*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

**, without mentioning**

*4**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
and*n = 3*does not mean that he has not found a*n = 4***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 ** a^{n} + b^{n} = c^{n}** is unsolvable in general, it follows:

**If Pierre de Fermat has discovered “really wonderful proof” of the theorem, then he noticed in equality а^{n} + b^{n} = c^{n} 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 **

** а^{n} + b^{n} = c^{n}**.

or in the form:

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

*c ^{n} – (а^{n} + b^{n}) = 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 **

*c ^{n} – (а^{n} + b^{n}) = 0*.

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

** c^{n} – (а^{n} + b^{n}) ≠ 0**.

We may combine both equalities ** c^{n} – (а^{n} + b^{n}) = 0 **and

**as follows:**

*c*^{n}– (а^{n}+ b^{n}) ≠ 0*c ^{n} – (а^{n} + b^{n}) = m*

*(1)*

where for** n = 2 **is satisfied

**, and for**

*m = 0***is satisfied**

*n > 2***.**

*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

**to**

*n = 2***changes also**

*n > 2***from**

*m***to**

*m = 0***.**

*m ≠ 0*Then question raises:

Doesn’t it mean that between ** n** and

*m =***there is a functional dependence**

*c*^{n}– (а^{n}+ b^{n})*f(n) =**c ^{n} – (а^{n} + b^{n})*

wherein independently of the choice of integers ** (a, b, c)** for each value of

**corresponds a single value of**

*n***, and vice versa — to each value of**

*f(n)***corresponds a single value of**

*f(n)***?**

*n**Example 1:*

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

It is satisfied:

** 3^{2} + 4^{2} = 5^{2 }**and

*5*^{2}**(**

*–***)**

*3*^{2}+ 4^{2}

*= 0*** 3^{3} + 4^{3} < 5^{3 }**and

*5*^{3}**(**

*–***)**

*3*^{3}+ 4^{3}

*= 34*** 3^{4} + 4^{4} < 5^{4 }**and

*5*^{4}**(**

*–***)**

*3*^{4}+ 4^{4}

*= 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

**is mutually uniquely.**

*f(n)**Example 2:*

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

It is satisfied:

** 9^{2} + 40^{2} = 41^{2 }**and

*41*^{2}**(**

*–***)**

*9*^{2}+ 40^{2}

*= 0*** 9^{3} + 40^{3} < 41^{3 }**and

*41*^{3}**(**

*–***)**

*9*^{3}+ 40^{3}

*= 4192*** 9^{4} + 40^{4} < 41^{4 }**and

*41*^{4}**(**

*–***)**

*9*^{4}+ 40^{4}

*= 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

**is mutually uniquely.**

*f(n)*

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

- Between
and*n*exists uniquely each line, regardless of the choice of integers*f(n)*.*(a, b, c)* - When changing the integers
, then it changes the values of*(a, b, c)*, corresponding to*f(n)*, but mutually unambiguous correspondence between*n*and*n*is retained.*f(n)* - When for
is satisfied*n = 2*and*c*^{n}– (а^{n}+ b^{n}) = 0increases, then the difference*n**c*^{n}increases and there is no way to get equality*– (a*^{2}+ b^{2})for*а*^{n}+ b^{n}= c^{n}, regardless of the choice of integers*n > 2*.*(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) = c ^{n} – (а^{n} + b^{n})*

*(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 ** а^{n} + b^{n} = c^{n}**, because (as it will be shown below) if used

*(2)*it may be obtained simply and “

**” of**

*really wonderful proof**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

*a ^{2}+ b^{2} = c^{2 }*

*(3)*

*Example:*

*3 ^{2} + 4^{2} = 5^{2}*

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

*а ^{2} + b^{2} < с^{2 }*

*(4)*

*а ^{2} + b^{2} = c^{2 } *

*(5)*

*а ^{2} + b^{2} > c^{2 } *

*(6)*

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

*c ^{2} – (а^{2} + b^{2}) > 0*

*c ^{2} – (а^{2} + b^{2}) = 0 *

*c ^{2} – (а^{2} + b^{2}) < 0 *

and combine in one relationship

*c ^{2} – (а^{2} + b^{2}) = 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 *а ^{2} + b^{2 }< c^{2}*,

*Part 2:***For the integers (a, b, c) is satisfied а^{2} + b^{2 }= c^{2},**

*Part 3:***For the integers (a, b, c) is satisfied а^{2} + b^{2 }> c^{2},**

*Part 4:* Proof of Fermat’s Last Theorem, provided that for the integers *(a, b, c)* the function *f(n) =* *c ^{n} – (a^{n} + b^{n})* 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) = c ^{n} – (а^{n} + b^{n}) *

*(8)*

is a function of ** n**, wherein the correlation between

**and**

*n***is mutually unambiguous, regardless of the choice of integers**

*f(n)***.**

*(a, b, c)*

*PART 1*

*FOR INTEGERS (a, b, c) IS SATISFIED RELATIONSHIP*** а^{2} + b^{2} < с^{2}**.

*Theorem 1.1*

**If for the integers (a, b, c) is satisfied а^{2} + b^{2} < с^{2}, then for n > 2 is satisfied **

*а ^{n} + b^{n} < c^{n }* and might not be satisfied

*а*.

^{n }+ b^{n }= c^{n}*Proof:*

We multiply both sides of the inequality ** а^{2 }+ b^{2 }< c^{2 }**with

**and we obtain**

*c*^{n-2}*а ^{2}.c^{n-2} + b^{2}.c^{n-2} < с^{2}.c^{n-2 } *

*(9)*

We multiply both sides of the inequality ** а^{n-2} < c^{n-2 }**with

**and obtain**

*a*^{2}*a ^{2}.a^{n-2} < a^{2}.c^{n-2}*

We multiply both sides of the inequality ** b^{n-2} < c^{n-2 }**with

**and obtain**

*b*^{2}*b ^{2}.b^{n-2} < b^{2}.c^{n-2}*

Since ** a^{2}.a^{n-2} < a^{2}.c^{n-2 }**and

**, when we replace in**

*b*^{2}.b^{n-2}< b^{2}.c^{n-2}*(9)*,

**с**

*а*^{2}.c^{n-2}**and**

*a*^{2}.a^{n-2}**с**

*b*^{2}.c^{n-2}**we obtain the stronger inequality**

*b*^{2}.b^{n-2 }*а ^{2}.a^{n-2} + b^{2}.b^{n-2} < с^{2}.c^{n-2 } *

*(10)*

Of *(10) *it derives:

- For
is satisfied*n > 2*.*a*^{n}+ b^{n}< c^{n} - Where
grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}< c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
is impossible to be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

*Example:*

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

It is satisfied:

** 3^{2} + 5^{2} < 7^{2} **and

*7*^{2}– (3^{2}+ 5^{2}) = 15** 3^{3} + 5^{3} > 7^{3} **and

*7*^{3}– (3^{3}+ 5^{3}) = 191** 3^{4} + 5^{4} > 7^{4} **and

*7*^{4}– (3^{4}+ 5^{4}) = 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 ** а^{2} + b^{2} < с^{2 }**correspondence between

**and**

*n***is mutually unambiguous.**

*f(n)*

*PART 2*

*FOR INTEGERS (a, b, c) IS SATISFIED RELATIONSHIP*** а^{2} + b^{2} = с^{2}**.

*Theorem 2.1*

**If for the integers (a, b, c) is satisfied а^{2} + b^{2 }= c^{2}, **

*n > 2* is satisfied *а ^{n} + b^{n} < с^{n}* and might not be satisfied

*а*.

^{n}+ b^{n}= с^{n}*Proof:*

Of the equality ** а^{2} + b^{2} = c^{2}** it follows

**;**

*а*^{2}< c^{2}**and**

*b*^{2}< c^{2}

*а < c*;*b < c*.Then in that case for ** n > 2** it is satisfied:

*a ^{n-2} < c^{n-2}*

*(11)*

*b ^{n-2} < c^{n-2}*

*(12)*

Multiplying both sides of* (11) *with ** а^{2}**, and both sides of

*(12)*with

**, it is obtained**

*b*^{2}*а ^{2}.a^{n}*

^{-2 }

*< а*^{2}.c^{n}

^{-2}^{ }

*(13)*

*b*^{2}*.b ^{n}*

^{-2}

*< b*

^{2}

*.c*^{n}

^{-2}*(14)*

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

*а ^{2}.a^{n}*

^{-2}

*+ b*

^{2}

*.b*^{n}

^{-2}

*< а*^{2}.c^{n}

^{-2}

*+ b*

^{2}

*.c*^{n}

^{-2}*(15)*

When we multiply both sides of the equality ** а^{2 }+ b^{2 }= c^{2 }**with

**we receive**

*c*^{n-2 }*а ^{2}.c^{n-2} + b^{2}.c^{n-2} = c^{2}.c^{n-2}*

*(16)*

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

*а ^{2}.a^{n-2} + b^{2}.b^{n-2} < c^{2}.c^{n-2}*

*(17)*

Of *(17) *it derives:

- For
is satisfied*n > 2*.*a*^{n}+ b^{n}< c^{n} - Where
grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}< c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
is impossible to be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

*Example:*

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

It is satisfied:

** 3^{2} + 4^{2} = 5^{2 }**and

*5*^{2}– (3^{2}+ 4^{2}) = 0** 3^{3} + 4^{3} < 5^{3 }**and

*5*^{3}– (3^{3}+ 4^{3}) = 34** 3^{4} + 4^{4} < 5^{4 }**and

*5*^{4}– (3^{4}+ 4^{4}) = 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 ** а^{2} + b^{2} = с^{2 }**correspondence between

**and**

*n***is mutually unambiguous.**

*f(n)** *

*Le**mma 2.1*

**If for the integers s and q is satisfied s > q, when n increases, then difference s^{n} – q^{n} is growing**.

*Proof: *

As ** s > q**, we mark:

*s ^{1} – q^{1} = p_{1}*

*s ^{2} – q^{2} = p_{2 }*

*s ^{3} – q^{3} = p_{3}*

*s ^{4} – q^{4} = p_{4}*

*. . . . . . . . . .*

*s ^{n} – q^{n} = p_{n}*

*. . . . . . . . . .*

For entered indications it may be written

*s ^{1} – q^{1} = s – q = p_{1}*

*s ^{2} – q^{2} = (s – q).(s + q) = p_{2}*

*s ^{3} – q^{3} = (s – q).(s^{2}+ s.q + q^{2}) = p_{3}*

*s ^{4} – q^{4} = (s – q).(s^{3}+ s^{2}.q + s.q^{2}+ q^{3}) = p_{4}*

*. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . *

*s ^{n} – q^{n} = (s – q).(s^{n-1} + s^{n-2}.q + s^{n-3}.q^{2} + . . . + s.q^{n-2}+ q^{n-1}) = p_{n }*

*(18)*

*. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .*

Given that ** s** and

**are integers and**

*q***, it follows:**

*s – q = p*_{1}> 0*0 < p _{1} < p_{2} < p_{3} < p_{4} < . . . < p_{n} . . .*

*Theorem 2.2*

**If for integers (a, b, c) for n = 2 is satisfied c^{n} – (а^{n} + b^{n}) = 0, **

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

** c^{n} – (а^{n} + b^{n}) grows from 0 and tends to infinity**.

*Proof:*

If in *(18)* we lay down ** s = c** and

**, for degree**

*q = a***we obtain:**

*n – 2**c ^{n}*

^{-2}

*– а*^{n}

^{-2}

*= (c*

*– а).(c*^{n}

^{-3}

*+ c*^{n}

^{-4}

*.а + c*^{n}

^{-5}

*.а*^{2}+ . . . + c.а^{n-4}+ а^{n-3}) = p_{an }*(19)*

If in *(18)* we lay down ** s = c** and

**for degree**

*q = b***we obtain:**

*n – 2**c ^{n-2} – b^{n-2} = (c – b).(c^{n-3} + c^{n-4}.b + c^{n-5}.b^{2} + . . . + c.b^{n-4} + b^{n-3}) = p_{bn } *

*(20)*

Multiplying both sides of*(19) *with ** а^{2}**, and both sides of

*(20)*with

**, we receive**

*b*^{2}*a ^{2}.c^{n-2} – a^{2}.а*

^{n-2}

*= a*^{2}.p_{an}*(21)*

*b ^{2}.c^{n-2} – b^{2}.b^{n-2} = b^{2}.p_{bn}*

*(22)*

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

*a ^{2}.c^{n-2} + b^{2}.c^{n-2 }– (a^{2}.а*

^{n-2 }

*+ b*^{2}.b^{n-2}) = a^{2}.p_{an }+ b^{2}.p_{bn}*c ^{n-2}.(a^{2} + b^{2}) – (а*

^{n }

*+ b*^{n}) = a^{2}.p_{an }+ b^{2}.p_{bn}Given that it is satisfied, as provided that ** а^{2} + b^{2} = c^{2}**, it follows

*c ^{n-2}.c^{2} ^{ }– (а*

^{n }

*+ b*^{n}) = a^{2}.p_{an }+ b^{2}.p_{bn}*c ^{n} ^{ }– (а*

^{n }

*+ b*^{n}) = a^{2}.p_{an }+ b^{2}.p_{bn}Since according to the *Lemma* *2.1*, ** p_{an}** and

**grow, where**

*p*_{bn}**grows, it follows that where**

*n***grows of primary value**

*n***and tends to infinity, then difference**

*n = 2***increases from primary value**

*c*^{n}– (а^{n}+ b^{n})**and tends to infinity.**

*0**Theorem 2.3*

**If for n = 2 is satisfied a^{n} + b^{n} = c^{n}, then**

**For***n > 2*is satisfied*a*.^{n}+ b^{n}< c^{n}**For***n < 2*is satisfied*a*.^{n}+ b^{n}> c^{n}**For***n > 2*and n < 2 is impossible to be satisfied the equality*а*.^{n}+ b^{n }= c^{n}

*Proof:*

*According to Theorem 2.1*, if for ** n = 2 **is satisfied

**, where**

*а*^{n}+ b^{n}= с^{n}**increases (for**

*n***) is satisfied**

*n > 2***and might not be satisfied**

*а*^{n}+ b^{n}< с^{n}**. In that case between**

*а*^{n}+ b^{n}= с^{n}**and**

*n***there is mutually unambiguous correspondence that is expressed by the function**

*c*^{n}– (а^{n}+ b^{n})

*f(n) =***.**

*c*^{n}– (a^{n}+ b^{n})For ** n < 2** only one single value for

**(**

*n***) is possible, corresponding with one single value of the function**

*n = 1*

*f(1) =***.**

*c*^{1}– (a^{1}+ b^{1})For ** n = 2** only one single value of the function

*f(2) =***is possible.**

*c*^{2}– (a^{2}+ b^{2})*Example:*

** 3^{1} + 4^{1} > 5^{1 }**and

*5*^{1}– (3^{1}+ 4^{1}) = -2** 3^{2} + 4^{2} = 5^{2 }**and

*5*^{2}– (3^{2}+ 4^{2}) = 0** 3^{3} + 4^{3} < 5^{3 }**and

*5*^{3}– (3^{3}+ 4^{3}) = 34** 3^{4} + 4^{4} < 5^{4 }**and

*5*^{4}– (3^{4}+ 4^{4}) = 288*. . . . . . . . . . . . . . . . . . . . . . . . . . . *

Insofar we explored the integers ** (а, b, с)**, where for their second degrees are satisfied relationships

*a*^{2}+ b^{2}< c^{2},**and we proved that in both cases:**

*a*^{2}+ b^{2}= c^{2}- For
, equality*n > 2*might not be satisfied in integers*а*^{n}+ b^{n}= с^{n}.*(a, b, c)* - Between
and*n*there is mutually unambiguous correspondence that is expressed by the function*c*^{n}– (а^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

* *

* *

*PART 3*

*FOR INTEGERS (a, b, c) IS SATISFIED RELATIONSHIP*** а^{2} + b^{2} > с^{2}**.

* *

From the relationship ** а^{2} + b^{2} > с^{2}**,

**cases are following:**

*9**a ^{2} > c^{2}; b^{2} > c^{2}*

*a ^{2} > c^{2}; b^{2} < c^{2}*

*a ^{2} < c^{2}; b^{2} > c^{2}*

*a ^{2} > c^{2}; b^{2} = c^{2}*

*a ^{2} = c^{2}; b^{2} > c^{2}*

*a ^{2} = c^{2}; b^{2} < c^{2}*

*a ^{2} < c^{2}; b^{2} = c^{2}*

*a ^{2} = c^{2}; b^{2} = c^{2}*

*a ^{2} < c^{2}; b^{2} < c^{2}*

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:

*a ^{2} > c^{2}; b^{2} > c^{2}*

*a ^{2} > c^{2}; b^{2} < c^{2}*

*a ^{2} > c^{2}; b^{2} = c^{2}*

*a ^{2} = c^{2}; b^{2} < c^{2}*

*a ^{2} = c^{2}; b^{2} = c^{2}*

*a ^{2} < c^{2}; b^{2} < c^{2}*

* *

*Case 1:*

** а^{2} + b^{2} > с^{2} **and

**.**

*a*^{2}> c^{2}; b^{2}> c^{2}*Theorem 3.1*

**If a, b and c are integers for which is satisfied а^{2} + b^{2} > с^{2} **and

*a*, where^{2}> c^{2}; b^{2}> c^{2}*n*increases, for*n > 2*is satisfied*а*and might not be satisfied^{n}+ b^{n}> c^{n }*а*.^{n}+ b^{n}= c^{n}*Proof:*

Of ** a^{2} > c^{2}; b^{2} > c^{2}** it follows:

**,**

*a*^{n-2}> c^{n-2}**, for**

*b*^{n-2}> c^{n-2}**.**

*n > 2*When we multiply both sides of the inequality ** а^{2 }+ b^{2 }> c^{2 }**with

**we receive**

*c*^{n-2}*a ^{2}.c^{n-2} + b^{2}. c^{n-2} > c^{2}.c^{n-2}*

*(23)*

When we multiply both sides of the inequality ** а^{n-2} > c^{n-2 }**with

**and obtain**

*a*^{2}*a ^{2}.a^{n-2} > a^{2}.c^{n-2}*

When we multiply both sides of the inequality ** b^{n-2} > c^{n-2 }**with

**and obtain**

*b*^{2}*b ^{2}.b^{n-2} > b^{2}.c^{n-2}*

Since ** a^{2}.a^{n-2} > a^{2}.c^{n-2 }**and

**, when we replace in**

*b*^{2}.b^{n-2}> b^{2}.c^{n-2}*(23)*,

**with**

*а*^{2}.c^{n-2}**and**

*a*^{2}.a^{n-2}**with**

*b*^{2}.c^{n-2}**we obtain the stronger inequality**

*b*^{2}.b^{n-2 }*а ^{2}.a^{n-2} + b^{2}.b^{n-2} > с^{2}.c^{n-2 } *

*(24)*

Of *(24) *it derives:

- For
is satisfied*n > 2*.*a + b*^{n}> c^{n} - Where
grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}> c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
is impossible to be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

*Example:*

** 4^{2} + 5^{2} > 3^{2} **and

*3*^{2}– (4^{2}+ 5^{2}) = – 32** 4^{3} + 5^{3} > 3^{3} **and

*3*^{3}– (4^{3}+ 5^{3}) = – 164** 4^{4} + 5^{4} > 3^{4} **and

*3*^{4}– (4^{4}+ 5^{4}) = – 800*. . . . . . . . . . . . . . . . . . . . . . . . . . .*

* *

*Case 2:*

*а ^{2} + b*

^{2}

*> с*^{2}; a

^{2}

*> c*

^{2}

*; b*

^{2}

*< c*

^{2}*Theorem** 3.2*

**If (а, b, с) are integers for which is satisfied а^{2} + b^{2} > с^{2}; a^{2} > c^{2}; b^{2} < c^{2}, where n **

**grows for** *n > 2* is satisfied *а ^{n} + b^{n} > с^{n}* and might not be satisfied

*а*.

^{n}+ b^{n}= с^{n}*Proof:*

Of ** a^{2} > c^{2}; b^{2} < c^{2}** it follows:

**,**

*a*^{n-2}> c^{n-2}**, for**

*b*^{n-2}< c^{n-2}**.**

*n > 2*When we multiply both sides of the inequality ** а^{2 }+ b^{2 }> c^{2 }**with

**we receive**

*c*^{n-2}*a ^{2}.c^{n-2} + b^{2}.c^{n-2} > c^{2}.c^{n-2}*

*(25)*

When we multiply both sides of the inequality ** а^{n-2} > c^{n-2 }**with

**and obtain**

*a*^{2}*a ^{2}.a^{n-2} > a^{2}.c^{n-2}*

When we multiply both sides of the inequality ** b^{n-2} < c^{n-2 }**with

**and obtain**

*b*^{2}*b ^{2}.b^{n-2} < b^{2}.c^{n-2}*

Since ** a^{2}.a^{n-2} > a^{2}.c^{n-2 }**(regardless the value of

**), when we replace in**

^{2}.b^{n-2}*(25)*,

**with**

*а*^{2}.c^{n-2}**and**

*a*^{2}.a^{n-2}**with**

*b*^{2}.c^{n-2}**we obtain**

*b*^{2}.b^{n-2 }*а ^{2}.a^{n-2} + b^{2}.b^{n-2} > с^{2}.c^{n-2 } *

*(26)*

Of *(26) *it derives:

- For
is satisfied*n > 2*.*a*^{n}+ b^{n}> c^{n} - Where
grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}> c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
is impossible to be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

*Example:*

** 7^{2} + 3^{2} > 5^{2} **and

*5*^{2}– (7^{2}+ 3^{2}) = – 34** 7^{3} + 3^{3} > 5^{3} **and

*5*^{3}– (7^{3}+ 3^{3}) = – 245** 7^{4} + 3^{4} < 5^{4 }**and

*5*^{4}– (7^{4}+ 3^{4}) = – 1857*. . . . . . . . . . . . . . . . . . . . . . . . . . . .*

* *

*Case 3:*

*а ^{2} + b*

^{2}

*> с*^{2}; a

^{2}

*> c*

^{2}

*; b*

^{2}

*= c*

^{2}*Theorem** 3.3*

**If (а, b, с) are integers for which is satisfied а^{2} + b^{2} > с^{2}; a^{2} > c^{2}; b^{2} = c^{2}, where n **

**grows for** *n > 2* is satisfied *а ^{n} + b^{n} > с^{n}* and might not be satisfied

*а*.

^{n}+ b^{n}= с^{n}*Proof:*

Of ** a^{2} > c^{2}; b^{2} = c^{2}** it follows:

**,**

*a*^{n-2}> c^{n-2}**, for**

*b*^{n-2}= c^{n-2}**.**

*n > 2*When we multiply both sides of the inequality ** а^{2 }+ b^{2 }> c^{2 }**with

**we receive**

*c*^{n-2}*a ^{2}.c^{n-2} + b^{2}.c^{n-2} > c^{2}.c^{n-2}*

*(27)*

When we multiply both sides of the inequality ** а^{n-2} > c^{n-2 }**with

**and obtain**

*a*^{2}*a ^{2}.a^{n-2} > a^{2}.c^{n-2}*

When we multiply both sides of the equality ** b^{n-2} = c^{n-2 }**with

**and obtain**

*b*^{2}*b ^{2}.b^{n-2} = b^{2}.c^{n-2}*

Since ** a^{2}.a^{n-2} > a^{2}.c^{n-2 }**and

**, when we replace in**

*b*^{2}.b^{n-2}> b^{2}.c^{n-2}*(27)*,

**with**

*а*^{2}.c^{n-2}**and**

*a*^{2}.a^{n-2}**with**

*b*^{2}.c^{n-2}**we obtain**

*b*^{2}.b^{n-2 }*а ^{2}.a^{n-2} + b^{2}.b^{n-2} > с^{2}.c^{n-2 } *

*(28)*

Of *(28) *it derives:

- For
is satisfied*n > 2*.*a*^{n}+ b^{n}> c^{n} - Where
grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}> c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
is impossible to be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

*Example:*

** 7^{2} + 5^{2} > 5^{2} **and

*5*^{2}– (7^{2}+ 5^{2}) = – 49** 7^{3} + 5^{3} > 5^{3} **and

*5*^{3}– (7^{3}+ 5^{3}) = – 343** 7^{4} + 5^{4} > 5^{4} **and

*5*^{4}– (7^{4}+ 5^{4}) = – 2401*. . . . . . . . . . . . . . . . . . . . . . . . . . . .*

* *

*Case 4:*

*а ^{2} + b*

^{2}

*> с*^{2}; a

^{2}

*= c*

^{2}

*; b*

^{2}

*< c*

^{2}*Theorem** 3.4*

**If (а, b, с) are integers for which is satisfied а^{2} + b^{2} > с^{2}; a^{2} = c^{2}; b^{2} < c^{2}, where n **

**grows for** *n > 2* is satisfied *а ^{n} + b^{n} > с^{n}* and might not be satisfied

*а*.

^{n}+ b^{n}= с^{n}*Proof:*

Of ** a^{2} = c^{2}; b^{2} < c^{2}** it follows:

**,**

*a*^{n-2}= c^{n-2}**, for**

*b*^{n-2}< c^{n-2}**.**

*n > 2*When we multiply both sides of the inequality ** а^{2 }+ b^{2 }> c^{2 }**with

**it results**

*c*^{n-2}*a ^{2}.c^{n-2} + b^{2}.c^{n-2} > c^{2}.c^{n-2}*

*(29)*

When we multiply both sides of the equality ** а^{n-2} = c^{n-2 }**with

**we obtain**

*a*^{2}*a ^{2}.a^{n-2} = a^{2}.c^{n-2}*

When we multiply both sides of the inequality ** b^{n-2} < c^{n-2 }**with

**and obtain**

*b*^{2}*b ^{2}.b^{n-2} < b^{2}.c^{n-2}*

Since ** a^{2}.a^{n-2} = a^{2}.c^{n-2 }**and

**, when we replace in**

*b*^{2}.b^{n-2}< b^{2}.c^{n-2}*(29)*,

**with**

*а*^{2}.c^{n-2}**and**

*a*^{2}.a^{n-2}**with**

*b*^{2}.c^{n-2}**we obtain**

*b*^{2}.b^{n-2 }*а ^{2}.a^{n-2} + b^{2}.b^{n-2} > с^{2}.c^{n-2 } *

*(30)*

Of *(30) *it derives:

- For
is satisfied*n > 2*.*a*^{n}+ b^{n}> c^{n} - Where
grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}> c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
is impossible to be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

*Example:*

** 5^{2} + 7^{2} > 5^{2} **and

*5*^{2}– (5^{2}+ 7^{2}) = – 49** 5^{3} + 7^{3} > 5^{3} **and

*5*^{3}– (5^{3}+ 7^{3}) = – 343** 5^{4} + 7^{4} > 5^{4} **and

*5*^{4}– (5^{4}+ 7^{4}) = – 2401*. . . . . . . . . . . . . . . . . . . . . . . . . . . .*

* *

*Case 5:*

*а ^{2} + b*

^{2}

*> с*^{2}; a

^{2}

*= c*

^{2}

*; b*

^{2}

*= c*

^{2}*Theorem** 3.5*

**If (а, b, с) are integers for which is satisfied а^{2} + b^{2} > с^{2}; a^{2} = c^{2}; b^{2} = c^{2}, where n **

**grows for** *n > 2* is satisfied *а ^{n} + b^{n} > с^{n}* and might not be satisfied

*а*.

^{n}+ b^{n}= с^{n}*Proof:*

Of ** a^{2} = c^{2}; b^{2} = c^{2}** it follows:

**, for**

*a*^{n-2}= b^{n-2}= c^{n-2}**.**

*n > 2*When we multiply both sides of the inequality ** а^{2 }+ b^{2 }> c^{2 }**with

**we receive**

*c*^{n-2}*a ^{2}.c^{n-2} + b^{2}.c^{n-2} > c^{2}.c^{n-2}*

*(31)*

When we multiply both sides of the equality ** а^{n-2} = c^{n-2 }**with

**we obtain**

*a*^{2}*a ^{2}.a^{n-2} = a^{2}.c^{n-2}*

When we multiply both sides of the equality ** b^{n-2} = c^{n-2 }**with

**and obtain**

*b*^{2}*b ^{2}.b^{n-2} = b^{2}.c^{n-2}*

When replacing in *(31)*, ** а^{2}.c^{n-2}** with

**and**

*c*^{2}.c^{n-2}**with**

*b*^{2}.c^{n-2}**we obtain strengthened inequality**

*c*^{2}.c^{n-2 }*c ^{2}.c^{n-2} + c^{2}.c^{n-2} > с^{2}.c^{n-2 } *

*(32)*

Of *(32) *it derives:

- For
is satisfied*n > 2*.*a*^{n}+ b^{n}> c^{n} - Where
grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}> c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
is impossible to be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

*Example:*

** 5^{2} + 5^{2} > 5^{2} **and

*5*^{2}– (5^{2}+ 5^{2}) = – 25** 5^{3} + 5^{3} > 5^{3} **and

*5*^{3}– (5^{3}+ 5^{3}) = – 125** 5^{4} + 5^{4} > 5^{4} **and

*5*^{4}– (5^{4}+ 5^{4}) = – 625*. . . . . . . . . . . . . . . . . . . . . . . . . . .*

Of the five cases reviewed

*a ^{2} > c^{2}; b^{2} > c^{2}*

*a ^{2} > c^{2}; b^{2} < c^{2}*

*a ^{2} > c^{2}; b^{2} = c^{2}*

*a ^{2} = c^{2}; b^{2} < c^{2}*

*a ^{2} = c^{2}; b^{2} = c^{2}*

given that ** а^{2} + b^{2} > с^{2}**, it follows:

- For
is satisfied the inequality*n < 2*.*a*^{n}+ b^{n}> c^{n} - For
might not be satisfied the equality*n > 2*.*а*^{n }+ b^{n }= c^{n} - Where
grows (for*n*), difference between*n > 2*and*c*^{n}*а*^{n}+ b^{n} - Between
and*n*there is mutually unambiguous correspondence that is expressed by the function*c*^{n}– (а^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

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

*Case 6:*

** а^{2} + b^{2} > с^{2} **and

*a*^{2}< c^{2}; b^{2}< c^{2}* *

*Theorem 3.6*

**If (а, b, с) are integers for which is satisfied а^{2} + b^{2} > с^{2}; a^{2} < c^{2}; b^{2} < c^{2}, where n **

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

**For***n**≤ k*is satisfied*a*.^{n}+ b^{n}> c^{n}**For***n > k*is satisfied*a*.^{n}+ b^{n}> c^{n}**For***n = 2*might not be satisfied*а*.^{n }+ b^{n }= c^{n}**The value of***n*(*n = k*) after which the inequality*а*reverses its direction and is dependent of the integers^{n}+ b^{n}> c^{n}*(а, b, с)*.

*Proof:*

According to *Lemma 2.1*

- Where
increases, for*n*difference between*n > 2*and*c*^{n}is increasing (*a*^{n}grows faster than*c*^{n}).*a*^{n} - Where
increases, for*n*difference between*n > 2*and*c*^{n}is increasing (*b*^{n}grows faster than*c*^{n}).*b*^{n}

That means:

Where ** n** grows (provided that

**and**

*а*^{2}+ b^{2}> с^{2}**), there is a value of**

*a*^{2}< c^{2}; b^{2}< c^{2}**(**

*n***), where inequality**

*n = k + 1***reverses direction, and acquires type**

*а*^{n}+ b^{n}> c^{n}**.**

*а*^{n}+ b^{n}< c^{n}*Example 1:*

** 7^{2} + 8^{2} > 9^{2} **and

*9*^{2 }– (7^{2}+ 8^{2}) = -32** 7^{3} + 8^{3} > 9^{3} **and

*9*^{3 }– (7^{3}+ 8^{3}) = -126** 7^{4} + 8^{4} < 9^{4} **and

*9*^{4}– (7^{4}+ 8^{4}) = 64** 7^{5} + 8^{5} < 9^{5} **and

*9*^{5}– (7^{5}+ 8^{5}) = 9474** 7^{6} + 8^{6} < 9^{6} **and

*9*^{6}– (7^{6}+ 8^{6}) = 151648*. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .*

From this example it follows:

*9*^{3 }– (7^{3}+ 8^{3}) < 9^{2 }– (7^{2}+ 8^{2}).*9*^{4}– (7^{4}+ 8^{4}) < 9^{5}– (7^{5}+ 8^{5}) < 9^{6}– (7^{6}+ 8^{6}) < . . .- Value of
, after which the inequality*n*reverses its direction, is*7*^{n}+ 8^{n}> 9^{n}.*k = 3* - For
is satisfied*n < 3*.*7*^{n}+ 8^{n}< 9^{n} - For
, where*n > 3*grows, inequality*n*strengthens as to each value of*7*^{n}+ 8^{n}< 9^{n}corresponds one single value of*n*.*9*^{n}– (7^{n}+ 8^{n}) - For
might not be satisfied the equality*n > 3*.*7*^{n}+ 8^{n}= 9^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*9*^{n}– (7^{n}+ 8^{n})*f(n) =*.*9*^{n}– (7^{n}+ 8^{n})

*Example** 2:*

** 19^{2} + 20^{2} > 21^{2}** и

*21*^{2}– (19^{2}+ 20^{2}) = -320** 19^{3} + 20^{3} > 21^{3}** и

*21*^{3}– (19^{3}+ 20^{3}) = -5598** 19^{4} + 20^{4} > 21^{4}** и

*21*^{4}– (19^{4}+ 20^{4}) = -95840** 19^{5} + 20^{5} > 21^{5}** и

*21*^{5}– (19^{5}+ 20^{5}) = -1591998** 19^{6} + 20^{6} > 21^{6}** и

*21*^{6}– (19^{6}+ 20^{6}) = -25279760** 19^{7} + 20^{7} > 21^{7}** и

*21*^{7}– (19^{7}+ 20^{7}) = -372783198** 19^{8} + 20^{8} > 21^{8}** и

*21*^{8}– (19^{8}+ 20^{8}) = -4760703680** 19^{9} + 20^{9} > 21^{9}** и

*21*^{9}– (19^{9}+ 20^{9}) = -40407651198** 19^{10} + 20^{10} < 21^{10}** и

*21*^{10 }– (19^{10}+ 20^{10}) = 308814720400** 19^{11} + 20^{11} < 21^{11}** и

*21*^{11}– (19^{11}+ 20^{11}) = 28987241644002** 19^{12} + 20^{12} < 21^{12}** и

*21*^{12}– (19^{12}+ 20^{12}) = 1046512592320480** 19^{13} + 20^{13} < 21^{13}** и

*21*^{13}– (19^{13}+ 20^{13}) = 30499394276862402*. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .*

From this example it follows:

*21*^{9}– (19^{9}+ 20^{9}) < 21^{8}– (19^{8}+ 20^{8 }) < 21^{7}– (19^{7}+ 20^{7}) < 21^{6}– (19^{6}+ 20^{6}) < 21^{5}– (19^{5}+ 20^{5}) < < 21^{4}– (19^{4}+ 20^{4}) < 21^{3}– (19^{3}+ 20^{3}) < 21^{2}– (19^{2}+ 20^{2})*21*^{10 }– (19^{10}+ 20^{10}) < 21^{11}– (19^{11}+ 20^{11}) < (19^{12}+ 20^{12}) < 21^{13}– (19^{13}+ 20^{13}) < . . .- Value of
, after which the inequality*n*reverses its direction, is*19*^{n}+ 20^{n}> 21^{n}.*k = 9* - For
is satisfied*n < 9*.*19*^{n}+ 20^{n}< 21^{n} - For
, where*n > 9*grows, inequality*n*strengthens as to each value of*19*^{n}+ 20^{n}< 21^{n}corresponds one single value of*n*.*21*^{n}– (19^{n}+ 20^{n}) - For
might not be satisfied the equality*n > 9*.*19*^{n}+ 20^{n}= 21^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*21*^{n}– (19^{n}+ 20^{n})*f(n) =*.*21*^{n}– (19^{n}+ 20 )

From both examples (and all other examples that we may review) it is seen that where ** n** increases, given that

**and**

*а*^{2}+ b^{2}> с^{2}**:**

*a*^{2}< c^{2}; b^{2}< c^{2}- The inequality
(for*a*^{n}+ b^{n}> c^{n}) leads to the inequality*n < k*(for*a*^{n}+ b^{n}< c^{n}), without passing through inequality*n > k*(for*а*^{n}+ b^{n}= c^{n}).*n = k* - For
is satisfied*n > k*.*a*^{n}+ b^{n}< c^{n} - For
*n*is satisfied*≤ k*.*a*^{n}+ b^{n}> c^{n} - The value of
, (*n*) after which the inequality*n = k*reverses its direction and is dependent of the selection of integers*а*^{n}+ b^{n}> c^{n}.*(а, b, с)* - Between
and*n*there is mutually unambiguous correspondence that is expressed by the function*c*^{n}– (а^{n}+ b^{n}).*f(n) = c*^{n}– (a^{n}+ b^{n})

We explored the inequality ** а^{n} + b^{n} < c^{n}** in

**.**

*Part 1*From *Theorem 1.1* and obtained above it follows:

- For
is satisfied*n > k*.*a*^{n}+ b^{n}< c^{n} - For
, where*n > k*grows, inequality*n*strengthens as to each value of*a*^{n}+ b^{n}< c^{n}corresponds one single value of*n*.*c*^{n}– (a^{n}+ b^{n}) - For
might not be satisfied the equality*n > k*.*а*^{n }+ b^{n }= c^{n} - Mutually unambiguous correspondence between
and*n*is expressed by the function*c*^{n}– (a^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

Only the inequality ** a^{n} + b^{n} > c^{n}**, for

*n***remained to be explored.**

*≤ k*Of ** а^{2} + b^{2} > с^{2}**, it follows:

*c ^{2} – (а^{2} + b^{2}) < 0 *

*(33)*

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

*c ^{2}.c – (а^{2}.c + b^{2}.c) < 0*

*(34)*

It is satisfied:

** c^{2}.c – (а^{2}.c + b^{2}.c) < c^{2} – (а^{2} + b^{2}) < 0**.

Of ** a^{2} < c^{2}; b^{2} < c^{2}** we obtain

**.**

*a < c; b < c*Because ** a < c; b < c** it follows

**.**

*a*^{2}**.**

*a < a*^{2}.c; b^{2}.b < b^{2}.cReplacing in *(34) a^{2}.c* with

**;**

*a*^{2}.a**with**

*b*^{2}.c**, we obtain**

*b*^{2}.b*c ^{2}.c – (а^{2}.a + b^{2}.b) < 0*

*(35)*

since ** а^{2}.a + b^{2}.b < а^{2}.c + b^{2}.c** it follows

*c*^{2}*.c** – (а ^{2}.a*

*+ b*

^{2}

*.b*

*) < c*

^{2}

*.c*

*– (а*^{2}.c

*+ b*

^{2}

*.c*

*) < c*

^{2}

*– (а*^{2}+ b

^{2}

*) < 0**c ^{3} – (а^{3} + b^{3}) < c^{2} – (а^{2} + b^{2})*

In the same way we obtain

*c ^{k}*

*– (а*^{k}

*+ b*^{k}

*) < . . . < c*

^{4}

*– (а*^{4}+ b

^{4}

*) < c*

^{3}

*– (а*^{3}+ b

^{3}

*) < c*

^{2}

*– (а*^{2}+ b

^{2}

*)*

*> 0**(36)*

Where ** n** increases in the closed interval

**, then the difference between**

*[2, k]***and**

*c*^{n}**is increasing and an equality might not be obtained. For**

*а*^{n}+ b^{n}**the inequality reverses its direction. We have studied this case and there when**

*n > k***increases by the value of**

*n***to infinity, the difference between**

*n = k + 1***and**

*c*^{n}**increases and an equality might not be obtained.**

*а*^{n}+ b^{n}As of each value of ** n** for

**and**

*n [2, k]***corresponds only one value of**

*n > k***, it follows that between**

*c*^{n}– (a^{n}+ b^{n})**and**

*n***there is mutual uniquely line, which is expressed by the function**

*c*^{n}– (a^{n}+ b^{n})**.**

*f(n) = c*^{n}– (a^{n}+ b^{n})From insofar obtained for** n > 2** and

**, where**

*n < k + 1***grows, in**

*n***and**

*а*^{2}+ b^{2}> с^{2}**, it follows:**

*a*^{2}< c^{2}; b^{2}< c^{2}- The inequality
(for*a*^{n}+ b^{n}> c^{n}) leads to the inequality*n < k*(for*a*^{n}+ b^{n}< c^{n}), without passing through inequality*n > k*(for*а*^{n}+ b^{n}= c^{n}).*n = k* - For
is satisfied*n > k*.*a*^{n}+ b^{n}< c^{n} - For
*n*is satisfied*≤ k*.*a*^{n}+ b^{n}> c^{n} - The value of
, (*n*) where the inequality*n = k*reverses its direction is dependent of the selection of integers*а*^{n}+ b^{n}> c^{n}. (there is nothing else in the expression*(а, b, с)*, from which*c*^{n}– (a^{n}+ b^{n})may be dependent).*k* - Between
and*n*there is mutually unambiguous correspondence that is expressed by the function*c*^{n}– (а^{n}+ b^{n}).*f(n) = c*^{n}– (a^{n}+ b^{n})

We explored the integers ** (а, b, с)**, where for their second degrees are satisfied relationships

*a*^{2}+ b^{2}< c^{2},**and we proved that in all cases:**

*a*^{2}+ b^{2}= c^{2}, a^{2}+ b^{2}= c^{2}- For
, equality*n > 2*might not be satisfied in integers*а*^{n}+ b^{n}= с^{n}

** (a, b, c)**.

- Between
and*n*there is mutually unambiguous correspondence that is expressed by the function*c*^{n}– (а^{n}+ b^{n})*f(n) =*.*c*^{n}– (a^{n}+ b^{n})

That means:

**Equality***а*^{n }+ b^{n }= c^{n}is possible in integers**(a, b, c)****for****n = 2.****Equality***а*^{n }+ b^{n }= c^{n}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) = c^{n} – (a^{n} + b^{n}).**

Existence of the function *f(n) =*** c^{n} – (a^{n} + b^{n} ) **and of the mutually unambiguous correspondence between

**and**

*n***is evident and does not need any proof.**

*f(n)*To prove the existence of function *f(n) =*** c^{n} – (a^{n} + b^{n} )** is tantamount to prove the existence of function

**.**

*f(x) = a.x*^{2}+b.x + cNevertheless, the proofs we have made in ** Part 1**,

**and**

*Part 2***were necessary to recognize the existence of the function**

*Part 3*

*f(n) =***and its amazing properties.**

*c*^{n}– (a^{n}+ b^{n})Here the variable is ** n**, and not the integers

**.**

*(a, b, c)*Integers ** (a, b, c) **are randomly selected. Where

**changes, function**

*n***changes, and not integers**

*f(n)***(the argument is**

*(a, b, c)***). 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**

*n***and to each value of the function**

*f(n)***corresponds one single value of the argument**

*f(n)***.**

*n*It is impossible and improper the function *f(n) =*** c^{n} – (a^{n} + b^{n})** to be regarded as an argument and the argument

**to be regarded as a feature.**

*n*Given the result the so far it follows

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

*Fermat’s Last Theorem*

**Equality а^{n }+ b^{n }= c^{n} is not possible in integers**

**(a, b, c)****for****n > 2.***Proof:*

That for ** n = 2** is satisfied

**it follows that for**

*f(2) = 0***might not be satisfied**

*n > 2*** f(n) = 0**, as the mutually unambiguous correspondence between

**and**

*n***is disturbed.**

*f(n)*It is not possible for two different values of ** n** (

**and**

*n = 2***) to collate same value of**

*n = k > 2***(**

*f(n)***и**

*f(2) = 0***).**

*f(k) = 0*A value of ** n** may be compared with different values of

**, only when changing the integers**

*f(n)***.**

*(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 equationmay be represented as a function that makes the proof of the theorem his*а*^{n}+ b^{n}= c^{n}.*“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*.

* *