a5+(−1)3/4)2 4√−1+2ja5

−1− 2j(a5+(−1)3/4)2 4√−1+2ja5

, (129)

B2 = −(1− j) log

((1− i)

√a5 + (−1)3/4√

2 4√−1 + 2ja5

),(130)

B3 = log

((1− i)

√a5 + (−1)3/4√

2 4√−1 + 2ja5

), (131)

B4 = −2 4√−1a5 sin (4 (−B3 + jB1)− 4jB3)

+a25 cos (4 (−B3 + jB1)− 4jB3)

−a25 − i cos (4 (−B3 + jB1)− 4jB3)− i,(132)

B5 = −a5 sinh2 (B1 +B2) + a5 cosh2 (B1 +B2)

−(−1)3/4 sinh2 (B1 +B2)

+(−1)3/4 cosh2 (B1 +B2) , (133)

B6 =a5 sinh

2 (B1 +B2)

B5+a5 cosh

2 (B1 +B2)

B5

−2(−1)3/4 sinh (B1 +B2) cosh (B1 +B2)

B5

+

√B4√2B5

, (134)

B7 = cosh (B1 +B2)− sinh (B1 +B2) , (135)

B8 = cos (2 (jB1 − (1 + j)B3)) , (136)

B9 = cosh(− log

(−√B6

)+ (1 + j)B3

), (137)

-

B10 = cosh (B1 +B2) , (138)

B11 = Cosh[2B3 + 2(−B3 + Log[− B6√])], (139)

B12 = sinh(B1 +B2), (140)

B13 = sinh(− log

(−√B6

)+ (1 + j)B3

), (141)

B14 =√B4B7 (B10B13 −B12B9) , (142)

B15 = −√2B2

10 − 4√2B12B10 −

√2B2

12, (143)

B16 = −2√2a35B10B12

−(−1)3/4√2a25B

210

−(−1)3/4√2a25B

212 − 6(−1)3/4

√2a25B10B12

−2j√2a25B10B12 +

4√−1

√2a5B

210

+j√2a5B

210 +

4√−1

√2a5B

212 + i

√2a5B

212

+6 4√−1

√2a5B10B12 + 4j

√2a5B10B12

+a25√B4 + (−1)3/4a5

√B4 + ja5

√B4

+B15 − 4√−1√B4, (144)

B17 = −2a5B210 − 2a5B

212 + 4(−1)3/4B12B10 −

√2√B4,

B18 = −2 4√−1a5 sin (4 (−B3 + jB1)− 4jB3)

+a25 cos (4 (−B3 + jB1)− 4jB3)

−a25 − j cos (4 (−B3 + jB1)− 4jB3)− i. (145)

Under normal circumstances, without the Abel-Ruffini the-orem, this result should solve (1). Instead, we get two values,complying to the following axiom and theorems.

Axiom 1: There exists two radical numbers a and b, suchthat a line that connects the points (a; f(a)) and (b; f(b))passes through the interior point (x; 0), a root of the quinticfunction

f(x) = a5x5 + a4x

4 + a3x3 + a2x

2 + a1x+ a0. (146)

Theorem 2: If a line that joins the points (a; f(a)) and(b; f(b)) passes through the interior point (xr; 0), for thefunction (146) , then the line is

x = m

(f(x)− f

(a− x

f(a)− f(x)f(x) + a− f(a)

f ′(a)

))+ c,(147)

or

x = m

(f(x)− f

(b− x

f(b)− f(x)f(x) + b− f(b)

f ′(b)

))+ c,(148)

7. A Numerical Experiment: the Case x5 + x2 + x + 1/π = 0


E-ISSN: 2224-2880 502 Volume 20, 2021

where c is the x-intercept and m is given[a−x

f(a)−f(x) −b−x

f(b)−f(x)

]f(x) + a− b− f(a)

f ′(a) +f(b)f ′(b)

f(

a−xf(a)−f(x)f(x) + a− f(a)

f ′(a)

)− f

(b−x

f(b)−f(x)f(x) + b− f(b)f ′(b)

) .(149)

Theorem 3: If a line that joins the points (a; f(a)) and(b; f(b)) passes through the interior point (xr; 0), for thefunction (146) , then xr is given by

xr =a− b− f(a)

f ′(a) +f(b)f ′(b)

f(a− f(a)

f ′(a)

)− f

(b− f(b)

f ′(b)

) (−f (a− f(a)

f ′(a)

))

+a− f(a)

f ′(a), (150)

or

xr =a− b− f(a)

f ′(a) +f(b)f ′(b)

f(a− f(a)

f ′(a)

)− f

(b− f(b)

f ′(b)

) (−f (b− f(b)

f ′(b)

))

+b− f(b)

f ′(a). (151)

Fig. 1. Illustration of Axiom 1.

Applying the the procedure described in the previous sectionon

x5 + x2 + x+ 1/π = 0, (152)

with a5 = 1/π.The values of a and b are determined from α and β, as

directed by Theorem 1, respectively. Here

α = −0.6011483825679623 + 0.08040632391894242j,(153)

so that

a = −0.6012227544458956, (154)

and

β = −0.6012290127318318 + 0.36310327461376773j,(155)

so that

b = −0.6012348965947528. (156)

Substituting these into (150) or (151) gives

xr = −0.6012235335340363. (157)

The program Mathematica’s command

N[x5 + x2 + x+

1

π== 0, {x}

][[1]],

gives

xn = −0.6012235335340362. (158)

The Abel–Ruffini theorem declares that the quintic equationcannot be solved in radicals. Here we introduced an axiom,suggesting that two radicals numbers can be found, such thata line through them passes through the radical root of thequintic equation. This we tested on a numerical example, andit passed without any errors. We can therefore declare to haveundoubtedly determined a formula for determining the radicalroots for the quintic equation.

We have also demonstrated that the notion that the quinticequation has a real root not to be entirely accurate. Saying oneof its roots has a zero imarginary component, is much closer.

[1] B.K. Spearmn and K.S. Williams. Characterazation of solvable quinticsx5 + ax + b = 0. The American Mathematical Monthly, 15:986–992,1994.

[2] N. Bragis. arXiv:1406.1948v4.[3] R. Kulkarni.[4] M. A. Faggal and D. Lazard. Solution of polynomial equations

with nested radicals. International Journal of Scientific and ResearchPublications, 4:1–10, 2014.

[5] J.M. Manale. On solutions of the nonlinear heat equation using mod-ified lie symmetries and differentiable topological manifolds. WSEASTransactions on Applied and Theoretical Mechanics, ISSN / E-ISSN:1991-8747/ 2224-3429, 15:132–139, 2020.

[6] J.M. Manale. On solutions of the ii-d navier-stokes equations us-ing puknachevs subalgebra (g1,g2). International Journal of AppliedEngineering Research ISSN 0973-4562, Research India Publications.http://www.ripublication.com, 13:16418–16423, 2018.

[7] J.M. Manale. On errors in eulers formula for solving odes). InternationalJournal of Mathematical and Computational Method, pages 1–3, 2020.

[8] D. Zille. Differential Equations with Boundary-Value Problems, Interna-tional Metric Edition. CENGAGE Learning Custom Publishing, 2018.

[9] J. Martha, L. L. Abell and P. Braselton. Introductory DifferentialEquations. Academic Press, 2018.

[10] R. Haberman. Applied Partial Differential Equations with Fourier Seriesand Boundary Value Problems: Pearson New International Edition.Pearson Education Limited, 2014.

[11] G. Peterson and J. Sochacki. Differential Equations and Their Appli-cations: An Introduction to Applied Mathematics. Springer New York,2013.

[12] S.G. Krantz. Differential Equations. Apple Academic Press Inc., 2014.[13] J.M. Manale. On errors in eulers complex exponent and for-

mula for solving odes. Journal of Physics: Conference Serieshttps://iopscience.iop.org/issue/1742-6596/1564/1, pages 1–7, 2020.

8. Conclusion

References

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E-ISSN: 2224-2880 503 Volume 20, 2021

On the Unsolvable Quintic Equation x5 + x2 + x + 1/π = 0

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