*"We understand the relation between houses and walls. But it would be hard to cross the gap between houses and bricks without having enough intermediate concepts such as that of the wall."*

(Marvin Minsky, Society of Mind, 1985)*"Problems cannot be solved at the same level of awareness that created them."*

(Albert Einstein, 1920)

theorem among many from which is possible to be obtained by analysis of Parcelatories, is the called

*Parcelatories' Powers*that corresponds to the following statement:

This theorem can be used to determine the veracity of the existence of integer results in several polynomial equations, such as:

a

^{2}+ b

^{2}= c

^{2}, where

**a**,

**b**and

**c**∈ Ν

a

^{3}+ b

^{3}= c

^{2}, where

**a**,

**b**and

**c**∈ Ν

a

^{2}+ b

^{2}= c

^{3}, where

**a**,

**b**and

**c**∈ Ν

a

^{2}+ b

^{2}+ c

^{2}= d

^{2}, where

**a**,

**b**,

**c**and

**d**∈ Ν

a

^{11}+ b

^{11}+ c

^{11}= d

^{7}, where

**a**,

**b**,

**c**and

**d**∈ Ν

or even

a

^{1001}+ b

^{1001}+ c

^{1001}+ d

^{1001}= e

^{2003}, where

**a**,

**b**,

**c**,

**d**and

**e**∈ Ν

And so on!!

The idea is powerful, because from a simple statement it is possible to see in advance whether there are or not positive integers results in equations that would take decades to be calculated with today's faster computers or even with future computers. Therefore it is necessary to show it with enough accuracy and clarity.

**DEMONSTRATIONS**

The equations with possible positive integer values, formalized by the statement, are an union of two sets of equations, namely:

The set called '*co-primes*' corresponds to the set of equations which exponent of parcels (u) and the exponent of the result (w) are co-primes. Two numbers are co-primes when they have only the unit as a common divisor (* gcd(u,w) = 1*).

In this set there are two subsets of equations: those which number of parcels (p) are smaller than the exponent of the parcels (u), or p < u; and which number of parcels (p) are greater or equal to the exponent of the parcels (u), or p >= u.

Similarly, the set called '*no co-primes*' corresponds to the set of equations which exponent of parcels (u) and the exponent of the result (w) are not co-primes.

In this set there are also two subsets: those which number of parcels (p) are greater or equal to the exponent of the parcels (u), or p >= u; and which number of parcels (p) are less or equal of the exponent of the parcels (u), or p < u.

So, we can say that the statement of Parcelatories' Powers corresponds to the union of the '*co-prime*' Set with the '*no co-prime*' Subset, which number of parcels (p) are greater or equal to the exponent of the parcels (u). This union of sets corresponds to the blue areas represented in the drawing above.

For a complete demonstration of the statement is necessary to demonstrate separately what occurs in the '*co-prime*' Set and in the '*no co-prime*' Set. The union of the demonstrations will be the closed proof.

*DEMONSTRATIONS OF THE EXISTENCE OF INTEGER RESULTS*

IN EQUATIONS WITH CO-PRIME EXPONENTSIN EQUATIONS WITH CO-PRIME EXPONENTS

*I Proposition:*This proposition corresponds to the Set of Equations called '

*co-primes*', commented earlier:

**First Demonstration**

Let's begin determining the veracity of the existence of natural solutions to an equation which assertion is:

*"The sum of a triad of the seventh power is equal to a biquadratic"*, or in states

a^{7} + b^{7} + c^{7} = d^{4}, where **a**, **b**, **c** and **d** ∈ Ν

*l*),

*id est*,

a = b = c = 3^{k}, where **k** ∈ Ν,

and

d = 3^{l}, where ** l** ∈ Ν.

So,

a^{7} + b^{7} + c^{7} = d^{4} becomes 3^{7k} + 3^{7k} + 3^{7k} = 3^{4l}

or

3 * 3^{7k} = 3^{4l}, or yet 3^{7k + 1}= 3^{4l}.

In which we can conclude that

7k + 1 = 4*l*

or

(7k + 1) / 4 = *l*, noting that **k** and **l** ∈ Ν.

In this case, it is quickly that k = 1 is one of the possible solutions to the equation, because it results in

*l*= 2.

3^{7k} + 3^{7k} + 3^{7k} = 3^{4l}

becomes

3^{7} + 3^{7} + 3^{7} = 3^{8},

leading us to conclude that the equation *"the sum of a triad of the seventh power is equal to a biquadratic"* is also true in the Natural Numbers Set!

Note that the exponent of the parcels (u=7) is co-prime of exponent of the result (w=4), because both only have the unit as the common divisor (** gcd(7,4) = 1**).

Below are plotted some possible solutions to this equation:

**Second Demonstration**A second example is the determination of natural solutions to an equation which assertion is described as follows:

*"The sum of two eighth power is equal to a third eighth power"*, or in states

a^{8} + b^{8} = c^{8}, where **a**, **b** and **c** ∈ Ν.

*l*), ie,

a = b = 2^{k}, where **k** ∈ Ν,

and

c = 2*l*, where ** l** ∈ Ν.

So,

a^{8} + b^{8} = c^{8} becomes 2^{8k} + 2^{8k} = 2^{8l}

or

2 * 2^{8k} = 2^{8l} or yet 2^{8k + 1} = 2^{8l}.

It follows that

8k + 1 = 8*l*

or

or yet

(k/8) + (1/8) = *l*.

*l*, because of the fractional term of the equation (1/8). This can be seen plotting somes values for k and

*l*:

*"the sum of two eighth power is equal to a third eighth power"*has NO solutions with all equal parcels in the Natural Numbers Set!

Note that the exponent of the parcels (u=8) is not co-prime with the exponent of the result (w=8), because both have the unit and themselves as common divisors (** gcd(8,8) ? {1,8}**).

**Third Demonstration**

A third example is the determination of natural solutions to an equation which assertion is described as follows:

*"The sum of a quartet of eighth power is equal to a twelfth power"*, or in states

a^{8} + b^{8} + c^{8} + d^{8} = e^{12}, where **a**, **b**, **c**, **d** and **e** ∈ Ν

*l*), ie,

a = b = c = d = 4^{k}, where **k** ∈ Ν,

and

e = 4^{l}, where ** l** ∈ Ν.

So,

a^{8} + b^{8} + c^{8} + d^{8} = e^{12}, becames 4^{8k} + 4^{8k} + 4^{8k} + 4^{8k} = 4^{12l}

or

4 * 4^{8k} = 4^{12l}, or yet 4^{8k+1} = 4^{12l}.

It follows that

8k + 1 = 12*l*

or

(8k + 1)/12 = *l*,

or yet

(2k/3) + (1/12) = *l*, noting that **k** and **l** ∈ Ν.

*l*, because of the fractional part of the equation (1/12). This can be seen plotting some values for k and

*l*:

Otherwise, we also can **group the parcels two by two**, consider the values of parcels equal to 2 raised to any integer (k), and the result also equal to 2 raised to any other integer (*l*), ie,

a = b = c = d = 2^{k}, where **k** ∈ Ν,

and

e = 2^{l}, where ** l** ∈ Ν.

So,

(a^{8} + b^{8}) + (c^{8} + d^{8}) = e^{12} becomes (2^{8k} + 2^{8k}) + (2^{8k} + 2^{8k}) = 2^{12l}

or

(2 * 2^{8k}) + (2 * 2^{8k}) = 2^{12l}, or (2^{8k+1}) + (2^{8k+1}) = 2^{12l}.

It follows that

8k + 2 = 12*l*

or

(8k + 2)/12 = *l*,

or yet

(2k/3) + (1/6) = *l*, noting that **k** and ** l** ∈ Ν.

*l*, because of the fractional part of the equation (1/6). This can be seen plotting some values for k and

*l*:

The same process can be achieved by applying grouping of **three by three parcels**, considering the values of parcels equal to 3 raised to any integer (k), and the result equals to 3 also raised to any other integer (*l*), ie,

a = b = c = 3^{k}, where **k** ∈ Ν,

and

d = 3^{l}, where ** l** ∈ Ν,

and

e = 3^{m}, where **m** ∈ Ν.

So,

(a^{8} + b^{8} + c^{8}) + d^{8} = e^{12} becomes (3^{8k} + 3^{8k} + 3^{8k}) + 3^{8l} = 3^{12m}

or

(3 * 3^{8k}) + 3^{8l} = 3^{12m}, or yet (3^{8k+1}) + 3^{8l} = 3^{12m}.

In which we can conclude that to have some simplification and continue on, would have

8^{k + 1} = 8^{l}

or

Note that this equation is identical to the final equation of the

*(*

**Second Demonstration****I**). And, as was seen before, for any value of k is impossible to obtain an integer value for

*l*.

*** * ***

In general, whereas the exponent of the parcels (u) is multiple of the exponent of the results (w), we have always the following possibility of homogeneous groups:

Table 1:

and the following possibilities of heterogeneous groups:

Table 2:

*** * ***

Returning to the

*, during the try to find groups of three by three parcels that satisfy the equation, there was a need to verify previously whether there are groups of two by two that could satisfy the equation.*

**Third Demonstration**However,

That is, there are NO solutions to the equation presented in the * Third Demonstration*. And note that the exponent of the parcels (u=8) is not co-prime with the exponent of the result (w=8), because both have the unit and themselves as common divisors (

*).*

**gcd(8,8) ? {1,8}**

So, the first kind of grouping determines the condition of natural solutions to any equation: *if the exponent of the parcels (u) is multiple of the exponent of the result (w), then there may be only natural solutions when the exponent of the parcels (u) is equal to unity*, and this only occurs when they are co-primes.

In other words,

*DEMONSTRATIONS OF THE EXISTENCE OF INTEGER RESULTS*

IN EQUATIONS WITH NO CO-PRIME EXPONENTSIN EQUATIONS WITH NO CO-PRIME EXPONENTS

*II Proposition:*

This proposition does not conflict with the previous one, because it defines the Set of Equations of which have parcels that can be grouped homogeneously those in which all values of each parcel are heterogeneous (a

^{1}? a

^{2}? a

^{3}... ? a

^{p}), or yet those in which groups can be heterogeneous.

It corresponds to the union of Subsets of the Sets of Equations called '*co-primes*' and '*no co-primes*' discussed above, in which the number of parcels is greater than or equal to the exponent of the parcels (p >= u) (blue filled area):

In the text On the Prism of the Parcelatories, was defined that

This definition shows that p and v are divisors of b^{w}. So,

b = p^{1/w} * v^{1/w}

this means that,

b = p^{1/w} or b = v^{1/w}.

But, for exist

p^{1/w} ∈ Ν or v^{1/w} ∈ Ν

the conditions below must be satisfied:

p = w or v = w.

How we know in advance only the number of parcels (p) and nothing about the values of the parcels (v), we can only use the first part of this preposition, or

Furthermore, it is observed that the definition of the Theorem of Parcelatories refers to the equation in which all parcels are equal, ie, the equation

p * v = b^{w}

refers to:

a_{1}^{kx} + a_{2}^{kx} + a_{3}^{kx} + ... + a_{p}^{kx} = b^{ly},

where we have

v = a_{1}^{kx} = a_{2}^{kx} = a_{3}^{kx} = ... = a_{p}^{kx}.

And in a more accurate analysis, we can see that

and that leads us back to the aforementioned equation

by which we also can state that

A careful observer might comment that been the value of all parcels equals to v, this case corresponds only the first kind of homogeneous grouping mentioned in the first table. However, the same principle can be applied both in homogeneous groups as heterogeneous groups, as has been seen before, the same equation applies in the two cases due to the symmetry of the quantity of items of the grouping (n). In other words, the general equation can be simplified as follows:

**First Demonstration**

There are, at least, two different ways to demonstrate this proposition. One of them is evidencing the inefficiency of the process of homogeneous grouping to determine the veracity of the existence of entire solutions to an equation which assertion is:

*"The sum of four cubes is equal to a sixth power"*, or

a^{3} + b^{3} + c^{3} + d^{3} = e^{6}, where **a**, **b**, **c**, **d** and **e** ∈ Ν

a = b = c = d = 4^{k}, where **k** ∈ Ν,

and

e = 4^{l}, where ** l** ∈ Ν.

So,

a^{3} + b^{3} + c^{3} + d^{3} = e^{6} becames 4^{3k} + 4^{3k} + 4^{3k} + 4^{3k} = 4^{6l}

or

4 * 4^{3k} = 4^{6l}, or yet 4^{3k+1} = 4^{6l}.

It follows that

6k + 1 = 6*l*

or

(6k + 1)/6 = *l*

or yet

(k/6) + (1/6) = *l*, noting that **k** and ** l** ∈ Ν.

In this case, for any value of k would be impossible to get an integer value for *l*, because of the fractional part of the equation (1/6).

But there are several solutions to this equation, with heterogeneous values or groups, as can be seen in the table plotted below:

That is, even for different exponents, not co-primes to each other (* gcd(3,6) ? {1,3}*), there are solutions to the above equation, because the number of parcels (p) is greater than the value the exponent of the parcels (u).

**Second Demonstration**

In this second demonstration let us determine the inefficiency of the process of homogeneous groups to determine the veracity of the existence of natural solutions to an equation which assertion is:

"*The sum of two squares is equal to a third square"*, or in states

a^{2} + b^{2} = c^{2}, where **a**, **b** and **c** ∈ Ν

Similarly, if it is used the process of grouping, we conclude that in this case there are no values that satisfy the equation, because the exponents are not co-primes. Otherwise, let we see:

a = b = 2^{k}, where **k** ∈ Ν,

and

c = 2** l**, where

**∈ Ν.**

*l*So,

a^{2} + b^{2} = c^{2} becomes 2^{2k} + 2^{2k} = 2^{2l}

or

2 * 2^{2k} = 2^{2l}, or yet 2^{2k+1} = 2^{l}.

It follows that

2k + 1 = 2*l*

or

(2k+1)/2 = *l*

or yet

(k/2) + (1/2) = *l*, noting that **k** and **l** ∈ Ν.

Again, for any value of k would be impossible to get any integer value for *l*, because of the fractional part of the equation (1/2).

However, how is well known, there are several solution with heterogeneous values to this equation, also known as Equation of Pythagoras' Theorem, as can be seen in the table plotted below:

That is, even for different exponents, not co-primes to each other (

*), there are solutions to the above equation when the number of parcels (p) is greater or equal to value of the exponent of the parcels (u).*

**gcd(2,2) Ν {1,2}**

In other words,

*CONCLUSION AND NEWS*Someone will wonder if the *Theorem of Parcelatórias Power* is a more simple, or even the real solution to the equation of Fermat's Last Theorem.

I prefer to stick the true meaning of this mathematical discovery: with a simple statement it is possible to determine beforehand whether a polynomial equation has or has not solution in the Set of Natural Numbers Ν.

Let's suppose that someone wants to know if the equation a^{4} + b^{4} + c^{4} = d^{3} has the solution Set of Natural Numbers Ν. The practical rule would be as follows:

**Step 1:** determine if the equation is similar to that of Fermat's last theorem, or ie, if a power is equal to the sum of others powers in which all exponents are equal.

*In this case, the answer is yes, because a cube is expressed as being equal to a sum of powers in which all exponents are equal to 4.*

**Step 2:** determine the number of parcels (p), the exponent of the result (w) and the exponents of the parcels (u).

*In this case, we have: p = 3w = 3u = 4*

**Step 3:** determine if the exponents (u and w) are co-primes.

*In this case, 4 and 3 are co-primes, because gcd(4, 3) = 1. *

**Step 4:** if they are co-primes, we can say that the equation has solution in the Set of Natural Numbers N. Otherwise continue to the next step.

*In this case, we can stop here because it was determined that the exponents are co-primes.*

**Step 5:** if they are not co-primes, dertemine whether the number of parcels (p) is greater or equal to the exponent of the plots (u).

*In this case, although it be an irrelevant information, we have 3 <= 4.*

As it can be seen in the spreadsheet attached HERE, the Set of Equations defined by the *Theorem of Parcelatories' Powers* reminds a projected urban center, like is the Manhattan' Island, in New York City, where all the avenues (u) and streets (w) are perfectly symmetrical with each other and which transversal, as the Broadway Avenue (u=w), separates two sets of equations: those that admit only one or no solution, and those that admit no, one or more solutions.

It is interessant to note that this last case, which occurs when u <= w, evidences the *Cabtaxi-like Numbers*, those powers that are equivalent to more than one kind of Parcelatory Power.

A *Cabtaxi Number* that I always comment is the number 18^{3}, which admits two possibilities for Parcelatories' Powers:

(2^{3} **+** 12^{3} **+** 16^{3}) **=** 18^{3} **=** (9^{3} **+** 12^{3} **+** 15^{3})

+ + = = + +

There was infinite cases like that, in what it can see that it is perfectly possible for a cube correspond to the sum of two distinct trinity from other different cubes. This among other topics are motives to more investigatins and determinations of news theorems.

Any way, irrespective of how many solutions can be obtained, joining the sets mentioned in this text, we get to the complete proposition of *Parcelatories' Powers*:

**REFERENCES**

[1] Dickson, Leornard Eugene. History of the Theory of Numbers, 1919-23.

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