_{1}

^{*}

This article presents a brief and new solution to the problem known as the “Fermat’s Last Theorem”. It is achieved without the use of abstract algebra elements or elements from other fields of modern mathematics of the twentieth century. For this reason it can be easily understood by any mathematician or by anyone who knows basic mathematics. The important thing is that the above “theorem” is generalized. Thus, this generalization is essentially a new theorem in the field of number theory.

Fermat’s last theorem (known historically by this title) has been an unsolved puzzle in mathematics for over three centuries. The theorem itself is a deceptively simple formulation in mathematics, while Fermat famously stated that the problem had been solved around 1637. His claim was discovered 30 years after his death, as a clear statement on the margin of a book, but Fermat died without leaving any evidence as to his claim. This claim eventually became one of the most famous unsolved problems of mathematics. Efforts made to prove it, led to substantial development in number theory, and over time Fermat’s Last Theorem gained legendary prominence as one of the most popular unsolved problems in mathematics [

Because this problem is easily understood by everyone (in terms of its wording), most incorrect proofs have been created from time to time of any other problem in the history of mathematics.

The “Fermat’s last theorem” was made known to me, before it is solved by Professor Andrew Wiles [

Double inequalities (1.5) and (2.5) are the keys to the solutions I present to you. Also, very important are the conditions (1.7) and (1.14) for the classical theorem and (2.10) and (2.19) respectively for the general theorem. First I completed the solution at the classical problem. This solution, led me to generalize the problem.

Fermat’s last theorem (classical problem)

If x, y, z are positive integers that differ from each other, then the following equation:

(where,) (1.1)

when

Proof of Theorem

We consider positive integers x, y, z that differ from each other and hypothesize that they verify the Equation (1.1) for a natural number

Taking into account the Equation (1.1) and the condition (1.2), on the basis of the above hypothesis, we have:

Also, is:

By combining conditions (1.3) and (1.4) we have:

Comment: The double inequality (1.5) is sufficient but not necessary, i.e. the converse is not always the case. For example, we consider that

and

If we substitute z with

We will prove that when the positive integers x, y, z verify the Equation (1.1) for a natural number

Indeed from Equation (1.6) we have:

So, given the above we have:

We distinguish the following cases:

Α.

We have:

Considering Bernoulli’s inequality is (for

By combining the conditions (1.9) and (1.10) we have:

Because of condition (1.11) we observe that condition (1.5) is not satisfied, so in this case Equation (1.1) has no positive integer solutions

Β.

Since in case A. Equation (1.1) does not have positive integer solutions, obviously if they exist, this will be in case B, when the condition

(1.12)

We will then prove that when positive integers x, y, z verify the Equation (1.1) for a natural number

*Is, (

Combining the Equation (1.6) and Equation (1.13) we have:

**The number

Therefore, when Equation (1.1) verified, it is true that:

Given the condition (1.14), we have:

Based on condition (1.15), we distinguish the following sub cases:

Β_{1}.

We have:

Β_{2}.

We have: _{1}.

***The inequality

the conditions

- y > λ n ⇔ λ y < 1 n or 1 y ≤ λ y < 1 n or 1 y < 1 n

- λ < y − 2 or y > λ + 2 ≥ 1 + 2 = 3 or y > 3 ⇔ 1 y < 1 3

On the basis of inequalities

condition is satisfied while the second condition is not satisfied, on the contrary for

number one or

satisfied. This means that the Equation (1.1) is not verified always for every natural number

number

Conclusion 1: From the above it is concluded that Equation (1.1), when

“Generalization of the ‘Fermat’s last theorem’’

If

, (where,) (2.1)

when

Proof of Theorem

We consider positive integers

Taking into account Equation (2.1) and the condition (2.2), on the basis of the above hypothesis we have:

Also,

By combining conditions (2.3) and (2.4) we have:

Comment: The double inequality (2.5) is sufficient but not necessary, i.e. the converse is not always the case. For example we consider that

and

If we substitute

We distinguish the following cases:

Α.

We have,

Considering Bernoulli’s inequality, it is (for

By combining the conditions (2.7) and (2.8) we have,

Because of condition (2.9), we observe that double inequality (2.5) is not satisfied, so in this case Equation (2.1) has no positive integer solutions

Β.

Since in case A. the Equation (2.1) does not have positive integer solutions, obviously if they exist, this will be in case B, when the condition

We will then prove that when positive integers

1) First, we consider that

From condition (2.11) we have, {

*It is proved in the same way, as previously proved the condition (1.17) (see Annex).

First way: If

Second way: If

By applying mathematical induction we have:

- For

So, for

- For

- We will prove and for

Combining the conditions (2.14) and (2.15) we have,

Suffice it to prove that:

or

If,

2) Second, we consider that

First way: Based on the immediately above condition we have: {

Hypothesizing that:

**The condition (2.18) proves on the same way as the condition (2.12) (see Annex).

So, if

Second way: From condition (2.17) is:

Note: The proof that

Thus, in all cases, when the Equation (2.1) is verified for a natural number

Given condition (2.19) we have:

Based on condition (2.20) we distinguish the following sub cases:

Β_{1}.

We have:

Based on condition (2.21) we have: Equation (2.1) has positive integer solutions when

Β_{2}.

We have: _{1}.

***The inequality

- x m − 1 > n m − 2 λ ⇔ λ x m − 1 < m − 2 n or 1 x m − 1 ≤ λ x m − 1 < m − 2 n or 1 x m − 1 < m − 2 n

- λ < x m − 1 − ( m − 1 ) or x m − 1 > λ + ( m − 1 ) ≥ 1 + m − 1 = m or x m − 1 > 3 ⇔ 1 x m − 1 < 1 m

On the basis of inequalities 1 x m − 1 < m − 2 n and 1 x m − 1 < 1 m we distinguish the following conditions: 1 x m − 1 < 1 m < m − 2 n and 1 x m − 1 < m − 2 n ≤ 1 m . We observe

that for n < m ( m − 2 ) the first condition is satisfied while the second condition is not satisfied, on the contrary for n ≥ m ( m − 2 ) , the second condition is satisfied while the first condition is not satisfied. Therefore, there is always at least one natural number n greater than the number one or n > 1 , so that the condition

or 1 x m − 1 < m − 2 n ≤ 1 m is not satisfied. This means that the

Equation (2.1) is not verified always for each natural number n > 1 and every m ≥ 3 and this is contrary to the sentence “if λ ≥ 1 regardless from the parameters n, m the Equation (2.1) has solutions for each natural number n > 1

and every m ≥ 3 ” which arises from the hypothesis that n m − 2 λ ≥ n ( m − 1 ) ( n − m + 2 ) , according to the logic by which the solution of the problem was structured in this paper. This is an absurd and that is why inequality n m − 2 λ ≥ n ( m − 1 ) ( n − m + 2 ) is rejected. Therefore we consider the inequality n ( m − 1 ) ( n − m + 2 ) > n m − 2 λ is acceptable and so we ended up in inequality x m − 1 ≥ n ( m − 1 ) ( n − m + 2 ) > n m − 2 λ . (For a different reasoning, for the same, see in the Annex)

Conclusion 2: From the above it is concluded that the Equation (2.1) when n < m 2 − 2 m have integer solutions, whereas when n ≥ m 2 − 2 m have no integer solutions. In the second case, for m = 3 , answer to Fermat’s Last Theorem is given.

1) From condition n ≥ m 2 − 2 m , if m = 3 we have: n ≥ 3 2 − 2 × 3 = 3 . We observe that solution of “Fermat’s Last Theorem” occurs. This, to me, is a very strong indication that the solution of the generalization of Fermat’s theorem is correct.

2) If ( n − m + 2 > 0 or n − m + 2 ≥ 1 ), is n ≥ m − 1 . In this case we have:

i) If n ≥ m 2 − 2 m , is λ < ( m − 1 ) ( m − 2 ) n − m + 2 ≤ 1 and the Equation (2.1) has no positive integer solutions.

ii) Whereas, if m − 1 ≤ n < m 2 − 2 m , can be ( m − 1 ) ( m − 2 ) n − m + 2 > λ ≥ 1 and so, Equation (2.1) can have positive integer solutions.

3) What happens if n − m + 2 ≤ 0 or n ≤ m − 2 or n m − 2 ≤ 1 ? In this case we have: x m − 1 λ > x m − 1 x m − 1 − ( m − 1 ) > 1 ≥ n m − 2 . We observe that, if x 1 > λ ≥ 1 or λ ≥ x 1 ≥ 1 , is x m − 1 λ > x m − 1 x m − 1 − ( m − 1 ) > 1 ≥ n m − 2 . So, in this case Equation (2.1) also can have positive integer solutions. For example, if x 1 = 3 , x 2 = 4 , x 3 = 12 , x 4 = 13 , λ = 1 , n = 2 and m = 4 . We have, n − m + 2 = 2 − 4 + 2 = 0 and 3 2 + 4 2 + 12 2 = 13 2 and ( 1 3 1 2 ) 2 ≅ 1 .174 < 4 − 1 < 18 .778 = ( 13 3 ) 2 .

4) The immediately above example and the example which follows, namely: 27 5 + 84 5 + 110 5 + 133 5 = 144 5 [

5) Equation (1.1) and Equation (2.1) make sense if n > 1 , because for n = 1 , they have infinite solutions or else always have solutions. Thus, in this case, the assumptions and terms had used in the above solutions do not apply.

6) In Equation (1.1), if 1 < n < 3 , is n = 2 and so it has solutions that we known since ancient times as Pythagorean Triads.

7) Below, are presented some solutions of Equation (2.1), have made by prominent researchers, from time to time, of course using always the more times computer. It is easy to find that these solutions are perfectly in line with the general theorem.

Solutions of the Equation (2.1), which have made by prominent researchers:

30 4 + 120 4 + 272 4 + 315 4 = 353 4 (R. Norrie, 1911),

7 5 + 43 5 + 57 5 + 80 5 + 100 5 = 107 5 (Sastry, 1934, third smallest),

27 5 + 84 5 + 110 5 + 133 5 = 144 5 (Lander & Parkin, 1966),

19 5 + 43 5 + 46 5 + 47 5 + 67 5 = 72 5 (Lander, Parkin, Selfridge, smallest, 1967),

2682440 4 + 15365639 4 + 18796760 4 = 20615673 4 (Noam Elkies 1986),

95800 4 + 217519 4 + 414560 4 = 422481 4 (R. Frye, 1988),

127 7 + 258 7 + 266 7 + 413 7 + 430 7 + 439 7 + 525 7 = 568 7 (M. Dodrill, 1999),

90 8 + 223 8 + 478 8 + 524 8 + 748 8 + 1088 8 + 1190 8 + 1324 8 = 1409 8 (S. Chase, 2000),

55 5 + 3183 5 + 28969 5 + 85282 5 = 85359 5 (Frye, 2004).

The new solution to Fermat’s Last Theorem, which presented here, is as brief and simple as its wording. It is achieved without the use of abstract algebra or elements from other fields of modern mathematics of the twentieth century. For this reason, it can be easily understood by any mathematician or by anyone who knows basic mathematics. This means that it has pedagogical value. At the same time, it is important, that the above “theorem” is generalized to an arbitrarily large number of variables. This generalization is essentially a new theorem in the field of the number theory, very useful to researchers of that field, because it gives answers to many open problems of the number theory. Also, it is important, that the solutions which were found by many prominent researchers in the past, are perfectly in line with the general theorem.

The author declares no conflicts of interest regarding the publication of this paper.

Poulkas, D.Chr. (2020) A Brief New Proof to Fermat’s Last Theorem and Its Generalization. Journal of Applied Mathematics and Physics, 8, 684-697. https://doi.org/10.4236/jamp.2020.84053

1) Prove that when λ < x 1 and λ = x m − 1 − ( m − 1 ) is λ = x 1 − 1 = x 2 − 2 = ⋯ = x m − 1 − ( m − 1 ) .

Is, ( λ < x m − 2 and λ = x m − 1 − ( m − 1 ) or x m − 1 − ( m − 1 ) < x m − 2 ⇔ x m − 1 − x m − 2 < m − 1 or x m − 1 − x m − 2 = 0 or x m − 1 − x m − 2 = 1 or x m − 1 − x m − 2 = i , where i = 2 , 3 , ⋯ , m − 2 . Condition x m − 1 − x m − 2 = 1 is accepted, while the others are easily rejected. If x m − 1 − x m − 2 = 0 ⇔ x m − 1 = x m − 2 (no true) and if x m − 1 − x m − 2 = i ⇔ x m − 1 = x m − 2 + i or λ = x m − 1 − ( m − 1 ) = x m − 2 + i − ( m − 1 ) = x m − 2 − [ m − ( i + 1 ) ] (no true), because x m − 2 − [ m − ( i + 1 ) ] > x m − 1 − ( m − 1 ) . Indeed, if we consider that, x m − 2 − [ m − ( i + 1 ) ] > x m − 1 − ( m − 1 ) ⇔ ( m − 1 ) − [ m − ( i + 1 ) ] > x m − 1 − x m − 2 ≥ 1 or i > 1 or i ≥ 2 (it is true, also the condition x m − 1 − x m − 2 = i is not true). So, we have: x m − 1 = x m − 2 + 1 , therefore λ = x m − 1 − ( m − 1 ) ⇔ λ = ( x m − 2 + 1 ) − ( m − 1 ) ⇔ λ = x m − 2 − ( m − 2 ) . We repeat the same procedure for the couple x m − 2 , x m − 3 and for all other similar pairs, thus proving the condition (2.12).

2) Prove that λ ≠ x m − 1 − ( m − 1 ) , when x i − 1 ≤ λ < x i and 2 ≤ i ≤ m − 1 , by applying mathematical induction

If λ = x m − 1 − ( m − 1 ) , combining Equation (2.6) Equation (2.18) we have:

( x 1 ) n + ⋯ + ( x i − 1 ) n + ( λ + i ) n + ⋯ + ( λ + m − 1 ) n = ( 2 λ + m − 1 ) n (a.1)

By applying mathematical induction we have:

- For m = 3 , from Equation (a.1) we have:

( x 1 ) n + ( λ + 2 ) n = ( 2 λ + 2 ) n (a.2)

Also, is: ( x 1 ) n + ( λ + 2 ) n ≤ λ n + ( λ + 2 ) n < ( λ + λ + 2 ) n = ( 2 λ + 2 ) n .

So, for m = 3 , the condition (a.2) is not applies, therefore, is:

( x 1 ) n + ( λ + 2 ) n ≠ ( 2 λ + 2 ) n .

- For m = k , we suppose that is true the following condition:

( x 1 ) n + ⋯ + ( x i − 1 ) n + ( λ + i ) n + ⋯ + ( λ + k − 1 ) n ≠ ( 2 λ + k − 1 ) n (a.3)

- We will prove and for m = k + 1 is true that:

( x 1 ) n + ⋯ + ( x i − 1 ) n + ( λ + i ) n + ⋯ + ( λ + k − 1 ) n + ( λ + k ) n ≠ ( 2 λ + k ) n (a.4)

Combining the conditions (a.3) and (a.4) we have,

( x 1 ) n + ⋯ + ( x i − 1 ) n + ( λ + i ) n + ⋯ + ( λ + k − 1 ) n ≠ ( 2 λ + k − 1 ) n + ( λ + k ) n .

Suffice it to prove that: ( 2 λ + k − 1 ) n + ( λ + k ) n ≠ ( 2 λ + k ) n ⇔

( λ + k + λ − 1 ) n + ( λ + k ) n ≠ ( λ + k + λ ) n ⇔ ( 1 + λ – 1 λ + k ) n + 1 ≠ ( 1 + λ λ + k ) n or

1 ≠ ( 1 + λ λ + k ) n − ( 1 + λ – 1 λ + k ) n ⇔ 1 ≠ r n − ( r − 1 λ + k ) n , ( r = 1 + λ λ + k ) (a.5)

If, 1 > r n − ( r − 1 λ + k ) n ⇔ 1 r n > 1 − ( 1 − 1 r ( λ + k ) ) n > 1 − 1 = 0 (true, so the

Condition (a.5) also is true and consequently and the condition (a.4)). So, in this case be λ ≠ x m − 1 − ( m − 1 ) .

3) Justification for selecting the inequality y ≥ 2 n n − 1 > λ n in B_{1} of problem 2

The inequality y ≥ 2 n n − 1 > λ n was written this way, with the following reasoning: Hypothesizing that is λ n ≥ 2 n n − 1 we have: λ n ≥ 2 n n − 1 ⇔ λ ≥ 2 n − 1 . The maximum value of 2 n − 1 occurs when n = 2 or ( 2 n n − 1 ) max = 2 2 − 1 = 2 and therefore is λ ≥ 2 . However, it is known from the Greek ancient times that if 1 < n < 3 or n = 2 , the Equation (1.1) has solutions and for λ = 1 . But, this is an absurd and for this reason in this case condition λ n ≥ 2 n n − 1 is rejected. Considering now, that n ≠ 2 and hypothesizing for n ≥ 3 that λ n ≥ 2 n n − 1 , then again we have: λ n ≥ 2 n n − 1 ⇔ λ ≥ 2 n − 1 . Its maximum value of 2 n − 1 occurs when n = 3 or ( 2 n n − 1 ) max = 2 3 − 1 = 1 and therefore is λ ≥ 1 , since it is 2 n − 1 > 0 for

each n ≥ 3 . This, according to the logic by which the solution of the problem was constructed in this article, means that the Equation (1.1) has solutions for all n ≥ 3 (because the value of λ greater or equal than number one or λ ≥ 1 for all n ≥ 3 ). However, this conclusion is in stark contrast to the conclusion in B_{1}, which resulted from a valid 100% inequality and is therefore an absurd. That is

why inequality λ n ≥ 2 n n − 1 again is rejected and so we consider that inequality 2 n n − 1 > λ n is acceptable and therefore we ended up in inequality y ≥ 2 n n − 1 > λ n .

4) Justification for selecting the inequality x m − 1 ≥ n ( m − 1 ) ( n − m + 2 ) > n m − 2 λ in B_{2} of problem 2

The inequality x m − 1 ≥ n ( m − 1 ) ( n − m + 2 ) > n m − 2 λ was written this way, with the following reasoning: For m = 3 , the previous inequality becomes y ≥ 2 n n − 1 > λ n . Thus, with the same explanation as in case B_{2} of the problem 1 it turns out that the inequality y ≥ 2 n n − 1 > λ n is rejected. Because, according to the logic by

which the solution of the problem was structured in this paper, the Equation (2.1) always has solutions for λ ≥ 1 , regardless of the parameters n and m, therefore the previous conclusion for m = 3 is an absurd. That is why and the general

inequality n m − 2 λ ≥ n ( m − 1 ) ( n − m + 2 ) is rejected and therefore we consider that the inequality n ( m − 1 ) ( n − m + 2 ) > n m − 2 λ is acceptable and so we ended up in inequality x m − 1 ≥ n ( m − 1 ) ( n − m + 2 ) > n m − 2 λ .

5) Prove that, when the Equation (2.1) has positive integer solutions, is λ < x m − 1 .

We have: λ < x m − 1 − ( m − 1 ) < x m − 1 ⇔ λ < x m − 1 .