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    Parcelatories' Powers - Equations Like Fermat's Last Theorem
    By Eduardo Sardeiro | July 7th 2009 11:44 AM | 21 comments | Print | E-mail | Track Comments
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    I'm a brazilian scientist and automation systems consultant, who studies intelligence in its different aspects, from the Artificial Intelligence...

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    "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)



    Word 'one'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:


    Parcelatories' Powers Statement


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


    a2 + b2 = c2, where a, b and c ∈ Ν

    a3 + b3 = c2, where a, b and c ∈ Ν

    a2 + b2 = c3, where a, b and c ∈ Ν

    a2 + b2 + c2 = d2, where a, b, c and d ∈ Ν

    a11 + b11 + c11 = d7, where a, b, c and d ∈ Ν

    or even

    a1001 + b1001 + c1001 + d1001 = e2003, 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:


    CoPrimes and No-CoPrimes Sets of Equations


    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 EXPONENTS

    I Proposition:

    Statement Of First Preposition

    This proposition corresponds to the Set of Equations called 'co-primes', commented earlier:



    Co-Primes Set of Equations



    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

    a7 + b7 + c7 = d4, where a, b, c and d ∈ Ν

    In a first attempt, we can consider all parcels equal to the number of parcels (p=3) raised to any integer (k), and the result equal to the number of parcels (p=3) raised to any other integer (l), id est,

    a = b = c = 3k, where k ∈ Ν,

    and

    d = 3l, where l ∈ Ν.

    So,

    a7 + b7 + c7 = d4 becomes 37k + 37k + 37k = 34l

    or

    3 * 37k = 34l, or yet 37k + 1= 34l.

    In which we can conclude that

    7k + 1 = 4l

    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.

    The equation

    37k + 37k + 37k = 34l

    becomes

    37 + 37 + 37 = 38,

    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:


    Possibles values of equation a^7 + b^7 + c^7 = d^4




    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

    a8 + b8 = c8, where a, b and c ∈ Ν.

    To follow the same technique used in the previous demonstration, we must consider all parcels equal to the number of parcels (p=2) raised to any integer (k), and the result equal to the number of parcels (p=2) raised to another any integer (l), ie,

    a = b = 2k, where k ∈ Ν,

    and

    c = 2l, where l ∈ Ν.

    So,

    a8 + b8 = c8 becomes 28k + 28k = 28l

    or

    2 * 28k = 28l or yet 28k + 1 = 28l.

    It follows that

    8k + 1 = 8l

    or


    Equation 1


    or yet

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

    In this case, for any value of k is impossible to obtain an integer value for l, because of the fractional term of the equation (1/8). This can be seen plotting somes values for k and l:



    Possibles values to k and l in the equation a^8 + b^8 = c^8



    This leads us to conclude that the equation "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

    a8 + b8 + c8 + d8 = e12, where a, b, c, d and e ∈ Ν

    If we follow the same technique used in the previous demonstration, we must consider all parcels equal to the number of parcels (p=4) raised to any integer (k), and the result equals to the number of parcels (p=4) raised to another any integer (l), ie,

    a = b = c = d = 4k, where k ∈ Ν,

    and

    e = 4l, where l ∈ Ν.

    So,

    a8 + b8 + c8 + d8 = e12, becames 48k + 48k + 48k + 48k = 412l

    or

    4 * 48k = 412l, or yet 48k+1 = 412l.

    It follows that

    8k + 1 = 12l

    or

    (8k + 1)/12 = l,

    or yet

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


    Again, for any value of k is impossible to obtain an integer value for l, because of the fractional part of the equation (1/12). This can be seen plotting some values for k and l:


    Possibles values to k and l in the equation a^8 + b^8 + c^8 + d^8 = e^8 (case 1)



    That is, there are NO solutions with all equal parcels for this equation in the Natural Numbers Sets.


    1 Note



    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 = 2k, where k ∈ Ν,

    and

    e = 2l, where l ∈ Ν.

    So,

    (a8 + b8) + (c8 + d8) = e12 becomes (28k + 28k) + (28k + 28k) = 212l

    or

    (2 * 28k) + (2 * 28k) = 212l, or (28k+1) + (28k+1) = 212l.

    It follows that

    8k + 2 = 12l

    or

    (8k + 2)/12 = l,

    or yet

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

    Also in this case, for any value of k is impossible to obtain an integer value for l, because of the fractional part of the equation (1/6). This can be seen plotting some values for k and l:


    Possibles values to k and l in the equation a^8 + b^8 + c^8 + d^8 = e^8 (case 2)


    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 = 3k, where k ∈ Ν,

    and

    d = 3l, where l ∈ Ν,

    and

    e = 3m, where m ∈ Ν.

    So,

    (a8 + b8 + c8) + d8 = e12 becomes (38k + 38k + 38k) + 38l = 312m

    or

    (3 * 38k) + 38l = 312m, or yet (38k+1) + 38l = 312m.


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

    8k + 1 = 8l

    or


    I Equation



    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:

    Homogeneus Groups


    and the following possibilities of heterogeneous groups:


    Table 2:

    Heterogeneous Groups

    * * *


    Returning to the Third Demonstration, 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.

    However,

    Homogeneous Groups Statement


    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,


    Statement Of First Preposition






    DEMONSTRATIONS OF THE EXISTENCE OF INTEGER RESULTS
    IN EQUATIONS WITH NO CO-PRIME EXPONENTS


    II Proposition:


    Statement Of Second Preposition


    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 (a1 ? a2 ? a3 ... ? ap), 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):


    CoPrimes and no CoPrimes SubSets


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


    Potentiation Definition


    This definition shows that p and v are divisors of bw. So,

    b = p1/w * v1/w

    this means that,

    b = p1/w or b = v1/w.

    But, for exist

    p1/w ∈ Ν or v1/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


    p greater or equal to w


    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 = bw

    refers to:

    a1kx + a2kx + a3kx + ... + apkx = bly,

    where we have

    v = a1kx = a2kx = a3kx = ... = apkx.

    And in a more accurate analysis, we can see that


    Inverse Equation


    and that leads us back to the aforementioned equation


    Main Equation


    by which we also can state that


    p greater or equal to w



    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:


    General Equation



    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

    a3 + b3 + c3 + d3 = e6, where a, b, c, d and e ∈ Ν

    Case it been used the process of grouping, we conclude that in this case there are not values that satisfy the equation, because the exponents 3 and 6 are not co-primes. Otherwise, let we see:

    a = b = c = d = 4k, where k ∈ Ν,

    and

    e = 4l, where l ∈ Ν.

    So,

    a3 + b3 + c3 + d3 = e6 becames 43k + 43k + 43k + 43k = 46l

    or

    4 * 43k = 46l, or yet 43k+1 = 46l.

    It follows that

    6k + 1 = 6l

    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:


    Possibles values to equation a^3 + b^3 + c^3 + d^3 = e^6



    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

    a2 + b2 = c2, 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 = 2k, where k ∈ Ν,

    and

    c = 2l, where l ∈ Ν.

    So,

    a2 + b2 = c2 becomes 22k + 22k = 22l

    or

    2 * 22k = 22l, or yet 22k+1 = 2l.

    It follows that

    2k + 1 = 2l

    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:


    Possibles values to equation a^2 + b^2 = c^2



    That is, even for different exponents, not co-primes to each other (gcd(2,2) Ν {1,2}), 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).


    In other words,


    Statement Of Second Preposition




    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 a4 + b4 + c4 = d3 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 = 3
      w = 3
      u = 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.


    City Of Parcelatories' Powers


    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 183, which admits two possibilities for Parcelatories' Powers:

    (23 + 123 + 163) = 183 = (93 + 123 + 153)


    Cube of 2 + Cube of 12 + Cube of 16 = Cube of 18 = Cube of 9 + Cube of 12 +Cube of 15



    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:


    Parcelatories' Powers Statement




    REFERENCES

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




    Comments

    I already knew that. Math, prices, parcels, avenues, streets, taxis, deliveries and Manhatan. Everything Is connected to Everything else! Great!

    esardeiro
    Greetings to all NewYorkers!
    Thank you! Now how does the Fermat's Last Theorem take place?

    esardeiro
    Some friends have made me similar questions. The answer is the renowned equation of Fermat's Last Theorem (a^u + b^u = c^w) has the exponent of the parcels (u) and the exponent of the result (w) always equals. This means that the equation should NEVER be resolved, since exponents are not coprimes, because they have to themselves and the unit as common divisors (gcd(u,w) ∈ {1,u,w}), But the equation has only 2 parcels (p=2) and this creates the only exception that occurs when the exponent of the parcels is less than or equal to 2 (u <= 2). You can understand this better with the Theorem of Parcelatories. It was the "wall" which needed to explain the Fermat's Last Theorem. Fermat must have thought something like this to deduce his last theorem.
    esardeiro
    Hi everyone!

    The worksheet CalculationOfEquations.xls is already available online. With it, you can calculate equations like Fermat's Last Theorem equation.

    I marked with red those equations which doesn't have any solution, with green those equations which have one or more solutions and with blue those equations which the numbers of parcels (p) is greater than or equal to the exponent of the parcels (x) and which can have one or more solutions, in agree with the Parcelatories' Powers Theorem.

    If you need more information or have questions, please, contact me.
    So I'm unclear. Have you claimed to have proved Fermat's last theorem? Or are you saying that Fermat's last theorem is an exception to your theorem and therefore not provable?

    Also, are there any papers about this material (or mentioned in any books)? I've had horrible luck trying to find any preprints or anything on this subject.

    esardeiro
    Hi Justin!
    For sure, the Fermat's last theorem is a case of the Parcelatories' Powers theorem. It's not an exception.
    Do you have any doubt about the mathematical logic used here? If so, please, let me know.
    I believed I have found solutions to Fermat's Last Theorem! Can you please tell me how the Parcelatories' Powers theorem explains them?

    1782^12 + 1841^12 = 1922^12
    3897^12 + 4365^12 = 4472^12

    I am very interested in the answer to this. I would like to understand how the Parcelatories' Powers theorem proves Fermat's last theorem, but I have not been able to use it to show these are not solutions!

    esardeiro

    There are some discussion forums about this supposed solutions do Fermat's Last Theorem. Here is one of them: 1782^12 + 1841^12 = 1922^12?.

    In both of this equations, it can be identify:



    a^x + b^x = c^y



    p (number of parcels) = 2
    x (exponent of the parcels) = 12
    y (exponent of the result) = 12



    x and y aren't coprimes because they have themselves (12) and the unit (1) as common divisors. And the number the parcels (p) is less the the exponent of the result (y)  (2 < 12).

    So, the Parcelatories' Powers theorem says that when this occurs, the equation doesn't have any solutions.

    Yes, there is a little difference between x and y, and u and w. I didn't describe in this article, but this is irrelevant in this case.

    I developed a web application to found possible mistakes in this theorem. How do you know, a theorem must be proved on all aspects, included Big Numbers. I invite you to join us to explore this theorem. The link is:

    Parcelatorial Project



    Best Regards,

    Thanks for the prompt reply!

    In that case, most of the reasoning above is done with specific examples and I see little appeal to proof. In addition, the proof for p>=u seems to break down when you invoke the equation pv = b^w (which honestly just seems to follow from the fact that you can write b^w = b^(w-a)*b^a). You say that:

    b=p^(1/w)*v^(1/w)

    and that implies

    b=p^(1/w) or b=v^(1/w)

    which is certainly a false implication (3^3 = 9 * 3 but 9^(1/3) and 3^(1/3) are both not integers). Furthermore, if a number p^(1/w) is a natural number, it certainly does not imply p=w (for example, p^(1/2) a natural number does not mean p=2). Either I vastly misunderstand what those lines mean (always possible), or they are wrong.

    The rest of the analysis has some leaps that I'm not convinced of as well, and it seems like most of the argument for your theorem again rests on specific examples... which is not proof.

    Furthermore, the theorem in itself is of the form:

    "For all u,w, and p in the natural numbers, if u and w are coprimes or if p>=u, then the equation a1^u + a2^u + ... + ap^u = b^w has a solution with a1,...ap,b in the natural numbers."

    This is not the same as

    "For all u,w, and p in the natural numbers, if u and w are not coprimes and if u>p, then the equation a1^u + a2^u + ... + ap^u = b^w does not have a solution with a1,...ap,b in the natural numbers."

    One is of the form A=>B and the other (not A) => (not B) does not follow (Given Aristotle is a man, "all men are mortal => Aristotle is mortal" does not imply "at least one man is immortal => Aristotle is immortal".)

    Now, Fermat's last theorem is of the form: "Given n>2, the equation a^n + b^n = c^n does not admit integer solutions". If we are to look at this equation with your theorem (which I have doubts to the theorem's validity as well) then we have that the exponents are not coprime and that n>2 instead of 2>=n as you require. Thus, ALL we can say is that the theorem does not tell us the equation admits natural number solutions. It may still admit such solutions, that is logically still allowed, but your theorem does not guarantee it.

    Just to hammer the point home, I can craft a theorem that says

    "Given real numbers x and y, if x and y are integers, then x+y is an integer". (TRUE)

    From this I cannot say

    "Given real numbers x and y, if x and y are not integers, then x+y is not an integer." (FALSE. Counter example x=1/2=y, then x+y=1).

    So to enumerate my issues:

    1) The proof of the theorem seems either shaky or based solely on examples.
    2) The theorem, even if true, does not apply to Fermat's last theorem on purely logical grounds.

    esardeiro

    Hello Justin!
    Only now I read your message.
    First of all, thanks for your participation and math observations. All this effort will be rewarding soon.
    There is a forum in Portuguese on the link Parcelatórias de Potências where this piece of text was also questioned by other mathematicians, here in South America.
    An important point to consider is the set to which we are working. I always mention the set of the Natural Numbers to define the issue as well.
    So when I say

    b = p^(1/w) * v^(1/w)
    and that implies
    b= p^(1/w) or b=v^(1/w)

    it's because I made clear initially that the variables b, p, w and v are necessarily integers. Any value not belonging to this set, it'll be out of the question.
    This kind of mathematical procedure is widely applied and probably it was used by Fermat in the 17th century. For us, who are accustomed to working with Real Numbers, it looks like "jumps", but this is only an impression.
    I'll continue in next message.

    esardeiro
    Continuing ...
    The Parcelatories' Powers theorem is an intersection of two propositions. When the first failure (becames false), the second must work, exactly in that order. Thus, the connector "OR" is more appropriate.
    If the set of Natural Numbers were considerated, there is no problem in this syllogism. Syllogisms exist to be used, don't they? I mean, I'm not using fallacies.
    So, if I'm telling you that my name is "Eduardo Sardeiro", is not that I want you to believe it! But I want that you prove this too by youself!
    This truth is to be a general one and not only my thuth. And we are here to find the truth, aren't we?
    I'll continue next.
    esardeiro

    I wrote the text with many examples to make it clear. But if you see, the Parcelatories' Powers theorem doesn't need examples.
    Also, the Parcelatories' Powers theorem doesn't need the Fermat's Last Theorem to be proved. How I said before, It's much more complete and prove another theorems too!
    There is a proof by infinite induction  that has not been presented yet, but any good mathematician can extract it from observations of the Excel spreadsheet that was available above.
    Big hug,

    Yes, the theorem says this is the case, but you could also say you were Napoleon and I might believe you! I have not seen a proof of this theorem, so please forgive my skepticism.

    Is there some logic in the proof of the Parcelatories' Powers theorem that shows that
    178198219^12 + 18409158^12 = 192198078^12
    is not a solution? If so I would greatly like to see it! I do not think the link you gave will work in this case.

    Or maybe a copy of the proof of the theorem is available?

    esardeiro

    Hi again, Justin!

    Wasn't the evidence I have presented, relating p (number of the parcels), x (exponent of the parcels) and y (exponent of the results) enough? And the text explains all the transformations.
    Maybe you should read the previous text, On the Prism Of the Parcelatories. It's a short introduction of the Parcelatories Theory and the begin of all.
    How I said, there is a proof by infinite induction too, that I hasn't been presented yet, but any good mathematician can extract it from observations of the first excel spreadsheet that was available above.
    I'll be presenting all of this when I publish the Theory of CabTaxi-like Numbers.
    Regards,

    Hi Eduardo! Great work!
    I'd like to understand this better too.
    Can you explain, detailed and step by step, if the equations 1782^12 + 1841^12 = 1922^12" and "3897^12 + 4365^12 = 4472^12" have solutions or not?

    esardeiro
    Ok Peter! Let's go!

    The Parcelatories' Powers Theorem says:
    "The equation a1u + a2u + a3u + ... + apu = bw, where a, b, p, u and w belong to the set of Natural Numbers, will have solutions if the exponent parcels (u) and the exponent of the result (w) are co-primes, or if the number of parcels (p) is greater than or equal to the exponent of the parcels (u)"


    And both equations above are of the kind:
    a112 + a212 = b12,


    where the number of parcels (p) is equal to 2, the exponent of the parcels (u) is equal to 12 and the exponent of the result (w) is also equals to 12.

    According to the statement, the FIRST STEP is to determine if there is the possibility of grouping the parcels homogeneously, or in other words, if the exponents are co-primes.

    Because there are only two parcels, the only form of homogeneous grouping must be a1 = a2.

    Thus, we would have something like this:
    a1 = a2 = 2i

    and

    b = 2j,






    where i and j are any whole numbers.

    Note that it was chosen a power of two with the intent to simplify the bases. It's possible only the number two because the equation has only two parcels.

    Such that
    (2i)12 + (2i)12 = (2j)12 (substitution)

    or

    212i + 212i = 212j (multiplication of exponents).






    Hence we have
    2 * (212i) = 212j (putting 2 in evidence).


    This is the same that

    (21) * (212i) = 212j,



    and so
    2(12i + 1) = 212j (sum of exponents).


    Now we can simplify the operation eliminating the bases, because they are equal:
    12i + 1 = 12j.


    According to our initial guidelines, i and j must be whole numbers. Now, the question is if we can find integers i and j which satisfy the equation above.

    In other words, we need to determine i and j to satisfy the equation below in the set of Integer Numbers:
    (12i + 1) / 12 = j (transporting 12 to the left side of equation).


    What is not possible, because simplified, it becomes:
    i + (1/12) = j,


    where you can see that there will always be a fraction (1/12) in this sum.

    That is, the verification by homogeneous group tells us that there aren't any solutions to this equation.

    The same information could be to get quickly verifying if the exponent are co-prime. In this case they aren't, because 12 and 12 have themself and one as divisors in common. And, if they have any divisor in common greater than one, then the division (12i + 1) / 12 never could be perfect.

    Until here, is it simple, isn't?


    BUT this theorem has the "or" connector that prevents us from finalizing the check here, being necessary to perform a SECOND STEP.

    We need to see if there is the possibility of heterogeneous grouping (with different grouping of parcels).

    Then comes the definition of power obtained by the Theory of Parcelatories:

    "Powers are parcelatorias perfect which follow the equation p * v = bw"


    In our case, we know that p = 2 and w = 12. We also know the value of b that is provided (1,922 and 4,472), but this isn't necessary.

    What matters is that the value of bw, if exists, should be multiple of p and v, after all they are matched in the above equation by multiplication.

    Going forward, we can also say that b, if exists, should be multiple of p1/w and v1/w.

    Here is the point where is made an abstraction, only useful for the set of Natural Numbers:
    "On equations as n1/m, if there is solutions in the set of Natural Numbers, n must be greater than or equal to m (n >= m)"


    This happens because the operation of extracting the root of a number is similar to an operation of division with equal factors. So, if we divide a number by an other greater than it (m > n), we will never have a whole number. In other words, the result of equations like n/(n+k), where k>0, doesn't belong to the set of Natural Numbers, because (n+k) is greater than n.

    This is obvious, isn't?


    It is also easy to see that for the case n = m, the only solution can be n = m = 1, what results in n1/m = 11/1 = 1.

    So, if there is a possibility of getting an entire solution for equations like n1/m, this solution will only happen when n >= m, precisely.

    In our case, we mention that the condition for b exists, it must be multiple of p1/w and v1/w. And, as we have seen, for this to happen, p must be greater than or equal to w, and v must also be greater than or equal to w, what doesn't happen with p, since 2<12.

    The same logic applies to all terms on both side of equation.

    This means you never have the perfect parcelatories (powers) in the formation of its parcels. It could exist groupings, but this was already discarded in the first step.

    Getting false responses in both steps, we can assert surely that the 178212 + 184112 = 192212, 389712 + 436512 = 447212 or any other equation similar to a12 + b12 = c12, don't have any solutions in the set of Natural Numbers.


    Many people calculate these equations in software such as MS-Excel which around Big Numbers, don't been possible to verify the accuracy of the operations. But if you calculate by means of specific operations, you can find that:
    1,78212 = 1,025,397,835,622,633,634,807,550,462,948,226,174,976
    1,84112 = 1,515,812,422,991,955,541,481,119,495,194,202,351,681
    1,92212 = 2,541,210,259,314,801,410,819,278,649,643,651,567,616

    1,78212 + 184112 = 2,541,210,258,614,589,176,288,669,958,142,428,526,657

    1,92212 - (178212 + 184112) = 490,414,587,369,321,906,673,569,032,245,976,176,705









    So, 178212 + 184112 is different of 192212.

    In the same way,
    3,89712 = 12,267,748,656,558,579,529,608,352,140,578,465,199,191,841
    4,36512 = 47,842,181,739,947,321,332,739,738,982,639,336,181,640,625
    4,47212 = 63,976,656,348,486,725,806,862,358,322,168,575,784,124,416

    3,89712 + 4,36512 = 60,109,930,396,505,900,862,348,091,123,217,801,380,832,466

    4,47212 - (3,89712 + 4,36512) = 35,574,433,083,388,741,803,131,386,842,060,870,982,448,784









    So, 389712 + 436512 is different of 447212.

    Of course we can use this theorem to validate the existence of solutions in any equations similar to the equation of Fermat's Last Theorem (when all the parcels have the same exponents, in spite of the exponent of the result). The mathematical rigor used here assures this.

    Some people ask me If I secure, why do I need to do tests with supercomputers? My answer is because I need to test another theorems not published yet. But, in this case, I'm entirely secure!

    I hope to made clear the use of the Parcelatories' Powers theorem. Anyway, read the Theory of Parcelatories to get a better overview. Any other question, please, let me know.

    Best Regards,













































































    Hi Eduardo! Richard here again with another question.
    The OR connector doesn't make somehow incomplete or inconsistent in this theorem?
    Thanks!

    esardeiro
    Hi Richard!

    The most popular mathematical theorems are those which have a single clause, ie, those which we can express in a single sentence. The Reductionists search just for that: to synthesize everything in simple statements.

    However there are many other mathematical theorems that are the result of unions, intersections, complements, or subsets of other theorems, usually called propositions.

    In the case of Parcelatories of Powers' Theorem, plotting the truth-table of its two propositions leads us to the table below that, as can be seen, corresponds exactly to the truth-table of the OR operator.

    truth Table

    Rgds.
    Eureka, Eduardo! I finally understood!
    And, by my own, whit the Parcelatories Powers' theorem I constructed a new set of equations, something like this:
    "The equation a^x + b^x + c^x = d^x has integer solutions if, and only if, x <= 3",
    that is similar to the Fermat's last equation, ie,
    "The equation a^x + b^x = c^x, has integer solutions if, and only if, x <= 2"
    I enjoy this possibility: now everyone can construct their own set of equations.
    Regards!

    esardeiro
    This is the idea, Richard! (lol)

    Let's call this equation as the Richard Maya's Last equation, or better, the Richard Maya's First equation.

    I believe there are lots of other kids of equations like it, with dozen of combinations.

    And everyone can try and explorate these ideas so much as possible. So, this door will still open for all interested mathematicians and curious about these subjects!

    So, feel free to make up new questions.

    Big hug!