Мathematics

THE SIMPLEST PROOF OF THE FERMAT’S LAST THEOREM

Vassil Manev

INTRODUCTION

The achievement of Andrew Wiles in proving the Fermat’s Last Theorem published in the work “Modular elliptic curves and Fermat’s Last Theorem[1] provokes the question:

“Is that proof Pierre de Fermat had in mind, when in 1637  on the field of the book of Diophantus [2] he has written in Latin [3]:

“t is not possible to separate a cube into two cubes, or a square of two squares, or any degree at all higher than the second of two identical degrees. I have discovered a truly wonderful proof, but the margin is too narrow to contain.”

Obviously, the extremely complicated indirect method of building the proof of the theorem used by Andrew Wiles, does not express the level of mathematical knowledge in the first half of the 17th century. Therefore, the question “What proof had in mind Pierre de Fermat?” will continue to excite the mathematical community to the moment of appearance of a proof of Fermat’s Last Theorem, that is to answer the question.

Is it possible that happen after over 358 years no other proof has been found? Alternatively, likely Pierre de Fermat has not found proof of his theorem, as it is assumed in the work [4], page 5:

“He claimed that he had found a remarkable proof. There is some doubt as to various reasons. First, this note was published without his permission, in fact by his son after his death. Second, in his later correspondence Fermat discusses the case for n = 3 and 4, without mentioning this alleged evidence. It seems as if this was an improvised comments that Fermat failed to erase“.

This comment is logically untenable because:

  • Pierre de Fermat did not know about the interest in him and his notes that started after his death.
  • He has been engaged in mathematics in his free time.
  • After his death there was no way his son to seek permission from him to publish his notes.
  • That in his correspondence, after having discovered his theorem, are discussed the cases for n = 3 and n = 4 does not mean that he has not found a wonderful proof of the theorem. In his correspondence everyone can discuss everything, depending on the situation.

 

Taking into account that Pierre de Fermat is a genius mathematician is entirely possible that he was not mistaken and found “really wonderful proof” of his theorem.

If the theorem proves with the knowledge of his time it will appear that everything written so far in connection with the proof of Fermat’s Last Theorem is incorrect.

 

As Diophantine equation an + bn = cn is unsolvable in general, it follows:

If Pierre de Fermat has discovered “really wonderful proof” of the theorem, then he noticed in equality аn + bn = cn not Diophantine equation everyone notice, but something else.

 

In the following content under integers we will mean 1, 2, 3, 4, 5, . . .

 

Pythagoras’s theorem may be expressed in the following form:

For n = 2, there are integers (a, b, c), for which is satisfied the equation

аn + bn = cn.

or in the form:

For n = 2, there are integers (a, b, c), for which is satisfied the equation

cn – (аn + bn) = 0                                                                                                                                              

Fermat’s Last Theorem may be expressed in the following form:

For n > 2, there are no integers (a, b, c), for which is satisfied the equation

cn – (аn + bn) = 0.

It follows from Fermat’s Last Theorem that for n > 2 is satisfied

cn – (аn + bn) ≠ 0.

We may combine both equalities cn – (аn + bn) = 0 and cn – (аn + bn) ≠ 0 as follows:

cn – (аn + bn) = m                                                                                                                                              (1)

where for n = 2 is satisfied m = 0, and for n > 2 is satisfied m ≠ 0.

It is noteworthy that in (1) where we have combined both equations (Pythagoras’ and Fermat’s) in one equation where the change of n from n = 2 to n > 2 changes also m from m = 0 to m ≠ 0.

Then question raises:

Doesn’t it mean that between n and m = cn – (аn + bn) there is a functional dependence

f(n) = cn – (аn + bn)

wherein independently of the choice of integers (a, b, c) for each value of n corresponds a single value of f(n), and vice versa — to each value of f(n) corresponds a single value of n?

Example 1:

Let’s assume that (a, b, c) = (3, 4, 5).

It is satisfied:

32 + 42 = 52 and 52 (32 + 42) = 0

33 + 43 < 53 and 53 (33 + 43) = 34

34 + 44 < 54 and 54 (34 + 44) = 288

. . . . . . . . . . . . . . . . . . . . . . . . . . .

From this example it is seen that for (a, b, c) = (3, 4, 5) for the integers it results:

f(2) = 0; f(3) = 34; f(4) = 288; . . . ; f(n) = m; . . .

and is evident that the correlation between n and f(n) is mutually uniquely.

Example 2:

Let’s assume that (a, b, c) = (3, 4, 5).

It is satisfied:

92 + 402 = 412 and 412 (92 + 402) = 0

93 + 403 < 413 and 413 (93 + 403) = 4192

94 + 404 < 414 and 414 (94 + 404) = 259200

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

From this example it is seen that for (a, b, c) = (9, 40, 41) is satisfied:

f(2) = 0; f(3) = 4192; f(4) = 259200; . . . ; f(n) = m; . . .

and is evident that the correlation between n and f(n) is mutually uniquely.

 

Of both examples (and all other examples that we can review) it may be seen:

  • Between n and f(n) exists uniquely each line, regardless of the choice of integers (a, b, c).
  • When changing the integers (a, b, c), then it changes the values ​​of f(n), corresponding to n, but mutually unambiguous correspondence between n and f(n) is retained.
  • When for n = 2 is satisfied cn – (аn + bn) = 0 and n increases, then the difference cn – (a2 + b2) increases and there is no way to get equality аn + bn = cn for n > 2, regardless of the choice of integers (a, b, c).

 

Given the result the so far it follows

Fermat’s Last Theorem may be expressed as a function in the following form:

ƒ(n) = cn – (аn + bn)                                                                                                                                           (2)

where n is a positive integer, for which is satisfied n ≥ 2.

 

Perhaps Pierre de Fermat had in mind the function (2), and not Diophantine equation аn + bn = cn, because (as it will be shown below) if used (2) it may be obtained simply and “really wonderful proof” of Fermat’s Last Theorem.

If so, that means Pierre de Fermat has reached the term “function” in 1637, before Leibniz (1670).

 

This work offers proof of Fermat’s Last Theorem, using the function (2) but expressed in such a way as to use the knowledge in mathematics from the time of Pierre de Fermat, in order to show two things:

  • Fermat’s Last Theorem can be proven with knowledge in mathematics from the time of Pierre de Fermat.
  • Pierre de Fermat has not lied, had not deceived himself and has not fantasized, as he wrote that he found “really wonderful proof” of his theorem.

 

There are integers (a, b, c), for which is satisfied the relationship

a2+ b2 = c2                                                                                                                                                                                                                                                 (3)

Example:

32 + 42 = 52

If (a, b, c) are three random integers, for them is always carried one of the three relationships:

а2 + b2 < с2                                                                                                                                                         (4)

а2 + b2 = c2                                                                                                                                                         (5)

а2 + b2 > c2                                                                                                                                                         (6)

Relationships (4), (5) and (6) we may write as follows:

c2 – (а2 + b2) > 0                                                                                                                                           

c2 – (а2 + b2) = 0                                                                                                                                                       

c2 – (а2 + b2) < 0                                                                                                                                             

and combine in one relationship

c2 – (а2 + b2) = m                                                                                                                                              (7)

where for m may be satisfied: m > 0; m = 0; m < 0.

Dependence (7) is a particular case of the relationship (1) because (7) is obtained by (1), for n = 2.

 

Plan of the proof:

Part 1: If for the integers (a, b, c) is satisfied а2 + b2 < c2,

Part 2: For the integers (a, b, c) is satisfied а2 + b2 = c2,

Part 3: For the integers (a, b, c) is satisfied а2 + b2 > c2,

Part 4: Proof of Fermat’s Last Theorem, provided that for the integers (a, b, c) the function      f(n) = cn – (an + bn) is satisfied.

Part 5: Conclusion:

 

If we use directly the function (2), then the proof of Fermat’s Last Theorem is expressed on several pages (this will be understood from the following content) but now our goal is different:

To show that in all cases the equality

ƒ(n) = cn – (аn + bn)                                                                                                                                           (8)

is a function of n, wherein the correlation between n and f(n) is mutually unambiguous, regardless of the choice of integers (a, b, c).

 

PART 1

FOR INTEGERS (a, b, c) IS SATISFIED RELATIONSHIP а2 + b2 < с2.

 

Theorem 1.1

If for the integers (a, b, c) is satisfied а2 + b2 < с2, then for n > 2 is satisfied

аn + bn < cn  and might not be satisfied аn + b= cn.

Proof:

We multiply both sides of the inequality а2 + b2 < c2 with cn-2 and we obtain

а2.cn-2 + b2.cn-2 < с2.cn-2                                                                                                                                    (9)

We multiply both sides of the inequality аn-2 < cn-2 with a2 and obtain

a2.an-2 < a2.cn-2

We multiply both sides of the inequality bn-2 < cn-2 with b2 and obtain

b2.bn-2 < b2.cn-2

Since a2.an-2 < a2.cn-2 and b2.bn-2 < b2.cn-2, when we replace in (9), а2.cn-2 с a2.an-2 and b2.cn-2 с b2.bn-2 we obtain the stronger inequality

а2.an-2 + b2.bn-2 < с2.cn-2                                                                                                                                 (10)

Of (10) it derives:

  • For n > 2 is satisfied an + bn < cn.
  • Where n grows, inequality an + bn < cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > 2 is impossible to be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Example:

Let’s assume that (a, b, c) = (3, 5, 7).

It is satisfied:

32 + 52 < 72 and 72 – (32 + 52) = 15

33 + 53 > 73 and 73 – (33 + 53) = 191

34 + 54 > 74 and 74 – (34 + 54) = 1695

. . . . . . . . . . . . . . . . . . . . . . . . . . .

From this example it is seen that for (a, b, c) = (3, 5, 7) for the integers it results:

f(2) = 15; f(3) = 191; f(4) = 1695; . . . ; f(n) = m; . . .

and is apparent that in the case for а2 + b2 < с2 correspondence between n and f(n) is mutually unambiguous.

 

PART 2

FOR INTEGERS (a, b, c) IS SATISFIED RELATIONSHIP а2 + b2 = с2.

 

Theorem 2.1

If for the integers (a, b, c) is satisfied а2 + b2 = c2,

n > 2 is satisfied аn + bn < сn and might not be satisfied аn + bn = сn.

Proof:

Of the equality а2 + b2 = c2 it follows а2 < c2; b2 < c2 and а < c; b < c.

Then in that case for n > 2 it is satisfied:

an-2 <  cn-2                                                                                                                                                        (11)

bn-2 <  cn-2                                                                                                                                                        (12)

Multiplying both sides of (11) with а2, and both sides of (12) with b2, it is obtained

а2.an-2 < а2.cn-2                                                                                                                                                (13)

b2.bn-2 <  b2.cn-2                                                                                                                                              (14)

When we sum left and right sides of both inequalities (13) and (14), it results

а2.an-2 + b2.bn-2 < а2.cn-2 + b2.cn-2                                                                                                                 (15)

When we multiply both sides of the equality а2 + b2 = c2 with cn-2 we receive

а2.cn-2 + b2.cn-2 = c2.cn-2                                                                                                                                  (16)

When replacing right side of (15) with its equal of (16), it results

а2.an-2 + b2.bn-2 < c2.cn-2                                                                                                                                 (17)

Of (17) it derives:

  • For n > 2 is satisfied an + bn < cn.
  • Where n grows, inequality an + bn < cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > 2 is impossible to be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Example:

Let’s assume that (a, b, c) = (3, 4, 5).

It is satisfied:

32 + 42 = 52 and 52 – (32 + 42) = 0

33 + 43 < 53 and 53 – (33 + 43) = 34

34 + 44 < 54 and 54 – (34 + 44) = 288

. . . . . . . . . . . . . . . . . . . . . . . . . . .

From this example it is seen that for (a, b, c) = (3, 4, 5) for the integers it results:

f(2) = 0; f(3) = 34; f(4) = 288; . . . ; f(n) = m; . . .

and is apparent that in the case for а2 + b2 = с2 correspondence between n and f(n) is mutually unambiguous.

 

Lemma 2.1

If for the integers s and q is satisfied s > q, when n increases, then difference sn – qn is growing.

Proof: 

As s > q, we mark:

s1 – q1 = p1

s2 – q2 = p2

s3 – q3 = p3

s4 – q4 = p4

. . . . . . . . . .

sn – qn = pn

. . . . . . . . . .

For entered indications it may be written

s1 – q1 = s – q = p1

s2 – q2 = (s – q).(s + q) = p2

s3 – q3 = (s – q).(s2+ s.q + q2) = p3

s4 – q4 = (s – q).(s3+ s2.q + s.q2+ q3) = p4

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

sn – qn = (s – q).(sn-1 + sn-2.q + sn-3.q2 + . . . + s.qn-2+ qn-1) = pn                                                                 (18)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Given that s and q are integers and s – q = p1 > 0, it follows:

0 < p1 < p2 < p3 < p4 < . . . < pn . . .

Theorem 2.2

If for integers (a, b, c) for n = 2 is satisfied cn – (аn + bn) = 0,

where n grows from 2 and tends to infinity, then difference

cn – (аn + bn) grows from 0 and tends to infinity.

Proof:

If in (18) we lay down s = c and q = a, for degree n – 2 we obtain:

cn-2 – аn-2 = (c – а).(cn-3 + cn-4.а + cn-52 + . . . + c.аn-4 + аn-3) = pan                                                          (19)                                                   

If in (18) we lay down s = c and q = b for degree n – 2 we obtain:

cn-2 – bn-2 = (c – b).(cn-3 + cn-4.b + cn-5.b2 + . . . + c.bn-4 + bn-3) = pbn                                                          (20)

Multiplying both sides of(19) with а2, and both sides of (20) with b2, we receive

a2.cn-2 – a2n-2 = a2.pan                                                                                                                                 (21)

b2.cn-2 – b2.bn-2 = b2.pbn                                                                                                                                 (22)

When we sum both sides of both inequalities (21) and (22), is obtained

a2.cn-2 + b2.cn-2 – (a2n-2 + b2.bn-2) = a2.pan + b2.pbn

cn-2.(a2 + b2) – (аn + bn) = a2.pan + b2.pbn

Given that it is satisfied, as provided that а2 + b2 = c2, it follows

cn-2.c2  – (аn + bn) = a2.pan + b2.pbn

cn  – (аn + bn) = a2.pan + b2.pbn

Since according to the Lemma 2.1, pan and pbn grow, where n grows, it follows that where n grows of primary value n = 2 and tends to infinity, then difference cn – (аn + bn) increases from primary value 0 and tends to infinity.

Theorem 2.3

If for n = 2 is satisfied an + bn = cn, then

  • For n > 2 is satisfied an + bn < cn.
  • For n < 2 is satisfied an + bn > cn.
  • For n > 2 and n < 2 is impossible to be satisfied the equality аn + bn = cn.

Proof:

According to Theorem 2.1, if for n = 2 is satisfied аn + bn = сn, where n increases (for n > 2) is satisfied аn + bn < сn and might not be satisfied аn + bn = сn. In that case between n and cn – (аn + bn) there is mutually unambiguous correspondence that is expressed by the function f(n) = cn – (an + bn).

For n < 2 only one single value for n (n = 1) is possible, corresponding with one single value of the function f(1) = c1 – (a1 + b1).

For n = 2 only one single value of the function f(2) = c2 – (a2 + b2) is possible.

Example:

31 + 41 > 51 and 51 – (31 + 41) = -2

32 + 42 = 52 and 52 – (32 + 42) = 0

33 + 43 < 53 and 53 – (33 + 43) = 34

34 + 44 < 54 and 54 – (34 + 44) = 288

. . . . . . . . . . . . . . . . . . . . . . . . . . .

 

Insofar we explored the integers (а, b, с), where for their second degrees are satisfied relationships a2 + b2 < c2, a2 + b2 = c2 and we proved that in both cases:

  • For n > 2, equality аn + bn = сn might not be satisfied in integers (a, b, c).
  • Between n and cn – (аn + bn) there is mutually unambiguous correspondence that is expressed by the function f(n) = cn – (an + bn).

 

 

PART 3

FOR INTEGERS (a, b, c) IS SATISFIED RELATIONSHIP а2 + b2 > с2.

 

From the relationship а2 + b2 > с2, 9 cases are following:

a2 > c2; b2 > c2

a2 > c2; b2 < c2

a2 < c2; b2 > c2

a2 > c2; b2 = c2

a2 = c2; b2 > c2

a2 = c2; b2 < c2

a2 < c2; b2 = c2

a2 = c2; b2 = c2

a2 < c2; b2 < c2

of which:

  • First case is limited to the second.
  • Fifth case is limited to the fourth.
  • Seventh case is limited to the sixth.

and only 6 cases remain for review:

a2 > c2; b2 > c2

a2 > c2; b2 < c2

a2 > c2; b2 = c2

a2 = c2; b2 < c2

a2 = c2; b2 = c2

a2 < c2; b2 < c2

 

Case 1:

а2 + b2 > с2 and a2 > c2; b2 > c2.

Theorem 3.1

If a, b and c are integers for which is satisfied а2 + b2 > с2 and a2 > c2; b2 > c2, where n increases, for n > 2 is satisfied аn + bn > cn and might not be satisfied аn + bn = cn.

Proof:

Of a2 > c2; b2 > c2 it follows: an-2 > cn-2, bn-2 > cn-2, for n > 2.

When we multiply both sides of the inequality а2 + b2 > c2 with cn-2 we receive

a2.cn-2 + b2. cn-2 > c2.cn-2                                                                                                                                 (23)

When we multiply both sides of the inequality аn-2 > cn-2 with a2 and obtain

a2.an-2 > a2.cn-2

When we multiply both sides of the inequality bn-2 > cn-2 with b2 and obtain

b2.bn-2 > b2.cn-2

Since a2.an-2 > a2.cn-2 and b2.bn-2 > b2.cn-2, when we replace in (23), а2.cn-2 with a2.an-2 and b2.cn-2 with b2.bn-2 we obtain the stronger inequality

а2.an-2 + b2.bn-2 > с2.cn-2                                                                                                                                 (24)

Of (24) it derives:

  • For n > 2 is satisfied a + bn > cn.
  • Where n grows, inequality an+ bn > cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > 2 is impossible to be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Example:

42 + 52 > 32 and 32 – (42 + 52) = – 32

43 + 53 > 33 and 33 – (43 + 53) = – 164

44 + 54 > 34 and 34 – (44 + 54) = – 800

. . . . . . . . . . . . . . . . . . . . . . . . . . .

 

Case 2:

а2 + b2  >  с2; a2 > c2 ; b2 < c2

Theorem 3.2

If (а, b, с) are integers for which is satisfied а2 + b2 > с2; a2 > c2; b2 < c2, where n

grows for n > 2 is satisfied аn + bn > сn and might not be satisfied аn + bn = сn.

Proof:

Of a2 > c2; b2 < c2 it follows: an-2 > cn-2, bn-2 < cn-2, for n > 2.

When we multiply both sides of the inequality а2 + b2 > cwith cn-2 we receive

a2.cn-2 + b2.cn-2 > c2.cn-2                                                                                                                                  (25)

When we multiply both sides of the inequality аn-2 > cn-2 with a2 and obtain

a2.an-2 > a2.cn-2

When we multiply both sides of the inequality bn-2 < cn-2 with b2 and obtain

b2.bn-2 < b2.cn-2

Since a2.an-2 > a2.cn-2 (regardless the value of 2.bn-2), when we replace in (25), а2.cn-2 with a2.an-2 and b2.cn-2 with b2.bn-2 we obtain

а2.an-2 + b2.bn-2 > с2.cn-2                                                                                                                                 (26)

Of (26) it derives:

  • For n > 2 is satisfied an + bn > cn.
  • Where n grows, inequality an + bn > cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > 2 is impossible to be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Example:

72 + 32 > 52 and 52 – (72 + 32) = – 34

73 + 33 > 53 and 53 – (73 + 33) = – 245

74 + 34 < 54 and 54 – (74 + 34) = – 1857

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

 

Case 3:

а2 + b2 > с2; a2 > c2; b2 = c2

Theorem 3.3

If (а, b, с) are integers for which is satisfied а2 + b2 > с2; a2 > c2; b2 = c2, where n

grows for n > 2 is satisfied аn + bn > сn and might not be satisfied аn + bn = сn.

Proof:

Of a2 > c2; b2 = c2 it follows: an-2 > cn-2, bn-2 = cn-2, for n > 2.

When we multiply both sides of the inequality а2 + b2 > c2 with cn-2 we receive

a2.cn-2 + b2.cn-2 > c2.cn-2                                                                                                                                 (27)

When we multiply both sides of the inequality аn-2 > cn-2 with a2 and obtain

a2.an-2 > a2.cn-2

When we multiply both sides of the equality bn-2 = cn-2 with b2 and obtain

b2.bn-2 = b2.cn-2

Since a2.an-2 > a2.cn-2 and b2.bn-2 > b2.cn-2, when we replace in (27), а2.cn-2 with a2.an-2 and b2.cn-2 with b2.bn-2 we obtain

а2.an-2 + b2.bn-2 > с2.cn-2                                                                                                                                 (28)

Of (28) it derives:

  • For n > 2 is satisfied an + bn > cn.
  • Where n grows, inequality an + bn > cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > 2 is impossible to be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Example:

72 + 52 > 52 and 52 – (72 + 52) = – 49

73 + 53 > 53 and 53 – (73 + 53) = – 343

74 + 54 > 54 and 54 – (74 + 54) = – 2401

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

 

Case 4:

а2 + b2 > с2; a2 = c2 ; b2 < c2

Theorem 3.4

If (а, b, с) are integers for which is satisfied а2 + b2 > с2; a2 = c2; b2 < c2, where n

grows for n > 2 is satisfied аn + bn > сn and might not be satisfied аn + bn = сn.

Proof:

Of a2 = c2; b2 < c2 it follows: an-2 = cn-2 , bn-2 < cn-2, for n > 2.

When we multiply both sides of the inequality а2 + b2 > c2 with cn-2 it results

a2.cn-2 + b2.cn-2 > c2.cn-2                                                                                                                                 (29)

When we multiply both sides of the equality аn-2 = cn-2 with a2 we obtain

a2.an-2 = a2.cn-2

When we multiply both sides of the inequality bn-2 < cn-2 with b2 and obtain

b2.bn-2 < b2.cn-2

Since a2.an-2 = a2.cn-2 and b2.bn-2 < b2.cn-2, when we replace in (29), а2.cn-2 with a2.an-2 and b2.cn-2 with b2.bn-2 we obtain

а2.an-2 + b2.bn-2 > с2.cn-2                                                                                                                                 (30)

Of (30) it derives:

  • For n > 2 is satisfied an + bn > cn.
  • Where n grows, inequality an + bn > cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > 2 is impossible to be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Example:

52 + 72 > 52 and 52 – (52 + 72) = – 49

53 + 73 > 53 and 53 – (53 + 73) = – 343

54 + 74 > 54 and 54 – (54 + 74) = – 2401

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

 

Case 5:

а2 + b2  >  с2; a2 = c2 ; b2 = c2

Theorem 3.5

If (а, b, с) are integers for which is satisfied а2 + b2 > с2; a2 = c2; b2 = c2, where n

grows for n > 2 is satisfied аn + bn > сn and might not be satisfied аn + bn = сn.

Proof:

Of a2 = c2; b2 = c2 it follows: an-2 = bn-2 = cn-2, for n > 2.

When we multiply both sides of the inequality а2 + b2 > c2 with cn-2we receive

a2.cn-2 + b2.cn-2 > c2.cn-2                                                                                                                                 (31)

When we multiply both sides of the equality аn-2 = cn-2 with a2 we obtain

a2.an-2 = a2.cn-2

When we multiply both sides of the equality bn-2 = cn-2 with b2 and obtain

b2.bn-2 = b2.cn-2

When replacing in (31), а2.cn-2 with c2.cn-2 and b2.cn-2 with c2.cn-2 we obtain strengthened inequality

c2.cn-2 + c2.cn-2 > с2.cn-2                                                                                                                                  (32)

Of (32) it derives:

  • For n > 2 is satisfied an + bn > cn.
  • Where n grows, inequality an + bn > cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > 2 is impossible to be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Example:

52 + 52 > 52 and 52 – (52 + 52) = – 25

53 + 53 > 53 and 53 – (53 + 53) = – 125

54 + 54 > 54 and 54 – (54 + 54) = – 625

. . . . . . . . . . . . . . . . . . . . . . . . . . .

 

Of the five cases reviewed

a2 > c2; b2 > c2

a2 > c2; b2 < c2

a2 > c2; b2 = c2

a2 = c2; b2 < c2

a2 = c2; b2 = c2

given that а2 + b2 > с2, it follows:

  • For n < 2 is satisfied the inequality an + bn > cn.
  • For n > 2 might not be satisfied the equality аn + bn = cn.
  • Where n grows (for n > 2), difference between cn and аn + bn
  • Between n and cn – (аn + bn) there is mutually unambiguous correspondence that is expressed by the function f(n) = cn – (an + bn).

 

The sixth (last) case remained for consideration that is more special than others thus deserving more attention.

 

Case 6:

а2 + b2 > с2 and a2 < c2; b2 < c2

 

Theorem 3.6

If (а, b, с) are integers for which is satisfied а2 + b2 > с2; a2 < c2; b2 < c2, where n

grows, for n > 2  there is a value of n (n = k) so:

  • For n ≤ k is satisfied an + bn > cn.
  • For n > k is satisfied an + bn > cn.
  • For n = 2 might not be satisfied аn + bn = cn.
  • The value of n (n = k) after which the inequality аn + bn > cn reverses its direction and is dependent of the integers (а, b, с).

Proof:

According to Lemma 2.1

  • Where n increases, for n > 2 difference between cn and an is increasing (cn grows faster than an).
  • Where n increases, for n > 2 difference between cn and bn is increasing (cn grows faster than bn).

That means:

Where n grows (provided that а2 + b2 > с2 and a2 < c2; b2 < c2), there is a value of n (n = k + 1), where inequality аn + bn > cn reverses direction, and acquires type аn + bn < cn.

Example 1:

72 + 82 > 92 and 92 – (72 + 82) = -32

73 + 83 > 93 and 93 – (73 + 83) = -126

74 + 84 < 94 and 94 – (74 + 84) = 64

75 + 85 < 95 and 95 – (75 + 85) = 9474

76 + 86 < 96 and 96 – (76 + 86) = 151648

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

From this example it follows:

  • 93 – (73 + 83) < 92 – (72 + 82).
  • 94 – (74 + 84) < 95 – (75 + 85) < 96 – (76 + 86) < . . .
  • Value of n, after which the inequality 7n + 8n > 9n reverses its direction, is k = 3.
  • For n < 3 is satisfied 7n + 8n < 9n.
  • For n > 3, where n grows, inequality 7n + 8n < 9n strengthens as to each value of n corresponds one single value of 9n – (7n + 8n).
  • For n > 3 might not be satisfied the equality 7n + 8n = 9n.
  • Mutually unambiguous correspondence between n and 9n – (7n + 8n) is expressed by the function f(n) = 9n – (7n + 8n).

Example 2:

192 + 202 > 212 и 212 – (192 + 202) = -320

193 + 203 > 213 и 213 – (193 + 203) = -5598

194 + 204 > 214 и 214 – (194 + 204) = -95840

195 + 205 > 215 и 215 – (195 + 205) = -1591998

196 + 206 > 216 и 216 – (196 + 206) = -25279760

197 + 207 > 217 и 217 – (197 + 207) = -372783198

198 + 208 > 218 и 218 – (198 + 208) = -4760703680

199 + 209 > 219 и 219 – (199 + 209) = -40407651198

1910 + 2010 < 2110 и 2110 – (1910 + 2010) = 308814720400

1911 + 2011 < 2111 и 2111 – (1911 + 2011) = 28987241644002

1912 + 2012 < 2112 и 2112 – (1912 + 2012) = 1046512592320480

1913 + 2013 < 2113 и 2113 – (1913 + 2013) = 30499394276862402

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

From this example it follows:

  • 219 – (199 + 209) < 218 – (198 + 208 ) < 217 – (197 + 207) < 216 – (196 + 206) < 215 – (195 + 205) < < 214 – (194 + 204) < 213 – (193 + 203) < 212 – (192 + 202)
  • 2110 – (1910 + 2010) < 2111 – (1911 + 2011) < (1912 + 2012) < 2113 – (1913 + 2013) < . . .
  • Value of n, after which the inequality 19n + 20n > 21n reverses its direction, is k = 9.
  • For n < 9 is satisfied 19n + 20n < 21n.
  • For n > 9, where n grows, inequality 19n + 20n < 21n strengthens as to each value of n corresponds one single value of 21n – (19n + 20n).
  • For n > 9 might not be satisfied the equality 19n + 20n = 21n.
  • Mutually unambiguous correspondence between n and 21n – (19n + 20n) is expressed by the function f(n) = 21n – (19n + 20 ).

 

From both examples (and all other examples that we may review) it is seen that where n increases, given that а2 + b2 > с2 and a2 < c2; b2 < c2:

  • The inequality an + bn > cn (for n < k) leads to the inequality an + bn < cn (for n > k), without passing through inequality аn + bn = cn (for n = k).
  • For n > k is satisfied an + bn < cn.
  • For n ≤ k is satisfied an + bn > cn.
  • The value of n, (n = k) after which the inequality аn + bn > cn reverses its direction and is dependent of the selection of integers (а, b, с).
  • Between n and cn – (аn + bn) there is mutually unambiguous correspondence that is expressed by the function f(n) = cn – (an + bn).

 

We explored the inequality аn + bn < cn in Part 1 .

From Theorem 1.1 and obtained above it follows:

  • For n > k is satisfied an + bn < cn.
  • For n > k, where n grows, inequality an + bn < cn strengthens as to each value of n corresponds one single value of cn – (an + bn).
  • For n > k might not be satisfied the equality аn + bn = cn.
  • Mutually unambiguous correspondence between n and cn – (an + bn) is expressed by the function f(n) = cn – (an + bn).

Only the inequality an + bn > cn, for n ≤ k remained to be explored.

Of а2 + b2 > с2, it follows:

c2 – (а2 + b2) < 0                                                                                                                                              (33)    

We multiply both sides of (33) with c and we obtain

c2.c – (а2.c + b2.c) < 0                                                                                                                                     (34)

It is satisfied:

c2.c – (а2.c + b2.c) < c2 – (а2 + b2) < 0.

Of a2 < c2; b2 < c2 we obtain a < c; b < c.

Because a < c; b < c it follows a2.a < a2.c; b2.b < b2.c.

Replacing in (34) a2.c with a2.a; b2.c with b2.b, we obtain

c2.c – (а2.a + b2.b) < 0                                                                                                                                    (35)

since а2.a + b2.b < а2.c + b2.c it follows

c2.c – (а2.a + b2.b) < c2.c – (а2.c + b2.c) < c2 – (а2 + b2) < 0

c3 – (а3 + b3) < c2 – (а2 + b2)

In the same way we obtain

ck – (аk + bk) < . . . < c4 – (а4 + b4) < c3 – (а3 + b3) < c2 – (а2 + b2) > 0                                                        (36)

Where n increases in the closed interval [2, k], then the difference between cn and аn + bn is increasing and an equality might not be obtained. For n > k the inequality reverses its direction. We have studied this case and there when n increases by the value of n = k + 1 to infinity, the difference between cn and аn + bn increases and an equality might not be obtained.

As of each value of n for n [2, k] and n > k corresponds only one value of cn – (an + bn), it follows that between n and cn – (an + bn) there is mutual uniquely line, which is expressed by the function f(n) = cn – (an + bn).

From insofar obtained for n > 2 and n < k + 1, where n grows, in а2 + b2 > с2 and a2 < c2; b2 < c2, it follows:

  • The inequality an + bn > cn (for n < k) leads to the inequality an + bn < cn (for n > k), without passing through inequality аn + bn = cn (for n = k).
  • For n > k is satisfied an + bn < cn.
  • For n ≤ k is satisfied an + bn > cn.
  • The value of n, (n = k) where the inequality аn + bn > cn reverses its direction is dependent of the selection of integers (а, b, с). (there is nothing else in the expression cn – (an + bn), from which k may be dependent).
  • Between n and cn – (аn + bn) there is mutually unambiguous correspondence that is expressed by the function f(n) = cn – (an + bn).

 

We explored the integers (а, b, с), where for their second degrees are satisfied relationships        a2 + b2 < c2, a2 + b2 = c2, a2 + b2 = c2 and we proved that in all cases:

  • For n > 2, equality аn + bn = сn might not be satisfied in integers

(a, b, c).

  • Between n and cn – (аn + bn) there is mutually unambiguous correspondence that is expressed by the function f(n) = cn – (an + bn).

That means:

  • Equality аn + bn = cn is possible in integers (a, b, c) for n = 2.
  • Equality аn + bn = cn is not possible in integers (a, b, c) for n > 2.

 

By that, the most complete, most accurate, most consistent and most general proof of Fermat’s Last Theorem, with the mathematical knowledge from the first half of the 17th century, has been completed.

 

 

 

PART 4

PROOF OF FERMAT’S LAST THEOREM, PROVIDED THAT FOR INTEGERS (a, b, c) is Satisfied The Function f(n) = cn – (an + bn).

 

Existence of the function f(n) = cn – (an + bn ) and of the mutually unambiguous correspondence between n and f(n) is evident and does not need any proof.

To prove the existence of function f(n) = cn – (an + bn ) is tantamount to prove the existence of function f(x) = a.x2 +b.x + c.

Nevertheless, the proofs we have made in Part 1, Part 2 and Part 3 were necessary to recognize the existence of the function f(n) = cn – (an + bn ) and its amazing properties.

Here the variable is n, and not the integers (a, b, c).

Integers (a, b, c) are randomly selected. Where n changes, function f(n) changes, and not integers (a, b, c) (the argument is n). In the modification of n the function changes, but so that to each of the argument n corresponds to a single value of the function f(n) and to each value of the function f(n) corresponds one single value of the argument n.

It is impossible and improper the function f(n) = cn – (an + bn) to be regarded as an argument and the argument n to be regarded as a feature.

Given the result the so far it follows

  • For n = 2 there are integers (a, b, c), for which is satisfied f(n) = 0.
  • For n > 2 there are no integers (a, b, c), for which is satisfied f(n) = 0.

 

Fermat’s Last Theorem

Equality аn + bn = cn is not possible in integers (a, b, c) for n > 2.

Proof:

That for n = 2 is satisfied f(2) = 0 it follows that for n > 2 might not be satisfied

f(n) = 0, as the mutually unambiguous correspondence between n and f(n) is disturbed.

It is not possible for two different values of n (n = 2 and n = k > 2) to collate same value of f(n) (f(2) = 0 и f(k) = 0).

A value of n may be compared with different values of f(n), only when changing the integers (a, b, c).

With this the proof of Fermat’s Last Theorem is complete.

Obviously, Pierre de Fermat was right, recording that the proof of his theorem might not fit in the box on the page.

 

PART 5

CONCLUSION:

 

It appeared that the simple evidence mentioned by Pierre de Fermat exists.

That means:

Pierre de Fermat has not lied, had not deceived himself and has not fantasized, as he recorded that he found “really wonderful proof” of his theorem.

Even more:

Pierre de Fermat was the first person on Earth reaching the term “function”.

 

Presented in this work, proof of Fermat’s Last Theorem allows us to note the following:

  • Proof of Fermat’s Last Theorem may be accomplished using mathematical knowledge from the first half of the 17th century (the time when Pierre de Fermat has lived) as in the above discussion the term “function” may be avoided.
  • Pierre de Fermat was aware that Diophantine equation аn + bn = cn may be represented as a function that makes the proof of the theorem his “really wonderful”.

Perhaps Pierre de Fermat did not publish his proof, because he wanted to link it with the term “function”, but since has failed to properly define for the rest of his life, the proof remained not published.

 

We hope that in this way we have managed to protect the authority of Pierre de Fermat and remove doubts in his genius that emerged after the proof of Andrew Wiles and the assertion that was spread in internet that only mediocre and failed in their profession people may search simple proof of Fermat’s Last Theorem.

Thus Pierre de Fermat was placed at a lower level of mediocre people because he is a lawyer by education and not only he was looking simple proof of his theorem, but even wrote “found really wonderful proof, but the box is too small to make it fit.

We, ordinary people, must learn to respect genius.

 

The author expresses his gratitude to:

Nenko Ivanov, Alexander Ivanov, Todor Penchev, Stefko Banev that during the initial discussions on the manuscript provided their useful comments and helped this work to be finished.

 

References    

[1]. Wiles Andrew, Modular Elliptic Curves and Fermat’s Last Theorem, Annals of Mathematics, 1995, 142 (3), 443 – 551.

[2]. Diophanti Alexandrini, Arithmeticum, 1621.

[3]. T. Heath, Diophantus of Alexandria, Cambridge University Press, 1910, reprinted by Dover, NY, 1964, 144-145.

[4]. Nigel Boston, The Proof of Fermat’s Last Theorem, University of Wisconsin-Madison, 2003.