Very Original Proofs of Two Famous Problems: “Are There Any Odd Perfect Numbers?” (Unsolved until to Date) and “Fermat’s Last Theorem: A New Proof of Theorem (Less than One and a Half Pages) and Its Generalization”

This article presents very original and relatively brief or very brief proofs about of two famous problems: 1) Are there any odd perfect numbers? and 2) “Fermat’s last theorem: A new proof of theorem and its generalization”. They are achieved with elementary mathematics. This is why these proofs can be easily understood by any mathematician or anyone who knows basic mathematics. Note that, in both problems, proof by contradiction was used as a method of proof. The first of the two problems to date has not been resolved. Its proof is completely original and was not based on the work of other researchers. On the contrary, it was based on a simple observation that all natural divisors of a positive integer appear in pairs. The aim of the first work is to solve one of the unsolved, for many years, problems of the mathematics which belong to the field of number theory. I believe that if the present proof is recognized by the mathematical community, it may signal a different way of solving unsolved problems. For the second problem, it is very important the fact that it is generalized to an arbitrarily large number of variables. This generalization is essentially a new theorem in the field of the number theory. To the classical problem, probably solved the problem, as he stated in his historic its letter, with a cor-respondingly brief solution. To win the bet on the question of whether Fermat was telling truth or lying, go immediately to the end of this article before the General Conclusions.


Are There Odd Perfect Numbers?
In number theory, a perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. For instance, 6 has divisors 1, 2 and 3 (excluding itself), and ( ) 6  − . This is known as the "Euclid-Euler" theorem.
The first four perfect numbers 6, 28, 496 and 8128 were the only ones known in the first Greek mathematics, and the mathematician Nikomachos announced 8128 as early as 100 AD [1]. Later, to this day, many other researchers have been able to add other numbers to the list of perfect numbers.
Most numbers are not perfect. Although there are a surprisingly large number of results in terms of their form, two very simple questions, as to their wording, remain unresolved to this day: 1) Are there odd perfect numbers? 2) Are there infinite perfect numbers? Both of these questions-problems seem so simple, but have resisted thousands of attempts to answer with proofs or counter-examples.
The first problem is one of the subject of present work while the second may be the subject of a subsequent our research.
It is unknown whether there is any odd perfect number, although various results have been obtained. In 1496, Jacques Lefebvre stated that Euclid's rule gives all the perfect numbers, thus implying that no odd perfect number exists. Euler stated: "Whether (…) there are any odd perfect numbers is a most difficult question". More recently, Carl Pomerance presented a heuristic argument suggesting that indeed no odd perfect number should exist. All perfect numbers are also Ore's harmonic numbers, and it has been conjectured as well that there are no odds. Also, numbers up to 10 1500 have been checked without success, making the existence of odd perfect numbers appear unlikely. According to date research, if any odd perfect numbers exist, it must satisfy the following conditions: 1500 10 N > [3], N is not divisible by 105 [4], N is of the form ( )  [5]. Euler showed that an odd perfect number, if it exists, must be of the form: These few, in order to motivate the reader to search for more information related to the perfect numbers, so that he can understand easily the originality of the proof that is presented in this article compared to the ones of other previous researchers.
Also we did not use the old definition of a perfect number, as described at the beginning, but the definition of the full sum of its natural divisors, that given lat-

Fermat's Last Theorem: A New Proof of Theorem and Its Generalization
The 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 is discovered 30 years after his death famously stated that the problem had been solved around 1637.
His claim was a clear statement on the sidelines of a book, but Fermat died without to leaving any proof of his claim. This claim eventually became one of the most famous unsolved problems of mathematics. Efforts to prove it, led to an increased interest in number theory and over time Fermat's latest theorem gained top spot as one of the most popular unsolved problems in mathematics  [7]. I was informed of its existence before it was solved by Professor Andrew Wiles [8]. For several years I never tried to solve it. At some point twelve years ago, I thought I would can to solve it in a different way from Professor Andrew Wiles, believing that he might have a brief solution. One morning in my office, I started to analyze the problem and within a short time I invented the double inequality (1.5) and at that moment with great enthusiasm I exclaimed like Archimedes: I solved Fermat's last theorem! One of the most important steps was the invention of the variable λ . I thought that a tracker should be found that runs through all natural numbers from one to infinity with the speed of light, performing instantaneous checks on all possible combinations that can verify the Equations (1.1) and (2.1). As will be seen below, this is perfectly achieved with the ratios: y λ for the problem 3.1 or The proofs presented in this article starts from a zero basis. In other words, I consider that Equations (1.1) and (2.1) when the exponent n equals the number 1 have infinite integer solutions (trivial) and of course this case is ignored, whereas when the exponent n is greater than number one or 1 n > it's examined for whom of the exponents Equations (1.1) and (2.1) can have integer solutions and for whom of the exponents do not have, (as is clear, is ignored even the same the Fermat's assumption).

Are There Any Odd Perfect
Therefore, in order to be perfect a positive odd integer it must to be satisfied Condition (3).
We distinguish the following cases: A In order the root d (as shown in the Condition (4)) to be a positive integer, it must to be a positive integer and the mathematical expression: We will prove that this is not possible, because it applies the below condition: . Indeed assuming that: , then equivalently we have: 2 2 2 4 4 2 1 2 1 n n n n n n − + < − − < − + ⇔ 2 5 2 0 n − + < < . This condition, because 3 n ≥ , n N ∈ (due to Condition (4)), it's true.
Therefore, 3 n ∀ ≥ , n N ∈ , it's true and that: It follows from Condition (5) that mathematical expression: Also, if we set: (1*) From now on by the expression, "where k appropriate integer", we will mean that k is given by the mathematical formula: By combining the Conditions (6) and (7) we have: where k appropriate integer, In order the root 1 d (as shown in the Condition (8)) to be a positive integer, it must to be a positive integer and the mathematical expression: We will prove that this is not possible, because it applies the below condition: . Indeed, assuming that: , then equivalently we have: Condition (9) from the right is satisfied. In order the same condition to be satisfied and from the left it must to be true that: So, assuming that: 2 4 5 0 n k − + + < and taking into account divisor 1 d as is defined above and since n k − + + < ⇔ 2 2 2 5 n k k − > + ⇔ 2 2 2 2 5 n n k k > − > + . Given the Condition (7) and because 1 3 d ≥ or Because Condition (10) or 2 4 5 0 n k − + + < implies Condition 3 n ≥ (which is true) and since Condition 3 n ≥ as true implies that: 2 4 5 0 n k − + + < either 2 4 5 0 n k − + + ≥ , we will continue the proof further, after the paragraph (3*), assuming that the second of the two previous conditions, is true, or: . From the last condition, equivalently we have: which it's true, therefore it's true and that: (3*) Condition 3 n ≥ can be reached and as follows: Assuming that 2 4 5 0 n k − + + < and because 0 k ≥ , we have: 2 4 5 0 n k − + + < ⇔ 0 4 2 5 k n ≤ < − ⇒ 0 2 5 n < − ⇔ 2 5 n > ⇔ 5 2 n > or n ≥ 3 which it's true, since due to Condition (8) it applies that 3 n ≥ , n N ∈ . Although this way is easier we preferred the previous way, which it is more extensive we to show that the choice, the divisor 1 d , to be the smallest of all the non-trivial divisors of N is correct. As is easily understood, from the previous analysis and especially from the Condition (3), the smaller non-trivial divisor 1 d leads to the smallest possible value of k, which increases the probability the condition 2 4 5 0 n k − + + < to be, true (just as we would like). Next assuming that: 2 4 5 0 n k − + + ≥ , where k appropriate integer, are infinitely many we will consider that this result it's an indication that the Condition 2 4 5 0 n k − + + ≥ is false and we will continue further our proof for 0 k > . So, given and Condition (3), we have: 2 4 5 0 n k − + + ≥ ⇔ and so its Therefore, Condition 2 4 5 0 n k − + + ≥ is false and it's true that: 2 4 5 0 n k − + + < or 2 4 5 0 2 2 n k k − + + < < + , where 3 n ≥ , n N ∈ and k appropriate integer, So, consequently it applies that: It follows from the Condition (12) that mathematical expression: Second (2 nd ) generalized proof (continues from Condition (9) onwards): The second generalized proof is presented here, because it's also original and undoubtedly confirms the correctness of the first proof.
According to Equation ( ρ ρ + that is the first member of the last equality and 2 4µ which is term of the second member, it must also to divides and the number 2n which also is term of the same member.
Therefore it should be valid that: 2 4 n λ = , where λ a positive integer and consequently 2 n λ = . Also, because We will prove that Equality (13) is not possible. For this, we work as follows: Therefore will be: n Let a positive integer such that a n µ = ′ − , then by substitution in the Equation (13) we have: 1 a n k n n n a n µ µ From the last equality of the two polynomials, if we equate the coefficients of the terms with the same degree, we have: is perfect square of a positive integer, then the mathematical is an odd number, must be perfect square of a positive even integer.
This means that and the mathematical expression: ( ) , must be perfect square of a positive integer. We will prove that this is not possible, because it valid the below condition: , then equivalently we have: Condition (14), from the right is satisfied. We will examine whether the same condition is satisfied and by the left.
At the first we observe that: 0 2 4 5 n k − + + ≥ , because according to our hypothesis it holds that ( ) (9) is satisfied and consequently it applies that: and so the mathematical expression: , which is contrary to our hypothesis. Then, given that From the last condition if 0 2 3 4 n΄ k΄ − + + < , equivalently we have: Therefore, Condition (14) or 0 2 3 4 1 n΄ k΄ k΄ − + + < < + it's true and consequently it applies that: It follows from the last condition that mathematical expression: ( ) cannot be a positive integer because it's contained between the consecutive integers ( ) So, it applies that: . This means that: Important comment (a posteriori that is after the completion of the proof): The fact that the condition 2 4 5 0 n k − + + ≥ is false is also substantiated as k ≥ × = ≥ , which is absurd.

Comments-Remarks
1) According to the above when, 2 1 n d d That is absurd because an odd integer cannot be equal to an even integer (since number d, is odd integer).
ii) If one pair of the m non trivial pairs of divisors of N and Given Condition (7) equivalently, we have: That is absurd because an odd integer cannot be equal to an even integer (since number d λ , is odd integer).
2) Condition (5)  , divisor 1 is calculated twice, can be considered as the only odd perfect number.

"Fermat's Last Theorem: A New Proof of Theorem and
Its Generalization"

A New Proof of Fermat's Last Theorem (Classical Problem)
If , , x y z are positive integers that differ from each other then the following equation:

Proof of Theorem
We consider positive integers , , x y z that differ from each other and assume that they verify the Equation (1.1) for a natural number 1 n > . Also, we assume without loss of the generality, that: Given Equation (1.1) and condition (1.2), based on the above hypothesis, we have: By combining the conditions (1.3) and (1.4) we have: Note: The double inequality (1.5) is necessary but not sufficient, i.e. the converse is not always the case. For example, we consider that 3, 4, 5 x y z = = = and 3 n = . We have: If we replace the number z with the sum y λ + or z y λ = + , where λ is a positive integer, then for the positive integers , , x y z that according to hypothesis we originally made, verify Equation (1.1) for a natural number Given Condition (1.7) and the original hypothesis it's true that: We distinguish the following cases: Considering Bernoulli's inequality for 1 n > , it applies: By combining the Conditions (1.9) and (1.10) we have: Due to condition (1.11) the double inequality (1.5) is not satisfied and therefore in this case Equation (1.1) has no positive integer solutions, for any natural number n greater than the number one or Since in case A. Equation (1.1) doesn't have positive integer solutions, obviously if they exist, this will be the case in case B when condition y n λ > is applied. So, we have: Now, we will prove that when positive integers , , x y z verify Equation (1.1) for a natural number 1 n > , number y is greater than the product of number 3 by the number λ or 3 y λ > . We work as follows: a) If 1 3 n < < or 2 n = , assuming that 3 y λ ≤ , we distinguish two areas: . Thus, given the latter areas, we have. a.1) If 3 3 y λ < ≤ , assuming that the trinity , , x y z verify Equation (1.1) for a natural number 1 n > , then because from 3 y < ⇔ 4 y ≥ and from 3 3λ < ⇔ 1 λ < or 2 λ ≥ , it applies the following: is not valid and therefore condition 3 , assuming also that the integers , , 1 x y y + , verify Equation (1.1), it applies that: ( ) This means that the number x is an odd integer and because due to Condition (1.8) it applies that: 1 x y z λ ≤ < < < , it follows that in this sub-case no trinity , , 1 x y y + that satisfying Equation (1.1) can be formed. Thus, the area is not valid and therefore condition 3 y λ ≤ is false.  3  2  3   1  1  1  1  1  1 3  2  3  3  3  1  1  1  3  3 .5) is not satisfied and Equation (1.1) has not positive integer solutions. Thus and in this case condition 3 y λ ≤ is false. Therefore, in all cases condition 3 y λ ≤ is false and it's true that: Next, we will prove that when positive integers , , x y z verify Equation (1.1) for a natural number 1 n > , number λ is less than the difference Indeed, given the Conditions (1.12), (1.13) and considering separately the in-tervals1 3 n < < and 3 n ≥ , we have: given by the author of the present article in a previous its paper [9] and is now listed on the Appendix 2 (the old proof continues from this point onwards after, as it was in its original publication, adding some clarifications).
Therefore, in this case, due to the last conditions double inequality (1.5) is sa- Also, if 3 n ≥ due to Condition (1.12), we have: Next, we will prove and we that: 3 3 3 x y z + ≠ , despite the fact that, Euler was the first which proved it. For this, assuming that the positive integers , , x y z Thus, condition (1.5) is not satisfied and Equation (1.1) has no positive integer solutions. Therefore, the area 4 2 3 3 y λ < ≤ < is not valid and is excluded.
, which that is also absurd. So, in all sub-cases, condition 1 λ ≤ is false because it implies that: 3 3 3 x y z + ≠ . Please, at this point, pay special attention to the following remark: What about the trinities , , x y z for which, it applies that: x y z + ≠ and 3 n > . An example of such a trinity is the fol- 3 n ≥ and all the trinities , , x y z except of the trinities for which it's true that: 3 3 x y z + ≠ , we have: Therefore, in sub-case B2, due to the last condition the double inequality (1.5) is not satisfied and also Equation (1.1) has no positive integer solutions.

Generalization of Fermat's Last Theorem (New Theorem)
Considering Bernoulli's inequality for 1 n > , it applies: is not satisfied and Equation (2.1) has not positive integer solutions. Thus, in this case, condition Therefore, in all cases condition Given condition (2.12) we have: (2.13) (9*) Be careful not to get confused. Condition (2.13), which corresponding to the Condition (2.20), has nothing to do with the present proof, it's simply written here to exist link to the continuation of proof of problem 3.2, as this recently given by the author of the present article in a previous its paper [9] and is now listed on the Appendix 2 (the old proof continues from this point onwards after, as it was in its original publication, adding some clarifications).
We distinguish the following sub-cases: Proof: At the first we will prove that: Also, it is true and the follows condition: Therefore, in this sub-case, due to the last conditions the double inequality (2.5) always is satisfied and so Equation Because the second member of the last equation is an even positive integer or ( ) ( ) which is absurd because the second member of the last equality is an odd integer or ( )

A Very Brief Solution to "Fermat's Last Theorem"
Here we present to you the shortest solution that has ever been achieved to "Fermat's last theorem", the most famous mathematical problem of the world.  − will be an odd integer. So and the numbers x and n x will also are odd integers. Therefore, it applies that: Note: A wonderful second proof that the condition 1 λ ≤ it's false, is given in the case B2) of the problem 3.2, a little above. Also, to the Annex 1 is presented a second proof why the area 4 2 3 3 y λ < ≤ < is excluded.

General Conclusions
The proofs of the two problems which presented in this article are brief and very simple as their wording. They are achieved without the use of abstract algebra or elements from other fields of modern mathematics of the twentieth century. For this reason, they can be easily understood by any mathematician or anyone who knows basic mathematics. That means they have pedagogical value. It is also worth mentioning that the proofs of the two problems were achieved without the use of a computer. When it comes to the conjecture of the odd perfect numbers, modern researchers have been trying to prove it by using computer-assisted methods. Although computers have changed the way we approach Mathematics, their overuse at the expense of mathematical thinking is an abuse and this should be seriously taken into consideration by the scientific community.
Also, it's very important, that Fermat's last theorem is generalized to an arbitrarily large number of variables. This generalization is essentially a new theorem in the field of number theory very useful for researchers of this field, because it can give answers to many open problems of the number theory.
Finally, the proofs presented here are completely original and were not based on the work of other researchers.

Conflicts of Interest
The author declares no conflicts of interest regarding the publication of this paper. Continuity of the proof, as it is had given recently by the author of this article and now it is set out here, adding some clarifications.
Based on Condition (1.15), we distinguish the following sub cases: