When I saw the cool but "unknown" sieved pattern of accumulation to powers, (see Table 2), I became very interested in the figurate content of series since accumulation is accumulation, sieve or no sieve.  When I read that Pascal and Fermat had also been interested in figurate numbers in the 17th century¹, my interest felt supported.  I would map out dots in accumulating sets of colors.  (Later I learned these are a kind of Ferrers' diagrams.)

    I put together a rule for figurate patterns—not only for series that are power values but for series of shell/nexus numbers for power values, as well as (partial) summations of the 1^st through x^th of values of a power, not to mention their (partial) summations and also the shells of shells.  I recognized the theme was neighboring figurate number multiples of Euler's Triangle row values.  The n^th row of Euler's Triangle always regarded values for the power of n (whether levels of nexus numbers, integers raised to a power level, or [partial] summations of power values).

    With research, I found that Worpitzky's Identity of 1883 defines how an xⁿ value equals the values of the n^th row of Euler's Triangle, each multiplying the x^th "neighboring area" of (n+1)^th-polytope-figurate numbers, (meaning the (n+1)^th diagonal of Pascal's Triangle,) in order of occurrence, to then be summed.  So this name exists for one level of the figurate pattern I had also identified.²

Fermat's Last Theorem Explained

Table 1.  Figurate numbers as values that define xⁿ values (Worpitzky's Identity of 1883).

Polytope/Figurate Numbers ("n-PT") of Pascal's Triangle (left)            Euler's Triangle (right)

Two neighboring triangular numbers, "weighted" by Euler's Triangle second row of 1 and 1, sum to an x² value.

Three neighboring tetrahedral numbers "weighted" by Euler's Triangle third row of 1, 4 and 1, sum to an x³ value.

Four neighboring pentatope numbers "weighted" by Euler's Triangle fourth row of 1, 11, 11 and 1, sum to an x⁴ value.

Five neighboring 6-PT/polytope numbers "weighted" by Euler's Triangle fifth row of 1, 26, 66, 26 and 1, sum to an x⁵ value.

Et cetera. 

    Remember, each x^th value in a polytope level equals the 1^st through x^th of the previous level, by definition.  It may legitimately be said, for example, that three neighboring sets of the 1^st through x^th of triangular numbers, (rather than tetrahedral numbers) each "weighted" by Euler's Triangle third row of 1, 4, and 1, sum to an x³ value.

    Keeping this in mind when adding two values of a power leads to an elementary proof of Fermat's Last Theorem.  This thinking also makes sense according to the terms of Fermat's comment by which we know the "theorem" (which never was explained).

If a cube were split, could it become two cubes?

Diagram 1.  How cubes add up via tetrahedral values then via triangular values.

(B-2)th values or the (a-1)^th through the (c-2)^th of (next-lower polytope values of) triangular numbers multiplied by 1

Plus (B-1)th values or the a^th through the (c-1)^th of triangular numbers multiplied by 4

Plus Bth values or the (a+1)^th through the c^th of triangular numbers multiplied by 1

would have to add up to b³ if a cube could split into two cubes because the amount defines the value of c³ - a³.

    Can the (a-1)^th through the (c-2)^th then the a^th through the (c-1)^th then the (a+1)^th through the c^th triangular numbers be three consecutive tetrahedral values? (They would have to in order to sum to a cube.  Every single b³ value is defined with three consecutive tetrahedral values "weighted" by 1, 4, and 1.)

    One of the three sets of triangular values might be a tetrahedral value, but never would all three, consecutive sets be consecutive tetrahedral values . . .

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See another essay for the rest of the elementary FLT argument.  The below does not really follow through.  It will be reworked at a later date.  --Cecilia

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By contradiction of logic, to split a cube into two cubes is impossible.

Figure 1.  For this graph of the figurate content of a c³ value (at the level of its shell/nexus numbers), integer a, b and c are 0 < a < b < c.  White values sum to a³, white and orange values sum to b³, and, white, orange and blue values sum to c³. 

Power of 2 comparison

The method of assuming (c-1)th - (a-1)th triangular values is the (b-1)th triangular value then the cth - ath integer values give the bth integer that adds to the first triangular value to add up to the bth triangular value sometimes functions for the power of 2.  But this is not the only allowed formation of square values since a value that is c² - a² may split into two parts (weighted by Euler Triangle row 2 multipliers that are 1 and 1) that are not triangular values (but are re-construable into two neighboring triangular values).

In comparison, higher powers must present as three or more neighboring figurate numbers (of the diagonals of Pascal's Triangle).  Each accelerating-in-values version of the (c-2)th - (a-2)th, ((c-1)th - (a-1)th), and cth - ath values (as well as the parallels in the higher powers) is multiplied by values of Euler Triangle's row n.  For the power of 3, multipliers are already 1,4, and 1 so have to be precisely tetrahedral values because, if the increasing values are not tetrahedral, a sum is given that is not an x³ value (since each xⁿ is defined by Worpitzky's identity of 1883 as three neighboring tetrahedral values multiplied by three or more, specific, non-same Euler Triangle values).

Power of 4

Diagram 2.  How power of 4 values add up via 5-PT values then via tetrahedral values.

To split a power of 4 into two powers of 4 is impossible.

Figure 2.  For this graph of the figurate content of a c⁴ value (at the level of its shell/nexus numbers), integer a, b and c are 0 < a < b < c.  White values sum to a⁴, white and orange values sum to b⁴, and, white, orange and blue values sum to c⁴. 

I.  What are the rates of increase between "B" ^th values of diagram 2?

II.  What "should" the increases be between values, in order to generate consecutive 5-PT values as are requisite for b⁴ definitions?

Can I. and II. be equivalent?

NO.  Notice that the three increases within I. are values that, in turn, increase by the (c-1)^th integer less the (a-1)^th integer, then the c^th integer less the a^th integer, whereas increases within values of II. are the (b-1)^th triangular value and the b^th triangular value.  Once again, (see how cubes split), this is not possibly equal because c - a cannot then equal both of two different, consecutive triangular values.  The rate of increase within B^th values does not match the 5-PT rate of increase.

To split a power of 4 into two power of 4 values is impossible.

Power of 5

Imagine the application of 1, 26, 66, 26, and 1 as multipliers to consecutive 6-PT figurate values, then also applying to 5-PT values in the same manner of diagrams 1 and 2.

I.  What are the increases between "B" ^th values of the power of 5?

II.  What "should" the increases between values be to generate consecutive 6-PT values as are requisite for b⁵ definitions?

Can I. and II. be equivalent?

NO.  Notice that the rate of increase within I., in turn, differs by

while the increases within II. is at this level:

which was already proven at the b⁴ level to NOT be equitable (since ultimately, c - a cannot equal both of two different, consecutive triangular values).  B^th values cannot be 6-PT values.

To split a power of 5 into two power of 5 values is impossible.

Et cetera

Nor may any power higher than the power of 2 be split into two smaller cases of the power because the above pattern holds for all powers, of 3 and larger.  It is clear that "B" sets of figurate values (resulting from the figurate values of cⁿ less the figurate values of aⁿ) can NEVER be consecutive (n+1)-PT figurate values as is required to define bⁿ values.

Cubum autem in duos cubos, aut quadrato-quadratum in duos quadrato-quadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere:  cuius rei demonstrationem mirabilem sane detexi.  Hanc marginis exiguitas non caperet.   

To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain.

--Pierre de Fermat

Table 2.  Sieved Accumulation to the Power:  the initial observation.

Et cetera.

(1)  Pengelley, David J.  The bridge between the continuous and the discrete via original sources in Study the Masters:  The Abel-Fauvel Conference, Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.  http://www.math.nmsu.edu/~davidp/bridge.pdf

(2)  Worpitzky's Identity of 1883 and its application which, incidentally, is part of a pattern of Euler row values as "weights" that extends across the recursively accumulating series that regard a power level.

Click for demonstration page and examples of Worpitzky extended.

Click for a logically reciprocal formulation that is the Euler Triangle, extended—as well as direct accumulations to power values from initial Euler Triangle row values.

— Cecilia@noticingnumbers.net  April 2005

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