Fermat’s Last Theorem Proof by Tom Ballard


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 is the writers belief that this is Fermats original proof. This version addresses some of the comments from reviewers and resolves their objections.

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 the existence of a Pythagorean equation in the model for squared numbers.

A model for higher powers of “n” will then be introduced. This model will be an extension of the model for squared numbers. Simple manipulations 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 the afore-mentioned equations in the packaging process is the essence of the proof.


Validating Fermats assertion that he had a proof is of utmost importance. It will never be known for sure that he had a proof in his own day and age until some credible effort can be put together to substantiate 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 Fermats general approach; however, we have a few tantalizing clues from the notes 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 have been Fermat’s methodology in proceeding to a statement of his last theorem. 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 a Pythagorean 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.


This will explain the methodology for creating P. triplets from even “root numbers”. To the knowledge of the writer, this method is entirely new. This has been confirmed by many math people. A simple model for squared numbers will be introduced. By manipulations of this model, 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 the factorability of the root number, a multiplicity of P. triplets can be generated 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.


Figure 1. Model for squared integer numbers

This model consists of a horizontal base of “ones” and an upper structure of “twos”, thereby forming a triangular shape. The vertical and diagonal dimensions of the triangle equal the horizontal dimension. The cumulative value of the contents of the triangle equals the square of the horizontal dimension. Figure 1 shows the numbers from 1 to 8 and their squares. Observe that the cumulative value is actually the sum of odd numbers.Throughout this treatise, the model will 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.


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), equal to 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 and form 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”. The following generalized equations are now obtained from the bottom of Figure 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 again 2(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 also observed 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 representing the sides of rhomboids. Figure 3 shows this process in generalized form. 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 upward and forms a rhomboid of “twos” whose sides are (z-x) and (z-y) and whose 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” will generate at least one P. triplet. If a given root number contains many prime factors, it will generate a multiplicity of P. triplets. The root number, 210, 2(3*5*7), generates 8 P. triplets.

The following process generates P. triplets in x, y, and z. Steps 1 thru 4 determine the rhomboid side dimensions, (z-y) and (z-x) as shown in Figure 3.

1. Select any even root number and express it in prime factors 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 general terminology. One side will be even since it contains 2 as a factor. 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 for squared numbers” and should be used with this model throuighout the proof. Later this will be an important consideration in the proof of the theorem.


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

root number r squared r squared/2 Even factor Odd 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 squared r squared/2 Even factor Odd 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 factor because x, y, and z would then have a common factor, 2. The result would be a 6, 8, 10 Pythagorean triplet which is a trivial doubling of the 3, 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 squared r squared/2 Even factor Odd 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 squared r squared/2 Even factor Odd 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 of primes in the root number goes up, the number of factors increases rapidly. Consider r = 2(1*3*5), which equals 30. This root number yields four P. triplets:

Even factors Odd 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 factors Odd 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 rhomboid changes shape with each different pair of factors. A long, slim rhomboid will result in large dimensions for z and either x or y.. As the rhomboid dimensions become more equal, the x and y values become more equal, and the z value becomes smaller. The rhomboid area is unchanged.

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

“r” in prime factors Numerical 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 given P. triplet can be shown to have an even root number, “r”, as shown in Figures 2 and 3, the P. triplet will eventually be generated as sequential even root numbers are selected and processed. The only limitations will be the capacity of the computational devices used.

This method is used in Dave ParkersÕ interactive Pythagorean triplet generator 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.


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 width and height remain equal to z. For simplicity, the twos and ones are not shown in the model . It is understood that the bottom of the model consists 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” is placed 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 triangles and a trapezoidal form of thickness, (z-y). The remaining space is a wedge-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 discusssion to a single “z” triangle involving r, z, (z-x), and (z-y). When this is done the 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 be rearranged [see Figure 1 (a)]. View (b) now has the form of Figure 3 which 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 that the vertical dimension of “z” .may be traced up from the bottom of the figure, 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 has the 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 for the cases where “n” is equal to 3 or 4. However, beyond this it will be a hopeless 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 an integral 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 preserved during the operations shown in Figures 4 thru 9. In all operations involving sections of the model for higher powers, the sections have been placed with the bottom side, consisting of “ones”, on a level plane with the other sections. Figures 4 thru 9 show this clearly. The exact filling 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 completely justified by the proof process wherein the consequences of this postulate 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 = evenTherefore, 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 conducted within the framework of the model for squared numbers. This model occurs in figures 9 and 10 as a result of the sequential filling of the voids (in compliance with the postulate) and provides the following requisites 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 z may be constructed in each “z”. triangle. The numbers, x, y, and z which 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 completely immaterial !! This is because the proof process has introduced a general characteristic of all of the higher powers of n, namely, values of (z-x), (z-y), and r which, in the model, exist for any value of n !! The advanced models simply become longer as the value of “n” increases. The operations shown in Figures 4 thru 9 result in sections with similar shapes 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 for advanced powers (n>2) of x, y, and z.


2 thoughts on “Fermat’s Last Theorem Proof by Tom Ballard

  1. Pingback: Longest Math problem Fermat’s Last Theorem

  2. A very good presentation any junior high school student (with good mathematics) would be able to comprehend! Please publish this thing to The Annals of Mathematics!

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