Fermat's Last Theorem Proof by Tom Ballard

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SYNOPSIS

This is an improved version of an earlier treatise. It features a

short-form proof that xn + yn = zn is impossible in integers for n >2. It isthe writers belief that this is Fermats original proof. This version

addresses some of the comments from reviewers and resolves theirobjections.

To begin, a model for squared numbers will be introduced and used to

devise a method to create all Pythagorean (x2 + y2 = z2) relationships.Equations will be derived from this process which indicate theexistence of a Pythagorean equation in the model for squarednumbers.

A model for higher powers of "n" will then be introduced. This modelwill be an extension of the model for squared numbers. Simplemanipulations of this model will show that the "end game" packaging of

quantities postulated to be xn and yn into spaces known to be xn and

yn requires that x, y, and z form a Pythagorean equation ! This is

totally incompatible with the postulation that xn + yn = zn where n >2.The proof is thus Reductio ad Absurdum. . The recogniton of theafore-mentioned equations in the packaging process is the essence ofthe proof.

FORWARD

Validating Fermats assertion that he had a proof is of utmostimportance. It will never be known for sure that he had a proof in hisown day and age until some credible effort can be put together tosubstantiate this. This treatise will provide that proof.. Also, a short-form proof may provide valuable adjuncts to the recent accomplishments of "Twentieth Century Mathematics".

We have no direct knowledge of what might have been Fermatsgeneral approach; however, we have a few tantalizing clues from thenotes he is said to have made in the margin of a text.

1. He was studying Diophantine analysis.

2. He conjectured about the possibility that the sum of two cubes could equal a third cube (in integers).

3. He rapidly asserted that this was impossible.

4. With equal rapidity he dismissed the possibility of an integral

solution for xn + yn = zn for any n > 2 !

5. He claimed to have a marvelous demonstration of this that was "Too long to include in the margin".

All of this bespeaks for a proof with the following characteristics:

1. It must be quite brief, perhaps only a few pages.

2. It was inspired by the Diophantine literature he was studying.

3. It required the recognition of a commonality in all xn + yn = zn

equations where n > 2. How else could he have proceeded so rapidly to his final statement?!!

4. It probably has several plateaus of logic which are simple in themselves, but obtuse in their application.

This treatise contains four sections which reconstruct what might havebeen Fermat's methodology in proceeding to a statement of his lasttheorem. These are:

1. Establishment of a model for squared integer numbers.

2. The use of this model to generate all Pythagorean (n =2)relationships and to identify equations which indicate the existence of aPythagorean in x, y, and z when the equations occur in the model for squared numbers.

3. The creation of an advanced model in integers for xn + yn = zn

where n > 2.

4. The use of this model to show that packaging postulated values of higher powers of x and y into spaces known to be actual values results in a Pythagorean in x, y, and z; which is absurd and thereby proves the theorem.

GENERATION OF ALL PYTHAGOREAN TRIPLETS

This will explain the methodology for creating P. triplets from even"root numbers". To the knowledge of the writer, this method is entirelynew. This has been confirmed by many math people. A simple modelfor squared numbers will be introduced. By manipulations of thismodel, it will be shown how a known Pythagorean triplet relates to an even "root number". Then an inverse process will be shown whereby

any even "root number" can generate P. triplets. Depending on thefactorability of the root number, a multiplicity of P. triplets can begenerated from one root number.

This method generates all valid P. triplets. Trivial P. triplets such as 6, 8, and 10 having a common factor, ( in this case, 2), are avoided.

MODEL FOR SQUARED NUMBERS

Figure 1. Model for squared integer numbers

This model consists of a horizontal base of "ones" and an upperstructure of "twos", thereby forming a triangular shape. The verticaland diagonal dimensions of the triangle equal the horizontaldimension. The cumulative value of the contents of the triangle equalsthe square of the horizontal dimension. Figure 1 shows the numbersfrom 1 to 8 and their squares. Observe that the cumulative value isactually the sum of odd numbers.Throughout this treatise, the modelwill be referred to as a "triangle". Sometimes it will be called an r, x, y,or z triangle depending on the horizontal dimension. Wherever

meaningful, it may also be called an r2, x2, y2, or z2 triangle.

Figure 1(a).

The left figure shows how any diagonal segment of the model will

represent a square. The overall triangle is 82, (64), and the shaded

segment is 42, (16). The right figure shows how a vertical trapezoidal segment of the model can be represented by a slanted segment.

CHARACTERISTICS OF PYTHAGOREAN RELATIONSHIPS

Consider a known P. triplet, x = 8, y = 15, and z = 17 as shown in Fig. 2

Figure 2. From Pythagorean equation, 82 + 152 = 172.

The contents of the overall "z" triangle are z2, (289). The dimension "y"below the figure is the dimension of the "y" triangle which is 15. The

contents of this triangle are y2, (225). Since x2 = z2- y2, the contents of

the outlined trapezoid with width, (z-y), must equal x2, (64).

Note the shaded rhomboid atop the trapezoid with height, (z-x), equalto 9, and width, (z-y), equal to 2. This rhomboid encloses a group of"twos" and has a total value of 2(9*2) = 36. For the trapezoid to form

an x2 triangle, the rhomboid contents must spill down as shown andform the small, shaded "r" triangle which abuts the lower portion of the

trapezoid. The result is an x2 triangle.

The small "r" triangle plus the lower portion of the trapezoid now

equals x2, (64). Note that "r", equals 6, and r2 equals 36 which is the

sum of the rhomboid contents, 2(9*2). "r" will always be even since r2

is composed from the rhomboid contents which are all "twos". Thefollowing generalized equations are now obtained from the bottom ofFigure 2.

(1) x = r + (z-y)(2) y = r + (z-x)(3) z = r + (z-x) + (z-y)

(4) r2 = 2(z-x)(z-y) (since r2 = rhomboid contents)

Figure 2(a). From Pythagorean equation, 82 + 152 = 172.

For additional clarity, the process is repeated in Figure 2(a) with the

trapezoid equal to y2. The trapezoid width is now (z-x). The rhomboid dimensions are still 9 by 2.

Again, the rhomboid contents spill down and form the small "r" triangle.It is the same size as before since the rhomboid dimensions are still(z-x) and (z-y). The numerical content of the "r" triangle is again2(z-x)(z-y), 36. The "r" triangle abuts the lower portion of the trapezoid

to form y2. The generalized equations observed in Figure 2 are alsoobserved from Fig. (2a).

The operations shown in Figures 2 and 2(a) have shown how the rhomboid contents in a known Pythagorean triplet "spill" down to form

triangles whose contents equal r2. P. triplets in x, y, and z can be created by a reverse process wherein an even number, r, is selected, squared to obtain (4), and factored to generate quantities representingthe sides of rhomboids. Figure 3 shows this process in generalizedform. The "ones" and "twos" have been omitted from Figure 3 for

simplicity. The factors of r2 are designated 2(z-x) and (z-y) to facilitate the explanation process using generalized terminology.

Figure 3. Method of generating P. triplets from even root numbers.

The shaded triangle, whose numeric content is r2, now spills upwardand forms a rhomboid of "twos" whose sides are (z-x) and (z-y) andwhose numeric content equals that of the "r" triangle. Now a trapezoid

equal to x2 in numeric content has been created. A slanted trapezoid

equal to y2 with bottom dimension, (z-x), has also been generated.

Note the interchangeability of the contents of the rhomboids and the r2

triangle Note, also, that (1), (2), and (3) can be observed at the bottom of Figure 3. Figure 3 is the Framework for a Pythagorean triplet.

P. triplets in x, y, and z can be calculated by using equations (1), (2),and (3) with actual numbers in place of the generalized values, r, (z-x), and (z-y). Henceforth "r", will be called the "root number" since it is the starting point for creating P. triplets. Every even value of "r" willgenerate at least one P. triplet. If a given root number contains manyprime factors, it will generate a multiplicity of P. triplets. The rootnumber, 210, 2(3*5*7), generates 8 P. triplets.

The following process generates P. triplets in x, y, and z. Steps 1 thru4 determine the rhomboid side dimensions, (z-y) and (z-x) as shown inFigure 3.

1. Select any even root number and express it in primefactors as shown.

Root number = r = 2(p1p2p3p4 . . . )

2. Square "r" to get the "root square" which is the contents of the "r" triangle and the rhomboid

Root square = r2 = 4(p1p2p3p4 . . . )2

3. Obtain the the value of the rhomboid area (which is (z-y)(z-x) in Figures 2 and 3).

Since the rhomboids are composed of "twos", the root square must be halved to obtain the rhomboid area.

Rhomboid area = (Root square)/2 = 2(p1p2p3p4 . . . )2

4. Obtain the rhomboid sides by separating the rhomboid area into 2 factors. These will be (z-x) and (z-y) in generalterminology. One side will be even since it contains 2 as afactor. The primes may be grouped in any combination.Only one is shown below.

Even factor = 2(p1p2 . . )2

Odd factor = (p3p4 . . )2 .

5. Generate x, y, and z from (1), (2), and (3) by inserting actual numbers.

(1) x = 2(p1p2p3p4 . . . ) + (p3p4 . . )2

(2) y = 2(p1p2p3p4 . . . ) + 2(p1p2 . . )2

(3) z = 2(p1p2p3p4 . . . ) + (p3p4 . . )2 + 2(p1p2 . . )2

In summary, equations (1), (2), and (3) are the result of r2 = 2(z-x)(z-y). These equations were generated from the "Model forsquared numbers" and should be used with this model throuighout theproof. Later this will be an important consideration in the proof of thetheorem.

EXAMPLES OF PYTHAGOREAN TRIPLET GENERATION

The most primitive root number, 2(1*1), which is 2, yields only one solution:

root number r squaredr squared/2Even factorOdd factor

2(1*1) 4(1*1)2 2(1*1)2 2(1)2 (1)2

x = 2(1*1) + (1)2 = 3

y = 2(1*1) + 2(1)2 = 4

z = 2(1*1) + 2(1)2 + (1)2 = 5

The root number, 2(2*1), which is 4, yields only one valid solution:

root number r squaredr squared/2Even factorOdd factor

2(2*1) 4(2*1)2 2(2*1)2 2(2)2 (1)2

x = 2(2*1) + (1)2 = 5

y = 2(2*1) + 2(2)2 = 12

z = 2(2*1) + 2(2)2 + (1)2 = 13

The prime, 2, in the even factor cannot be transposed to the odd factorbecause x, y, and z would then have a common factor, 2. The resultwould be a 6, 8, 10 Pythagorean triplet which is a trivial doubling of the3, 4, 5 solution.

The root number, 2(3*1), which is 6, yields two valid solutions since the prime, 3, may be transposed from the even to the odd factor.

Root number r squaredr squared/2Even factorOdd factor

2(3*1) 4(3*1)2 2(3*1)2 2(3)2 (1)2

x = 2(3*1) + (1)2 = 7

y = 2(3*1) + 2(3)2 = 24

z = 2(3*1) + 2(3)2 + (1)2 = 25

Transposing the factor 3, the following valid solution results:

Root number r squaredr squared/2Even factorOdd factor

2(3*1) 4(3*1)2 2(3*1)2 2(1)2 (3)2

x = 2(3*1) + (3)2 = 15

y = 2(3*1) + 2(1)2 = 8

z = 2(3*1) + 2(1)2 + (3)2 = 17

The above examples are the smallest P. triplets. As the number ofprimes in the root number goes up, the number of factors increasesrapidly. Consider r = 2(1*3*5), which equals 30. This root numberyields four P. triplets:

Even factorsOdd factors x y z

2(1)2 (3*5)2 255 32 257

2(3)2 (5)2 55 48 73

2(5)2 (3)2 39 80 89

2(3*5)2 (1)2 31 480 481

The root number, 2(3*5*7), which equals 210, yields eight P. triplets:

Even factorsOdd factors x y z

2(1)2 (3*5*7)2 11235 212 11237

2(3)2 (5*7)2 1435 228 1453

2(5)2 (3*7)2 651 260 701

2(7)2 (3*5)2 435 308 533

2(3*5)2 (7)2 259 660 709

2(3*7)2 (5)2 235 1092 1117

2(5*7)2 (3)2 219 2660 2669

2(3*5*7)2 (1)2 211 22260 22261

Note that the values of x, y, and z vary widely because the rhomboidchanges shape with each different pair of factors. A long, slimrhomboid will result in large dimensions for z and either x or y.. As therhomboid dimensions become more equal, the x and y values becomemore equal, and the z value becomes smaller. The rhomboid area isunchanged.

As the number of prime factors in "r" increases, the number of P. triplets doubles with each additional prime as follows:

"r" in prime factorsNumerical value of

"r"No. of P. triplets

generated

2(1*3*5*7*11) 2310 16

2(1*3*5*7*11*13) 30030 32

2(1*3*5*7*11*13*17) 510510 64

2(1*3*5*7*11*13*17*19) 9699690 128

This method generates all valid Pythagorean triplets. Since any givenP. triplet can be shown to have an even root number, "r", as shown inFigures 2 and 3, the P. triplet will eventually be generated assequential even root numbers are selected and processed. The onlylimitations will be the capacity of the computational devices used.

This method is used in Dave ParkersÕ interactive Pythagorean tripletgenerator which appears in his website at: http://members.aol.com/parkerdr/math/. The P. triplet generator treats higher powers of individual prime numbers as prime numbers thereby avoiding trivial solutions.

MODEL FOR HIGHER POWERS

Consider the model for squares in Figure 1. A model for z3 is easily

formed by stacking z2 triangles horizontally as shown in Figure 4. Thus

z3 is a wedge-shaped figure whose width, height, and length are all

equal to z. For higher powers of z, the length is z(n-2) while the widthand height remain equal to z. For simplicity, the twos and ones are notshown in the model . It is understood that the bottom of the modelconsists of "ones" and the rest consists of "twos".

Figure 4. Model for higher powers of "z". (cubed value shown)

Figure 5. Relationships between x3, y3, and z3 in the advanced model

Figure 5 shows how y3 may represented as a "wedge" in the advanced

model. If x3 + y3 = z3 is postulated to be true in integers, then x3 must be equal to the L-shaped form in Figure 5. Figure 6 shows this form separately.

Figure 6. x3 (postulated) form separated from the model.

Next, consider a separate z3 wedge with the wedge representing x3

removed, thereby leaving a void known to equal x3. Now suppose that

the form in Fig. 6, postulated to be x3, is placed snugly on a level base

within this x3void as shown in Figure 7

.

Figure 7. x3 L-shaped form placed within x3 void.

Note that parts of the L-shaped body (postulated to be x3 ) protrude

above, to the side, and to the front of the void representing x3 ).Suppose that these protrusions are sliced away and the "detritus" isplaced aside temporarily. The result is shown in figure 8.

Figure 8. x3 void partially filled in.

Figure 8 shows that the x3 void has been partially filled by x2 trianglesand a trapezoidal form of thickness, (z-y). The remaining space is awedge-shaped void with dimensions "r". This void is thus composed of

identical r2 triangles. Note that the foregoing process has retained the "ones" on the bottom of the advanced model, thereby maintaining the integrity of the advanced model.Now, in keeping with the postulate that

x3+ y3 = z3 in integers, the remaining "detritus" must exactly fit into the

void of r2 triangles to complete filling of the x3 void. The exact fit, of course, includes all of the "twos" and "ones" in their proper places.Note the dimensions, z, x, r, z-x, and z-y which have been identified as

a result of the sequential filling of the x3 void.

Since the r2 triangles are exactly alike, we may confine the discusssionto a single "z" triangle involving r, z, (z-x), and (z-y). When this is donethe following equations emerge from Fig.8..

x = r + (z-y)z = r + (z-x) + (z-y)

If the foregoing exercise is repeated with a postulated y3 L-shaped

form filling a known y3 void, Figure 9 is obtained

Figure 9. y3 void partially filled in.

Again, in keeping with the postulate that x3+ y3 = z3, the remaining

detritus must exactly fit into the void of r2 triangles to complete filling of

the y3 void. The following equations emerge:y = r + (z-x)z = r + (z-x) + (z-y)

The value of "r" is the same as that found in Figure 8 since (z-x) and(z-y) have simply changed positions in the model. Again, we may confine the discussion to a single "z" triangle since the r, z, (z-x), and(z-y) dimensions are exactly the same in all of the triangles.

The two-stage filling of the x3 and y3 voids has yielded the following equations in integers which relate to a Pythagorean in x, y, and z.

(1) x = r + (z-y)(2) y = r + (z-x)(3) z = r + (z-x) + (z-y)

Figure 10. Side view of figure 9 showing rearrangement of sections..

View (a) of Figure 10 shows the side view of Figure 9 after the filling of

the voids. View (b) shows how the shaded sections of view (a) can berearranged [see Figure 1 (a)]. View (b) now has the form of Figure 3which is the "framework" for the construction of a P. triplet.

Note that equations (1), (2), and (3) can be clearly seen in view (b).The "y" and "x" triangles overlap at the bottom of the figure. Note thatthe vertical dimension of "z" .may be traced up from the bottom of thefigure, thru "r", z-y, and (z-x), thereby yielding z = r + (z-x) + (z-y).Note, also, the rhomboid of "twos" at the top of the figure which hasthe numerical content of 2(z-x)(z-y). The figure of view (b) will be revisited later in the proof.

Assuming that the "detritus" will fit exactly into the r3 void commits this proof to the "reductio ad absurdum" approach wherein the consequences of doing this will be examined. The result will be an enormous paradox which will prove the theorem.

The alternative to the above approach would be cutting and pasting

the "detritus" to show that it cannot fit into the r3 void . This could be called the "direct" approach to a proof. This might conceivably work forthe cases where "n" is equal to 3 or 4. However, beyond this it will be ahopeless slog into infinity.

It should be noted that the "Reductio ad Absurbum" approach shows

what must be as the consequence of assuming that xn + yn = zn exists in integers. Many reviewers have become diverted by examining "what if" situations before the proof has run its course. At this point, the following observations apply:

The integral nature of the model has been preserved.

Some reviewers have maintained that the proof has departed from anintegral nature, but this is not true. The quantities x, y, z, r, and (z-x)and (z-y) are all integers . This has been meticulously preservedduring the operations shown in Figures 4 thru 9. In all operationsinvolving sections of the model for higher powers, the sections havebeen placed with the bottom side, consisting of "ones", on a level planewith the other sections. Figures 4 thru 9 show this clearly. The exactfilling of the "r" voids has been done in like manner as the continuation

of the postulate that xn + yn = zn exists in integers. This is completelyjustified by the proof process wherein the consequences of thispostulate will be examined..

The dimension, r, is an even number by the following logic:

Consider the postulate, xn + yn = zn, where x, y, and z are integers with no common factor, 2.

Consider equation (1) , which is x = r + (z-y). This yields r = (x+y) - z.

1. If z is odd in xn + yn = zn , then xn + yn (and x + y) must also be odd.

Therefore, r in r = (x+y) - z is: r = (odd ) - odd =

even

2. If z is even in xn + yn = zn, then xn + yn (and x + y) must also be even.

Therefore, r in r = (x+y) - z is: r = (even) - even = even

Therefore, r and each "r2" triangle will always be even for

any postulated xn + yn = zn in integers.

Each r2 triangle is equal to a rhomboid of "2s". The content of the rhomboid will be a squared number.

This is true because when the "detritus" was postulated to completely fill the "r" voids in Figures 8 and 9 , the "detritus" had to be even and

equal to all of the even r2 triangles in the void. This is a most important

consequence of the continuation of the postulate that xn + yn = zn .

Since "r" is an even integer, and the contents of each r2 triangle is a

square number, r2 can be factored and used in the construction of a P. triplet as shown in Figure 3.

Equations (1), (2), and (3) are the result of generating a P. triplet

from r2 = 2(z-x)(z-y)

In this proof, all operations involving (1), (2), and (3) will be conductedwithin the framework of the model for squared numbers. This modeloccurs in figures 9 and 10 as a result of the sequential filling of thevoids (in compliance with the postulate) and provides the followingrequisites for the construction of a P. triplet.

1. All of the quantities involved, x, y, z, and r are integers.

2. The quantity "r" is an even integer.

3. When (1), (2), and (3) are used with r2 = 2(z-x)(z-y) to

construct a P. triplet in the model for squared numbers, the result is an exact duplicate of the "z" triangles in Figures 8 and 9. The integers, x, y, and z form a P. triplet.

Equations (1), (2), and (3) are inexorably related to the model for squared numbers where they originated. P. triplets are generated in

Figure 10 from r2 = 2(z-x)(z-y). When equations (1), (2), and (3) are used outside of this model, that is, in the purely arithmetical sense, results occur which can mislead reviewers

As a consequence of the assumption that xn + yn = zn in integers, the following has resulted:

1. The filling of the r3 voids has created individual "r" triangles which are all alike.

2. These "r" triangles are all even if xn + yn = zn in integers (as shown earlier).

3. Each of the "z" triangles in Figures 8 and 9 is a model for squared numbers.

4. Equations (1), (2), and (3) in integers appear in each of the "z" triangles.

5. Using Figure 10 (which is identical to Figure 3) a P.

triplet can be generated from (1), (2), and (3), and (4) [r2 = 2(z-x)(z-y)] which is identical to the "z" triangles in figures 8 and 9.

Now in keeping with the geometry of Figure 3, a P. triplet in x, y, and zmay be constructed in each "z". triangle. The numbers, x, y, and zwhich were postulated to form a cubed relationship in the advanced

model must now form a x2 + y2 = z2 relationship in the "z" triangles of

the same model! This is clearly absurd.. If x2 + y2 = z2, then z3 would

greatly exceed x3 + y3 in the postulate. The cubed relationship is impossible in integers !!

The value of "n" in the postulation, xn + yn = zn is completelyimmaterial !! This is because the proof process has introduced ageneral characteristic of all of the higher powers of n, namely, valuesof (z-x), (z-y), and r which, in the model, exist for any value of n !! Theadvanced models simply become longer as the value of "n" increases.The operations shown in Figures 4 thru 9 result in sections with similarshapes and characteristics for n > 3 as were observed for n = 3 !!!

Thus, the proof process will always yield the absurdity that x2 + y2 = z2

no matter what value of n is postulated in xn + yn = zn .

Therefore xn + yn = zn is impossible in integers for n > 2 !!!

In a nutshell, this proof works because the proof process requires that

x, y, and z in the relationship, xn + yn = zn, must fit into a Pythagorean"framework". This framework is a "z" triangle within the model foradvanced powers (n>2) of x, y, and z.