Notes on a Particular Class of Perfect Cuboids

VALERIU DRAGANCOMOTI National Research Institute for Gas Turbines

Computational Fluid Dynamics Department220 D Iuliu Maniu Bd., sector 6, cod 061126, OP 76, CP174 Bucharest

ROMANIAdrvaleriu@gmail.com

Abstract: The current paper deals with the problem of the perfect cuboid, in particular with cuboids having only twoof its edges even numbers. It is shown that in this particular class, a discussion can be made and some conclusionscan be drawn to limit the existence of the perfect cuboid. In this paper Euclids formalism is used to generatePythagorean triangles that define the faces of the cuboid. Because the pairs, (a; b), (c; d), (u; v), used to generatethe triangles were not assumed to be relative primes, the triangles are not necessarily primitives. A computerizedalgebraic solver is used to solve the equations which were then compared with the numerical solutions obtainedindependently. Further generalization is required in order to extend the conclusions of this paper to a broader classof possible cuboids.

Key–Words: Perfect cuboid, Euler brick, Euclid equation, Pythagorean triangles

1 Introduction

The problem of proving or disproving the existence ofa perfect cuboid is one of the oldest open problems inmathematics. Both numerical and analytical methodshave been used, during the years with good indica-tion that such a cuboid cannot exist [1]. Reference [2]refers to semi-perfect cuboids i.e. cuboids with oneof the sides or diagonals non-integers. As remarked in[2], it is known that an infinite number of semi-perfectcuboids can exist [3]. Reference [2] provides a gen-eralization of a solution given by Bromhead [4] anddisproves the existence of a perfect cuboid within therespective class.

Shapirov (2011) [5] argues that the perfect cuboidproblem is linked to a class of univariate polynomialswith three integers. Shapirov further argues that due tothe restrictions imposed; the polynomials should be ir-reducible, disproving the existence of perfect cuboids.However, his study only provides numerical valida-tion for 10000 combinations of the three integers. Asindicated by [6], numerical search for perfect cuboidswith all odd sides came up empty in spite of the largenumber of examples tested (i.e. up to 1010).

Our case refers to perfect cuboids that have theirfaces described by Pythagorean triangles generatedthrough Euclids formalism. While other methods forgenerating such triplets exist [7]-[8], this method issimple and easier to use. Furthermore, the perfectcuboid considered in this paper must have at two evensides and the other side odd.

Further observations regarding the relations be-tween the Pythagorean triangles formed by the sidesand diagonals of the cuboid lead to properties that areproven to be inconsistent unless one of the sides (ordiagonal) is equal to zero.

Since the pairs of numbers used for the rea-soning are not relative primes, the proof applies toall Pythagorean triangles obtained through Euclidsmethod.

2 Problem FormulationOne of the known properties of primitive perfectcuboids, should they exist, is that they must have twoeven edges and precisely one odd edge. It is there-fore sufficient and also convenient to study the caseof a primitive cuboid - since it provides us with moreparticular conditions to begin the argument. The nextsubsection offers proof of the above mentioned prop-erty of a perfect cuboid and also introduces the nota-tions which will be used later on.

2.1 Preliminary discussionIf a perfect cuboid with all three sides even numbersexists, then by repeated divisions by 2, the resultingcuboids would also have to be perfect. This is be-cause the faces diagonals would be hypotenuses ofPythagorean triangles with two even catheti and aretherefore even themselves. Furthermore, the space di-agonal is also the hypotenuse of a Pythagorean trian-

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Figure 1: If all three sides (L1, L2, L3)of thecuboid are even numbers, then all diagonals(Df1 , Df2 , Df3 , Ds) are even as well

gle (e.g. VDB) which has both the catheti even num-bers hence the space diagonal is also even.

Therefore, it is possible to reach, after enough di-visions by 2, a perfect cuboid with at least one oddside. Hence, if a perfect cuboid with all even sidesexists, there exists also a perfect cuboid with at leastone odd side. This is because the expression under thesquare roots must be natural or irrational (but from thehypothesis we know that it cannot be irrational).

The problem is therefore reduced to proving thata perfect cuboid with two even sides and one odd sidecannot exist.

All faces of a perfect cuboid are defined byPythagorean triangles. From Euclids set of equations,each of the triangles defining the faces of the cuboidmust contain at least one even side. Since two facesmay have a common side which can be the even oneand because the third even side is perpendicular tothem and therefore cannot coincide with the other two- we conclude that: a perfect cuboid must have at leasttwo even sides. Furthermore, if a perfect cuboid ex-ists, one of its faces will be defined by two even sidesand, as a consequence, have the respective face diag-onal also an even number.

Assuming a perfect cuboid exists, having all itsedges even numbers, the following will apply:

Let E1 = 2nL1, E2 = 2nL2 and E3 = 2nL3 bethe edges of the perfect cuboid, where n,L1, L2, L3.are natural numbers. The face diagonals of the cuboidwill then be of the form

Df =√(2nL1)

2 + (2nL2)2 = 2n

√L21 + L2

2 (1)

and the space (volume) diagonal will be

Dv =√(2nL1)

2 + (2nL2)2 + (2nL3)

2

= 2n√L21 + L2

2 + L23. (2)

Since all diagonals are natural numbers by defini-

Figure 2: The assigned values of the sides and diago-nals for the demonstrational

tion, it is implied that√L21 + L2

2 ∈ N (3)

and √L21 + L2

2 + L23 ∈ N. (4)

This is because the square root of a natural numberwhich is not a perfect square cannot be a rational num-ber, √

k = w2√j1 · j2 · . . . · ji (5)

where k is natural and is not a perfect prime,w is natu-ral and j1 · · · ji are distinct prime numbers. Therefore,if the above expressions are not irrational, they mustbe natural numbers.

Knowing this we deduce that for any perfectcuboid which has all its edges even numbers ofthe form E=2nL, there exists another perfect cuboidwhich has at least one of its edges odd.

2.2 Demonstration for a primitive cuboidIn Fig. 2, a rectangular pyramid is presented, compris-ing of all the sides and diagonals of a cuboid, this isdone for an easier observation and has no bearing onthe demonstration itself. Note that the triangles VAB,VBD and VBC are Pythagorean and have the samehypotenuse, [VB].

Let the side [VD] be odd. Using the followingnotations:

[CD] = [AB] = 2ab (6)

[AV ] = a2 − b2 (7)

[V B] = a2 + b2 = c2 + d2 = u2 + v2 (8)

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Figure 3: The ABD and A’B’D’ Pythagorean triangles

[BD] = 2uv (9)

[V D] = u2 − v2 (10)

[BC] = [AD] = 2cd (11)

[V C] = c2 − d2 (12)

we convene that: a > b; c > d;u > v, in order for thesides to remain natural numbers.

Note that, regardless of the side chosen to be odd,the same general arrangement is reached, only withdifferent notations (the fact remains that either b ora; c or d; u or v; must be odd and the other one even).

We observe that, because [VD] is odd and [AD]is even, [AV] will also have to be odd; a similar argu-ment can be made for [VC] and [VB] which are alsoodd.

Figure 3 presents the DAB triangle, the one whichcontains the three even sides of the three above men-tioned Pythagorean triangles. Triangle DAB is itselfa Pythagorean triangle since all its three sides are nat-ural numbers as well as a right angle. It is also clearthat all its sides are divisible by 2, therefore anothertriangle, D’A’B’, exists such that[

A′B′] = 2xy (13)[A′D′] = x2 − y2 (14)[D′B′] = x2 + y2 (15)

where x > y. But, since [AB]=2[A’B’], we know that

2xy = ab (16)

Because [VA] is odd, we know that exactly one of aor b can be even, and if, for simplification, x and y areco-primes, we are faced with two possibilities:

a = 2x; b = y (17)

or

a = x; b = 2y. (17′)

We also know that the following equalities must ap-ply:

2cd = 2(x2 − y2

)(18)

2uv = 2(x2 + y2

)(19)

If the first variation, Ec.17, applies, let p and q be nat-ural numbers with the property that:

[V D]2 = [AV ]2 − [AD]2 (20)

[V D]2 = (a2 − b2)2 − (2cd)2 (21)

[V D]2 =(4x2 − y2

)2−[2(x2 − y2

)]2= p2 (22)

Rearranging,

12x4 − 3y4 = p2. (23)

And hence,

y =4

√12x4 − p2

3(24)

[CV ]2 = [V D]2 + [CD]2 = q2 (25)

p2 + (4xy)2 = q2 (26)

Substituting,

q2 = 12x4 − 3

(12x4 − p2

3

)+ 16x2

√12x4 − p2

3(27)

which reduces to

16x2

√12x4 − p2

3+ p2 − q2 = 0. (28)

Note that if the second variation, Ec. (17’), is ap-plied, with the right notations, we reach an equationwhich has the exact shape. In this latter case, the sub-stituted term should be y as a function of p and x, lead-ing to (28’):

16y2

√12y4 − p2

3+ p2 − q2 = 0, (28′)

Since the shape of Ecs.28 and (28’) are virtuallyidentical – apart from the notations – it is enough toprovide proof for one of the two variations. From thispoint forward, only the first variation will be used.

Using an algebraic solver [10] or a numericalsolver [11], it is possible to solve Ec.28 for p. Otherexamples where such solvers were used in complex

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mathematical and engineering problems can be foundin [12-13] also, other methods exist that use simulatedannealing [14] for the same purposes. The followingfour solutions verify Ec.28:

p = −√q2 − 16

3x2√100x4 − 3q2 − 128

3x4 (29)

p =

√q2 − 16

3x2√100x4 − 3q2 − 128

3x4 (30)

p = −√q2 +

16

3x2√100x4 − 3q2 − 128

3x4 (31)

p =

√q2 +

1

63x2

√100x4 − 3q2 − 128

3x4 (32)

Since the expression (29) and (31) are negative,the only two remaining possibilities for p are (32) and(30). Because q2 must be a natural number, the fol-lowing discussions are required in order to insure thatthe quantity under the square root is natural.

3 Problem Solution3.1 Discussion on Ec.(32)If we consider Ec.32 as the value for p, we arrive atthe conclusion that, in order for p to remain natural,the following condition must be satisfied

16x4

3

√100− 3q2

x4− 8

∈ Z (33)

If, for simplification we consider that x is not di-visible by 3, because 16x4 is clearly a natural number,the following expression must also be an integer√

100− 3 q2

x4 − 8

3(34)

This leads to the condition that there exist an integer,k ∈ Z, such that

100− 3q2

x4= (8 + 3k)2 (35)

or, equivalently(q

x2

)2

=100− (8 + 3k)2

3. (35′)

Solving this equation for q yields the two solutionsthat verify equation (35’):

q = x2√− (k + 6) (3k − 2) (36)

Table 1. The results for the values of k that satisfycondition (35’)

k (3k+8)2 q2/x4=[100-(3k+8)2]/3

-1 25 25-2 4 32-3 1 33-4 16 28-5 49 17

q = −x2√− (k + 6) (3k − 2). (36′)

Because both q and x are natural numbers, the expres-sion under the square root must be positive. Thus, wearrive at the conclusion that:

3k<2, if k>0 (37)

or,

if k < 0, k > −6. (37′)

From Ec.37, it is obvious that no integer can sat-isfy the condition imposed. However, Ec. (37’) re-quires the following discussion. Because of the condi-tion imposed through Ec.35, the equivalent expression(35’) must be a square rational number.

From Table 1, the only value of k that leads to arational square number is k=-1. This is because theother possible values are either prime numbers or canonly be factored in odd powers of primes.

Therefore, the following relation is deduced

q = 5x2 (38)

or, more conveniently,

q2 = 25x4 (38′)

Also,

16x2

3

√100− 3q2

x4− 8

= −16x2 (39)

Calculating p2 from Ec.32

p2 = 25x4 − 16x2 (40)

But, since by definition, from Ec.26,

p2 + (4xy)2 = q2 (41)

orp2 = q2 − 16x2y2 (42)

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Or, furtherp2 = 25x4 − 16x2y2 (43)

Equalizing Ec.40 with Ec.43, and baring in mind thatboth y and x are positive integers, we conclude that

y = 1 (44)

Equalizing Ec.23 with Ec.43, knowing that y=1, wearrive at the following

12x4 − 3 = 25x4 − 16x2 (45)

or13x4 − 16x2 + 3 = 0 (46)

which has the following solutions

x = ±1 (47)

x = ±√39

13(48)

The only natural value of x is x=1, which implies

x=y=1. (49)

Since the pair (x; y) must generate the Pythagoreantriangle D’A’B’, this means that one of its sides willhave to be zero (i.e. the pair x=y cannot be used togenerate a proper triangle). Thus, one of the sides ofthe cuboid will have to be zero in order for it to exist.

3.2 Discussion on Ec (30)Moving on to expression in Ec.(30), we observe thesimilar condition to the one in paragraph 2.1. is re-quired in order for p to be natural, again consideringthat x is not divisible by 3.

−16x2

3

√100− 3q2

x4+ 8

∈ Z (50)

The same reasoning as before can be made that an in-teger, m exists such that√

100− 3 q2

x4 + 8

3= m (51)

Rearranging, we obtain

100− 3q2

x4− (3m− 8)2 = 0 (52)

The only solutions for q, that satisfy this equation arelisted below

q = −x2√− (m− 6) (3m+ 2) (53)

Table 2. The results for the values of m that satisfycondition (55)

m (3m-8)2 q2/x4=[100-(3m-8)2]/3

1 25 252 4 323 1 334 16 285 49 17

q = x2√− (m− 6) (3m+ 2) (54)

Because, by definition, q and x natural numbersand m is an integer,

if m > 0, then m < 6 (55)

or,

if m < 0, then |m| < 2/3, (55′)

in order to keep the quantity under the squared rootpositive.

Since for Ec. (55’) there are no integer numbersthat are satisfactory, we will focus on Ec.(55). Theonly positive matches for m in this case are listed be-low:

Because(q

x2

)2

=100− (3m− 8)2

3(56)

is a square rational number, the only value for mwhich is satisfactory is

m=1, (57)

since the other values do not produce squared ratio-nal numbers – because their factors are odd powers ofprimes. Therefore

q

x2= 5 (58)

henceq = 5x2 (59)

and also

−16x2

3

√100− 3q2

x4+ 8

= −16x2 133

(60)

Introducing this into Ec.30, we deduce that

p2 = q2 − 16x213

3(61)

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But, from Eq.(26) we know that

p2 = q2 − 16x2y2. (62)

Because x, y, p and q are - by definition - positive, it isobvious that the only real solution for y is

y =√13/3 (63)

which cannot be a natural number. This denies the ini-tial hypothesis that both x and y are natural numbers,therefore invalidating Ec.30 as a possible solution forp.

It can therefore be stated that neither of the fourpossible variations of p, i.e. Eqs.(29-32) is consistentwith the hypotheses of the problem –which, in thisparticular case, does not have a solution.

4 Further workThe current section refers to a possible generaliza-tion of the previous demonstration in which we as-sumed that x and y are co-primes. For the generalizeddemonstration this assumption must be eliminated anda new parameter, n, must be introduced. There areeight possible combinations for the a and b pair, asseen below:

a = 2nx; b = y/n (64)

a = nx; b = 2y/n (65)

a = x/n; b = 2ny (66)

a = 2x/n; b = ny (67)

a = 2ny; b = x/n (68)

a = ny; b = 2x/n (69)

a = y/n; b = 2nx (70)

a = 2y/n; b = nx (71)

Knowing that, by definition a > b and x > y, wecan observe that variation (69) cannot exist whereasvariation (70) can only exist if n = 1 and x < 2y,hence reducing the problem to the one previouslystudied. Moreover, the cases (63)-(66) can be pairedwith (67)-(70) which lead to similar equation shapesfor the final equation.

Looking at the pair described by (66) we can writethe following

[V D]2 = x4(

1

n4− 4

)+y4

(16n4 − 4

)= p2 (72)

therefore,

p2 − q2 + 16x2

√√√√p2 + x4(4− 1

n4

)16n4 − 4

= 0 (73)

y4 =p2 + x4

(4− 1

n4

)16n4 − 4

(74)

Since a>b let k>1 be a rational number such thata = kb. Hence the expression for y becomes

y4 =

(x

2kn2

)4

(75)

Merging the two above equations we write the follow-ing:

p2/x4 =

(1

2kn2

)4

(16n4 − 4) +1

n4− 4 (76)

which can be re-arranged as:

p2/x4 = −(4k4n4 − 1)(4n4 − 1)

4k4n8(77)

which is a negative number.This invalidates the pair described by Ec. (66).

Consider now the pair described by Ec.(67). The fol-lowing expression can be written:

p2 − q2 + 16x2

√√√√p2 + x4(4− 1

n4

)n4 − 4

= 0 (78)

where

y4 =p2 + x4

(4− 1

n4

)n4 − 4

(79)

Let k > 1 be a rational number such that a = kb,we write

y4 =

(2x

kn2

)4

, (80)

therefore

p2

x4=

(2

kn2

)4

(n4 − 4) +16

n4− 4 (81)

which can be re-arranged

p2

x4= −4(k4n4 − 4)(n4 − 4)

k4n8(82)

Note that in order for this expression to be correct,n = 1 - which reduces to the problem of x and y beingco-primes. It must be said that one of the fundamentalproperties of primitive Pythagorean triangles is thatexactly one of its cathetes is a multiple of 4 [15]. This

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means that since we know that the even cathetes oftriangles VAB, VBD and VBC are all contained bya single face of the cuboid, they form a Pythagoreantriangle which is itself divisible by 4, or in the aboveformalism, the minimum value for n is n = 2.

A comment must be made, that the variations de-scribed by Ec.(68) and (69) which lead to virtual iden-tical relations as Ec.(66) and (67) respectively. Hencethe findings for Ec.(66) also apply to Ec.(68) and thefindings regarding Ec. (67) apply to (69). However,not the same can be stated for Ecs (69) and (70) whichalthough identical in form with Ecs. (63) and (64) re-spectively, were dismissed for different reasons.

Finally, observe the variation (64), by re-expressing the rational number k as a fraction,

k = g/t (83)

thus

y =2n2

g/t(84)

one can write the equation giving q2

q2 − p2 − 16x2(2n2x

g/t

)2

= 0 (85)

which reduces to

q2 − p2 − 64x4n4

(g/t)2= 0 (86)

since p is known to be an odd number-from the hy-pothesis, let p = 2P − 1.

q2 − (2P − 1)2 − 64x4n4

(g/t)2= 0 (87)

solving this equation for n leads to the following foursolutions

n =

(14 + i

4

)√g 4√4P 2 − 4P − q2 + 1

x√t

(88)

n = −

(14 + i

4

)√g 4√4P 2 − 4P − q2 + 1

x√t

(89)

n =

(14 −

i4

)√g 4√4P 2 − 4P − q2 + 1

x√t

(90)

n = −

(14 −

i4

)√g 4√4P 2 − 4P − q2 + 1

x√t

(91)

Figure 4: The ABD and A’B’D’ Pythagorean triangles

Note that, also from the hypothesis, 4P 2−4P−q2+1is negative so the positive expressions above must bere-expressed as

n =

(14 + i

4

)√i√g 4√q2 − 4P 2 + 4P − 1

x√t

(92)

and

n =

(14 −

i4

)√i√g 4√q2 − 4P 2 + 4P − 1

x√t

(93)

respectively.It must be said, however, that the expression(

1

4− i

4

)√i =

1

2√2

(94)

is still a real number.Another approach might be dwelling on the prop-

erty described by Ref.[15], the segment lengths fromfigure 3 must be re-expressed as seen in figure 4.

xy =ab

4(95)

Again, by simplifying the case and imposing thecondition that a and b are coprimes, we can start iden-tifying the terms in the following variations:

a = 2x; b = 2y (96)

ta = x; b = 4y (97)

a = 4x; b = y (98)

a = 4y; b = x (99)

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Note that the last two variations,

a = 2y; b = 2x (100)

a = y; b = 4x (101)

are impossible since x > y and a > b, by definition.Analyzing expression(96) we write

[AV ] = a2 − b2 = 4(x2 − y2) (102)

however, from the geometric conditions,

[BC] = [AD] = 4(x2 − y2) (103)

Hence, the height of the cuboid, [VD] is expressed

[V D] =√[AV ]2 − [BC]2 (104)

we observe that the height must be zero.Expression (97) leads to

[AV ] = x2 − 16y2 (105)

from which we deduce that

x > 4y (106)

Also, we observe that the expression for [VD] can bewritten

[V D]2 = −x4 (16− 1) + 16y4(16− 1) (107)

or[V D]2 = 15(16y4 − x4) (108)

which implies that

x < 2y (109)

which is a contradiction.Therefore expression (97) must be excluded.

Variation (98) may be reduced to

p2 = [V D]2= 15(16x4 − y4) (110)

and

[V C]2 = p2 + 64x2

√16x4 − p2

15(111)

In the last case, corresponding to variation (99), theequations also reduce to

p2 = [V D]2 = 15(16y4 − x4) (112)

and

[V C]2 = p2 + 64x2

√p2

240− x4

16(113)

or

[V C]2 = p2 + 16x2

√p2

15− x4 (114)

An observation which is applicable to both (97) and(98), is that p must be divisible by 15 in order for theexpression to remain integer.

More study is required for the generalized demon-stration, however, this approach may prove useful dueto the fact that it inserts more mathematical relationsbetween the terms.

5 ConclusionsThe paper provides a demonstration referring to a par-ticular class of cuboids, proving that under certainlimits no perfect cuboid can be constructed. In theinitial discussion an argument is made that, if a per-fect cuboid with all even edges exists, then even af-ter successive divisions by 2 of its edges it will re-tain its properties (i.e. it will still remain perfect).It is further argued that, by using Euclid’s model forPythagorean triangles, a perfect cuboid cannot haveless than two even sides. This leads to the particularcase of a cuboid having two even edges and one oddedge as the only possibility under the Euclid formal-ism.

The second section establishes the only config-uration for a perfect cuboid which has an odd sideand two even sides. Through certain observations allthe sides and diagonals of the cuboid are expressed asfunctions of two variables, x and y, which are naturalnumbers. The [VD] face diagonal is expressed in twopossible non-negative solutions, Eqs.(3) and (32), thatverify the conditions that arise from the setup.

Having disproved the only two possible solutionsfor q, expressions (32) and (30), we conclude thatno rectangular parallelepiped can be constructed suchthat all its sides and all its diagonals are positive inte-gers.

Since for the pairs (a, b), (c, d) and (u, v) therewas no condition that stated they have to be relativeprimes, the Pythagorean triangles that could have beenconstructed with them are not restricted to primitivetriangles. Some restrictions are however implied bythe fact that the cuboid must have an odd side. Thisis not a serious restriction since the initial discussionestablishes that if a perfect cuboid with all its sideseven numbers exists, it can always be reduced to aperfect cuboid having one odd side, by successivelydividing the sides by 2.

Because one of the premises is that x and y arecoprimes and that x (or, y) is not divisible by 3, thedemonstration is limited. In order for it to be com-plete, a generalization will need to be provided.

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In the Further work section, a generalization at-tempt is made, consisting in introducing a term n,which is a natural number. Another property of thenumber n is that - in turns - x and y are divisible by it,depending on the variation. The number of possiblecombinations leads to eight variations, however thosecan be reduced to only four, as follows:

Variation 1, a = 2nx and b = y/n

q2 = p2 + 16x2

√√√√4x4 (4n4 − 1)− p2

4− 1n4

(115)

Variation 2, a = nx and b = 2y/n

q2 = p2 + 16x2√x4 (4n4 − 1)− p2

4− 16n4

(116)

Variation 3, a = x/n and b = 2ny

q2 = p2 + 16x2

√√√√x4(4− 1

n4

)+ p2

4 (4n4 − 1)(117)

Variation 4, a = 2x/n and b = ny

q2 = p2 + 16x2

√√√√x4(4− 16

n4

)+ p2

n4 − 4(118)

Variation 3 is invalidated due to the fact that theexpression p2/x4 is a negative number. Also, varia-tion 4 requires that the factor n = 1, hence fallingunder the discussions in the previous sections of thepaper.

Although not a complete answer to the problem ofthe perfect cuboid, the present article provides someindication that such a mathematical construction maybe impossible.

References:

[1] R. K. Guy, Is there a Perfect Cuboid? FourSquares whose Sums in Pairs are Square. FourSquares whose Differences are Square, D18in Unsolved Problems in Number Theory, 2nded. New York: Springer-Verlag, pp. 173-181,(1994).

[2] W. J. A. Colman, On certain semi-perfectcuboids, (1986)

[3] Leech John, The rational cuboid revisited,Amer. Math. Monthly 84, (1977)

[4] T. B. Bromhead, On square sums of squares,Math. Gazette 44, (1960)

[5] Sharipov Ruslan, Perfect cuboids and irre-ducible polynomials, (2011)

[6] Eric W. Weisstein, Perfect Cuboid, FromMathWorld–A Wolfram Web Resource.http://mathworld.wolfram.com/ Perfect Cuboid.html, Butler, pers. comm.to Eric Weissten,(2004)

[7] Fibonacci, Leonardo Pisano,The Book ofSquares (Liber Quadratorum), An annotatedtranslation into modern English by L. E.Sigler, Orlando, FL: Academic Press. ISBN978-0-12-643130-8, (1987)

[8] L. E. Dickson, History of the Theory of Num-bers, Vol.II. Diophantine Analysis, Carnegie In-stitution of Washington, Publication No. 256,12+803pp

[9] J. Rukavicka, Dickson’s method for generatingpythagorean triples revisited, European Journalof Pure and Applied Mathematics, ISSN 1307-5543, Vol. 6, No. 3, (2013), pp.363-364,

[10] ** http://reference.wolfram.com/mathematica/guide/EquationSolving.html -(retrievedApr.2014)

[11] **Mathcad 14 User Guide- Chapter 9 Solving,pp.103-108, (2007)

[12] Krivulin Nikolai, An algebraic approach to mul-tidimensional minimax location problems withChebyshev distance, WSEAS TTransactions onMathematics, Issue 6, Volume 10, June (2011)

[13] Nakabayasy Tamiyo, Nakabayasgu Keiji, KakoFujio, Application of computer algebra approachto solve Engineering problems - A case study:Power supply stabilization loop circuit, AppliedMathematics in Electrical and Computer Engi-neering, ISBN: 978-1-61804-064-0

[14] Hime Aguiar, Diophantine equations and fuzzyadaptive simulated annealing, WSEAS TTrans-actions on Mathematics, Volume 13, (2014)

[15] Sierpinski, Waclaw, Pythagorean Triangles,Dover, ISBN 978-0-486-43278-6,(2003),

WSEAS TRANSACTIONS on MATHEMATICS Valeriu Dragan

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