On the history of attempts for proving Fermatís last theorem and the fundamental role of prime numbers
(regular) and its properties leading to solving Diophantine equation




19.1. Diophantine equations

Generally, solving the Diophantine equations is one of the most difficult problems in mathematics and many special cases of them that have been solved, there is no general and decisive information except that limited cases. There is no Challenger in this area and if every one is interested in this subject, is welcome here. One of the most important kinds of Diophantine equations that have caused the motivation of human to a lot of new results is concerned to Fermatís last theorem:


It is obvious that instead of investigating the equation (1) for every natural "n", it is enough to check equation (1) only for every prime number of "n", namely, because in case:


( p: prime number)   

Here we express historical process of decisive proof of Fermatís last theorem.

19.2. An introduction to the chronology of Fermatís last theorem

When Pierre Fermat propounded his famous theorem in margin of Diophantusís   book in 350 years ago that equation  hasnít answer for integer number  and  and  and claimed that he has an easy method for proving it (and since there was not enough space there, he ignored describing that). Many of greatest mathematicians have worked on it and if we ignore amateur persons who always have bothered the others, only a little improvement obtained about it. One of the basic advantages of Fermatís last theorem was that the mathematicians solved very interior problems of mathematics specially in number theory, and new branches was made in this way that the most important of them are Algebraic numbers and geometry arithmetic. But, Fermatís last theorem was not solved until in 1995 Andrew Wales presented a difficult solution for it in more than 200 pages.

Many of mathematicians became sorry of solving Fermatís last theorem, because, Fermatís last theorem was a motive for discovering a lot of properties of numbers. But we must wait for an easier solution for Fermatís last theorem and from this aspect; mathematician's work is not finished.

We want to clarify the method in which the solution of Fermatís last theorem is propounded by Wiles as it possible and in next part; we mention other possible solutions that are resulted from theorems of prime numbers like Wilsonís theorem, Fermatís small theorem and theorem of "" function of Euler.

In fact, we want to show application of theorems about prime numbers in solving Diophantine equations and also we want to present some answers for solving general Diophantine equation:


In fact, we are going to verify and solve equation (2) that is an extension of equation (1) and it is really an extension of Fermatís last theorem.

19.3. Chronology of Fermatís last theorem 

Pierre De Fermat (1608-1665), one of greatest mathematicians lived in 17th century. He made bases of analytic geometry (almost contemporary with Descarte) and found a general method for searching maximum and minimum (that is completed to calculation of infinite similes). After all, the most famous results of Fermat are in the field of number theory.

Theorical-calculational results of Fermat were not printed. They were found in the letters and his scattered papers after his death.

Most of Fermatís reasons for his results have not reached to us. These reasons and proofs have been arranged by mathematicians after him specially by Euler. Some of Fermatís propositions clarifies that even himself had not a convincing reasoning for them. More over, it is clarified that some of his prepositions have error.

For example, Fermatís guessed that every number in "" form is a prime number for every integer non-negative number "n", although Euler proved that this number is composite for "".

But in all of cases that Fermat claimed proof of a proposition, mathematicians could prove validity of proposition in some cases.

The most interesting and unique Fermatís proposition is "Fermat's great theorem" or "Fermatís last theorem". This theorem is:

When "n" be an integer number greater than "2", below equation:

can not have an answer for "x", "y" and "z" except "0". More over, we know that such numbers exist for. For example, 3, 4 and 5.In Fermatís papers, proof of this theorem found for  and it is interesting that this is the only perfect proof that has remained from Fermat.

But for general form and for "", Fermat has written in book margin of "Arithmetic" of Diophantus:

"In fact, "he" has found an interesting proof for it but the book margin is too small that it can be written in it".

Despite a lot of attempts of mathematicians, that as according to Dickson said, In his book "History of number theory" has engaged mathematicians for more than "350" years, this proof has not been obtained so that some people became doubtful about existing of such proof.

As we will see, except, for non values of "n", the validity of Fermatís last theorem has not been proved by primary methods. Because of this, many of mathematicians attempted to prove that Fermatís last theorem can not been proved by primary methods.

In 1908, Wolfskehl, a German who was interested in mathematics, put 100,000 marks prize for person that can solve Fermatís last theorem. Immediately, hundreds and thousands persons who wanted the reward wrote a lot of letters to scientific organizations and publications that they could prove Fermatís last theorem. Just in Guttingen of Germany, more than one thousand solutions were sent to mathematics society in Guttingen, "3" years after propounding Wolfskehlís prize.

This illness had spread in all over Europe and even some people who had a little information about mathematics took part in this race. Every body presented their solutions that often were trivial and even funny that they had written them for the reward.

After the First World War that all over Europe encountered financial crisis, Wolfskehlís prize lost its value and "FERMATISTS" lost their motivation of profit (Mathematicians had given this name to some people who attempted to solve the Fermatís theorem with weak scientific possibility and useless method).

Because of this and naturally, the flood of "Fermatistís proofs" subsided but not absolutely. Flow of letters to mathematics and scientific centers was still continued. The writers of these letters hoped to become famous at least, although even a little truth and correct reason had not been found in their letters. Some writers of these letters propounded that they didnít attended to their personal honor but they wanted to do a useful work for science and this is an example of depravity and trifling in history of science.

Fermatís last theorem was valuable for mathematicians specially because during the attempts to prove that, some new methods raised in mathematics and specially it helped and improved a branch of mathematics named "Algebraic number theory". This truth that Fermatís theorem was unsolved means that, existence of new and exact methods is necessary in mathematics. It seems that the primary solution can not be found for Fermatís theorem (it means that Fermatís theorem can not be proved with known methods), But attempts to find the new methods for proving the Fermatís theorem was valuable and useful for mathematics.

Apparently it is necessary to abandon searching for the primary solution for Fermatís last theorem. Any way ether it is true or false, entering to primary solution of Fermatís last theorem isnít advisable specially for young and inexperienced mathematicians. One of the purposes of this chapter of the book is showing this point that Fermatís theorem depends on how difficult and deep problems of number theory and how it has disappointed every body who had thought that he solved this theorem and put him among Fermatists (We ignore that in some cases, they didnít find even deficit or later result).

It is better to remind that attempt to find a counterexample blindly is also disappointing.

In 1856, Harier Renrett reminded that if natural numbers,  and  exist that adapt in below equality:

They must adapt with below inequality[1]:


In fact, if we suppose that, then:

That it results:

With the same method, we can prove. Therefore:

It means and. Because of existing symmetry, it also results:  and

We know that Fermatís theorem is proved for every "n" smaller than 100,000 (1978). Therefore, for its examples, we must use numbers more than 10 500000.

As we said, primary proof of Fermatís last theorem doesnít exist for none of numbers. Even in case "" that Euler verified it in 1768, it is clear that we must use numbers in below form:


("a" and "b" are integer numbers)

Fermat was unfamiliar with such numbers and it is clear that he couldnít use them.

In fact, Eulerís proof has some defects, because he used numbers in form (1) without discussion about complex numbers and expressing their bases, and he considered the properties of integer number for them, for example, he has used the simplest rules of divisibly theory for integer number without proof for the numbers of (1).

The first person who arranged the calculus of numbers of (1) and made a certain base for Eulerís reasoning was Gauess apparently. In 1825, Dirichlet and Legendre proved Fermatís theorem for "" almost contemporarily. Dirichletís proof was printed in 1828 that is very complicated. In 1921 Pelemelle simplified the proof to some extent.

For the next prime number, namely, Fermatís theorem had been proved by Lame in 1839. Immediately after Lame, his proof became simple and complete by Lebesgue.

In 1847, Lame declared that he could prove Fermatís theorem for all of prime numbers . Lameís method was different from Eulerís cognition and it was based on arithmetic properties of numbers in below form:


That are integer numbers and "" is n-th root of number "1":

But Liouville found a serious deficit in Lameís reasoning.

Deficit of this reasoning was that without reasoning Lame had supposed that numbers in form (2) are decomposed to multiple of prime factors uniquely like normal integer numbers. (That these factors arenít decomposable to multiple of other factors). Lame had forced to accept his error.

As these events occurred in France, in Germany, a young mathematician, named Kummer was searching seriously about Fermatís theorem. At first he thought that he could find the perfect proof of Fermatís theorem and in 1843, send his letter to Dirichlet. This proof was based on such numbers in form (2) and Kummer repeated Lameís error and supposed that these numbers are decomposed to prime factor only by one method.  Dirichlet remembered immediately that this truth needs to proof and gave back Kummerís letter to him.

Kummer found out soon that theorem about unique factorization to prime factors isnít correct for numbers in form (2) and stopped his attempts to prove it. Kummer didnít stop his searching and found an exit way that not only famed his name but also it led to a new branch in modern algebra. Kummer odd new numbers "abstract numbers" to numbers (2) and called them as "ideals" so that the property of being unique in decomposing to prime factors was correct.

We express an example. It is clear simply that in range of numbers in below form:


That "a" and "b" are integer numbers, number "21" is decomposed to multiple of prime factors by two methods:

Although numbers in form (3) arenít numbers in form (2), for every value of "n" (for numbers (2), the same example is possible only for).

But we can use Kummerís method about it. For this purpose, it is necessary to add "ideal numbers" A,B,C and D to numbers (3) and suppose:

It is clear that unique factorization is obtained for number "21" to multiple of prime factors (ideals).

Of course being "ideal" of new numbers has difficulties, but we can over come these difficulties. In 1847, Kummer could prove Fermat's theorem for all prime numbers  that adapt with conditions named (A) and (B). As it is clear from Kummerís letter to Liouville, Kummer thought that this condition is established for all prime numbers, but he found out soon that some exceptions exist (like number 37). Hilbert simplified Kummerís proof in 1894.

But this conclusion of Kummer wasnít favor, because he didnít find even one prime number "" yet so that Fermatís theorem is correct for it. Nevertheless, it was a basic point to improvement and we express it in more details.

In 1851, with hard and fine analyze, Kummer could complete seriously the results which he had obtained in 1847.He proved that condition (B) can be resulted from condition (A). (This proposition is famous as "Kummerís   pre-theorem").

And therefore, he simplified condition (A) as it was possible and the changed it that it can be used easily. This condition in its first arranging was with this request that prime number "" isnít a divisor of "h" that was defined hardly. Kummer decomposed number "h" to multiple of two factors () and he obtained clear formulas for  and  ( that of course was diffidently ); Then he proved that number "h" is divisible by "" if and only if prime number "" doesnít count  form of the first number of Bernoulli , it means that it is not a divisor of them .(Of course with condition that these fraction arenít reducible) . Kummer called such numbers as "regular"[3]. Bernoulliís numbers are these rational numbers:

that they can be obtained respectively, according to a simple rule. Then, Kummer condition can be test for every "" without any special difficulty.

Specially it is cleared that only numbers: 37, 57 and 67 arenít regular between prime numbers.

This improvement was very interesting that Kummer obtained it in completing his researches in 1847 but unfortunately proving these results is so hard that we canít mention it here.

It is possible to show that why Bernoulliís numbers exist with condition of being regular.

Kummer thought in all of his life that the number of regular numbers is infinite and a lot of mathematicians were agreeing with him in this field. But this truth has not been proved yet. Besides it, in 1915, Jensen proved with a simple method that conversely, the number of "irregular" prime numbers is infinite.

After 1851, Kummer attempted to search prime numbers that arenít regular and tried to prove Fermatís theorem for "irregular" prime numbers like for numbers 37, 59 and 67 hardly.

Therefore, validity of Fermat's theorem became proved for all of prime numbers. Although in years of first quarter of 20-th century Merteness and Vandiver found some lacks in Kummerís reasoning but all of them can be reform easily. In 1893, Mirimanoff could prove Fermatís theorem with easier method for  power.

In 1850, France science academy put "3" thousand Frank prize for proving Fermat's theorem. Giving the reward had been retarded until in 1857, it was given to Kummer (apparently, at first, Kummerís name wasnít even among the candidates).

After Kummer, no serious work had been done for proving Fermatís theorem till in 1929, Vandiver proved that Fermat's theorem is correct when prime powers of "":

I. Second factor "" from "h" isnít divisible by "".

II.  forms of Bernoulliís term:

arenít divisible by "" .

Testing the condition (2) isnít difficult with modern computers. But for condition (1), although it has been tested about all of numbers, no prime numbers has been found yet for "" that it doesnít adapt with this condition.

The condition (2) is also established for numbers. Therefore, Fermatís theorem is correct for all prime numbers .

Euler knew that for verification below equation:


("" is a prime number greater than "3")

It is necessary to distinguish the case that none of numbers "", "y" or "z" arenít divisible by "" from the one that at least one of these numbers is divisible by "".

If we have permission to a little carelessness, we can name this proposition that equation (4) can not be established for nonĖdivisible numbers by "" as first case of Fermatís theorem and call this proposition that equation (4) can not be established for numbers that one of them is divisible by "", as the second case of  Fermatís theorem.

Besides the general case of Fermatís theorem, primary proof can be found about its first case for most of numbers "". Sophie Germain (1776-1831), the first woman mathematician of today worked on this problem at early 19-th century. She proved that the first case of Fermatís theorem is correct for prime numbers "" if number "" be also a prime number.

In spite of these, Germainís wish became hope less and the method which she selected to prove wasnít useful even for the first case of Fermatís theorem.

Germain didnít print her results but wrote them in a letter to Lagendre, the French mathematician. In 1828 Lagendre published a detailed thesis about Fermatís theorem and mentioned the Germain theorem in it and resulted from them a sequence of results. For example, he proved that the first case of Fermatís theorem is correct for prime powers of "" when at least one of these five numbers is prime:


Besides it, the first case of Fermatís theorem is correct for all of prime numbers smaller than 197.Lagendreís theorem doesnít respond for power 197.

After Lagendre, many mathematicians attempted to improve his results. Apparently, Venet, German mathematician obtained certain theorem in 1893. (This theorem can not be improved by primary method).

Venet considered an integer number with symbol "" for every  and proved with general method of Germain that the first case of Fermatís theorem is correct for every prime power of "" if  exists then:

1. Number  be a prime number and not divisible by ""

2. Number  isnít divisible by "p".

Number "" has three equivalent definitions:

I. with below assumption:

II. [4] is "resultant" of polynomial ""and "":


We remind that despite attempts of mathematicians that very clever and punctilious scientists were among them, no another primary method has been presented for proving Fermatís theorem or at least for its first case.

(In spite of this, still Germainís results aren't clear up to end. For example, still we donít know that whether the number of prime powers of "" that these results use for them, is infinite or not).

Non-primary method for the first case of Fermatís theorem was propounded by Kummer. He proved in 1858 that the first case of Fermatís theorem for prime powers of "" is correct when at least one of these two Bernoulliís number ( and ) isnít divisible by "".

In 1905, Manov extended this result and proved that it is enough that one of the forms of four Bernoulliís numbers isnít divisible by "" and this includes all values of .

In 1909, Wieferich proved with use of Mirrimanovís method that the first case of Fermatís theorem is correct for all of prime powers of "" if  isnít divisible by "". This result is very interesting, because with help of it, for example we can judge about prime numbers smaller than or equal to "200183" except two numbers "1093" and "3511".

Laters, Mirrimanov and Frobenius simplified Wieferichís proof thatĖWieferichís condition can be changed from base "2" to "3", so that the first case of Fermatís theorem is correct for every prime power of "" that at least one of these two numbers or isnít divisible by "".

In 1912, Furtwangler proved Wieferich and Mirrimanov- Frobeniusís criterion briefly and in some lines with a method famous as "bilateral rule" of Eisenstein.

This work was a start for a series of examinations for persons who obtained situation in the new field in number theory ( like about groups family) and they proved in 1941 that base "2" can be changed  in Wieferichís criterion with every prime number not greater than "43". This result could prove validity of first case of Fermatís theorem for every powers of "" smaller than "253747889".

In 1934, Wandiver proved that the first case of Fermatís theorem is correct for a prime power of "" that its second factor "", isnít divisible by "". This theorem is interesting because as we said before, no prime power has been known till now that doesnít adapt with this condition. But since "" isnít divisible by "", researches are done only for  with this method.

* Here this question propound that how we can search about proof of Fermatís last theorem?

At first, we remind, if three numbers (x, y and z) be integer numbers and adapt in below equation:


(we donít put aside the case "" now), then every three numbers in () form adapt in this equation for arbitrary and integer number "". Conversely, if this triplet numbers () be an answer of equation (1), then is also answer of equation. Therefore, for finding all answers of equation (1) (except trifle case ), it is enough to calculate answers of  so that  "x", "y", "z" be coprime two-to-two (it means that they donít have divisor except unit) and for proving that equation (1) hasnít answer in set of integer numbers, it is enough to use counter demonstration and we suppose that answer of that includes numbers which are coprime to each other two-to-two , exists.

Moreover, it is clear that if in answer of  from equation (1) two numbers from "x", "y" and "z" have a common factor like , without any doubt, the third number is also  divisible by "". Therefore, in every equation in form (1), we can consider only one answer in form which "x", "y" and "z" are coprime            two-to-two. We call such answer as "primitive answer". Then it is clear, if Fermatís theorem is correct for a power like "n", it is also correct for every power in "an" form namely multiple of "n", because if below equation:


Has answer  in set of integer numbers. Then, answer  is obtained for equation (1). Therefore, it is enough to prove Fermatís theorem for  and (as we said before, Fermat himself proved this case "") that "" is a prime number greater than "2".

Here, we mention pretheorem that it has important effect in all reasoning about Fermatís theorem.


20.2.1. Pretheorem

We suppose "a", "b" and "c" are natural numbers (namely integer and positive) so that:

1. Below equality be established:

2. Numbers "a" and "b" be coprime. Therefore, natural numbers "x" and "y" exist so that.

It means that if multiplication of two natural numbers that are coprime is equal to n-th power of a natural number, then every one of two factors of multiple is equal to n-th power of a natural number.

If "n" be an odd number. It is clear that this pretheorem is correct even for integer and non-zero (positive or negative) numbers "a", "b" and "c".

Proof. We express pretheorem completely. Suppose:

is factorization of numbers "a" and "b" to multiple of prime factors. Here , are different prime numbers, as the same method and are different prime numbers. More over, since "a" and "b" are coprime none of numbers of is equal with none of numbers of .

Therefore, below formula is decomposition of  to multiple of powers of different prime numbers:

In other side we know (this is basic arithmetic theorem) that factorization of every natural number to multiplication of powers of different prime numbers is a unique factorization (if we donít take in account the changing of factorís row as two different factorization). It means that only with one method, we can obtain a natural number as multiple of powers of different prime numbers.

Therefore this factorization must be adapted with factorization of "c" when it has power "n" and it proves that all of powers are divisible by "n". Then "a" and "b" are equal with n-th power of a natural number.

We express this proof to emphasis on importance of basic arithmetic theorem (Is known for reader certainly).

For every primitive answer  of below equation:

("" is a prime number and greater than  "2")                   (3)

 is equal with multiplication of two integer numbers:



That  is equal to:

Is coefficient of binomial extension (that sometimes is shown by symbol).

It is resulted from equality (4) that every common prime divisor "p" of "a" and "b" must be a divisor of below number:

And therefore if "", it is a divisor of . But when "p" be a divisor of "a" and , then it is a divisor of "" that it is impossible because according to the condition,  and  are coprime.

Now, if  isnít divisible by , then  isnít also divisible by , it means  isnít a divisor of "a" or "b". Therefore "a" and "b" are coprime and then according to pretheorem (for  and ), there are integer numbers  and , so that for them:


Now, attend that equation (3) can be written in below form:

Since in this equality numbers () have symmetric role, we can write the same equalities for three numbers  and for the three numbers.

Therefore, for integer numbers,andthat are coprime and not divisible by "", more over they adapt in below equation("" is an add prime number):


Pairs of integer numbers and  exist so that every two numbers of a pair are coprime and we have:


These formulas are famous as "Abel formulas", although Germain knew them and Lagendre printed them for the first time.

We can obtain the same formulas (of course some complicated) for the case that one of numbers,and  is divisible by "". But after 50 years researches about these formulas, no real formula has been found and so, we ignore to mention them here.

Guass propounded verification of theoretical Ė numeral problems about divisibility of numbers and easy and simple symbolization in this field.

We consider "n" as an arbitrary natural number. According to what Gauss said, "a" and "b" are congruent with module "n" when their subtraction "" is divisible by "n". So:

It is clear that congruence relation is an equivalent relation and therefore,  set of all integer numbers can be divided to groups of congruent numbers. We show the set of these groups with symbol.

Congruencies can be added or multiplied to each other like equalities. Also we can simplify the congruence to its common factor if this factor and "n" be coprime.

We know that below congruence is established for every integer number "a" that is not divisible by"":


This proposition is called Fermatís small theorem.


If we multiply congruence (8) in "a":


It is clear that this congruence is also established for   . Therefore, congruence (9) is established for every integer number "a" and Euler arranged Fermatís small theorem as this method.

In Algebraic language congruence (9) means that every member of  field is a binomial root of "".

Congruence (9) can be proved without using theory of groups and with the same method.

Since prime number "" aliquots number "" (namely "" is divisible by "") and since  doesnít aliquot , so all coefficients of binomials:

are divisible by "". Therefore, for every two integer numbers "x1≠ "and"":

The same result exists for every polynomial by using mathematics induction method:


If in this formula , formula (9) is obtained (for ) .

Now, we can propound Germainís researches directly. We suppose   as a primitive answer from equation (3) includes numbers not divisible by "".  We choose arbitrary prime number "p" that is congruent in relative to module "" with unit, namely in below form:

That "m" is an integer number. We suppose none of,  and  are divisible by "p". Since, therefore an integer number like "" exists so:             [5]

If we multiply equality (3) in and change it as congruence, it is obtained:


That  and aren't divisible by "p".

We call integer number "as degree of "" in relative to module "p" if a number like "" not be congruence with "0" to module "p" exists so:

More over, we call two numbers "" and ""from "" degree as neighbor in relative to module "p", when:

 The congruence that we proved above, according to this naming means that if no one of ,  and  are divisible by "p", then neighborhoods from "" degree will exists in relative to module "p".

Now, we suppose that one of numbers,  and  are divisible by "p"      (and according to primality of these numbers only one of them). For clarifying the situation, we suppose  to be divisible by "p". Then one (and only one) of factors in Abel formula  (see relation 7) namely  and   that are coprime is divisible by "p".

Suppose  is divisible by "p". So  isnít divisible by "p" then:

In other word, it is resulted from formulas (7) of Able:



 Therefore, if "" not be congruence with "0" to module "p", then again same neighborhoods of "" degree exist in relative to module "p". At last, suppose  is divisible by "p". So in relation (4) (that), all terms of right side


except the last term , are divisible by "p" and therefore, we have:

but according to Abel's formula:




That "" is a number that for it:

(It is clear that  or  isnít divisible by "p").

Therefore, it is proved that number "l" is from degree "l" for  in relative to module "p".

Validity of below theorem also is proved.

19.4. Sophie Germainís theorem

Suppose that an integer number like "m" exists for prime number  so that

1. is a prime number

2. There isnít any neighborhood between numbers from degree "" in relative to module "p".

3. Number "" is not from degree "" in relative to module "p". Then, for power "", the first case of Fermatís theorem is correct.  For testing the specific cases of conditions (2) and (3) of this theorem, it should be noted that below congruence is established for every "" with degree "" in relative to module "":    


In fact, if that "" is not congruence with "0" to module "p", then according to Fermatís small theorem:

19.4.1. Problem

Prove the inverse of this theorem. Every answer of "" from congruence (11) is from degree -th in relative to module "p".

It seems easily that for every prime number "p", only two unequal numbers exist for "" that adapt in below congruence:



"1" and ""

It is clear that "1" and "" adapt in this congruence, but this point that this congruence hasnít another root can be proved based on an Algebraic theorem. A polynomial of degree "n" can not have more than "n" root in one field. Also we can use from this point that below number is divisible by prime number "p" only when that "" or "" is divisible by "p":

 Because numbers "1" and "" arenít neighbor of -th degree in relative to module , clearly, it is resulted that condition (2) from Germainís theorem is established for "" certainly.

Since can not be divisible by prime number, therefore condition (3) is established for"".

Then, if power "" (that is an add prime number) has this property that number "" is prime also then the first case of Fermatís theorem is correct for "".

We know Germain obtained this result by her self (A lot of writers called it as Germainís theorem).

Therefore, Germainís general theorem resulted in Lagendreís theorem. Remind that Lagendreís theorem says that for prime power "" if "" is a prime number that "m" is one of five numbers "2,4,5,7 or 8" then the first case of Fermatís theorem is correct.

Here, we prove it only for "" because with growing "m", induction becomes difficult.

Suppose that condition (1) from Germainís theorem is established for "". Namely number "" is a prime number.

We test that condition (2) from this theorem is also established.

According to what we said before, we prove that there isnít any neighborhood between roots of below congruence:


Since, then roots of congruence (12) are roots of two below congruencies:


We know that the first congruence has two roots, "1" and "".  The second congruence has two cases. It hasnít root or has two roots exactly:


In the first case, congruence (12) has two roots "1" and "" that they arenít neighbor clearly. Then condition (2) from Germainís theorem is established. In second case, congruence (12) has "4" roots:

Two roots of them can be neighbor only for:


If , then  is divisible by  that it is impossible for . And if , then:

Is divisible by, it means  is divisible by  that again it isnít possible for . Therefore, condition (2) from Germainís theorem is also established in the second case.

Now, we test the condition (3). If this condition is not established, then  namely  that it is resulted:

It means is divisible by "p".

Because we know "", it is possible only for "" but equation "" has answer "" that isnít a prime number and then this case is also impossible. It means that the condition (3) of Germainís theorem is also established.

19.4.2. Remark

Ventís theorem led to Germainís theorem. We must only prove that if prime number in "" form doesnít aliquot Ventís number namely"". Then condition (3) from Germainís theorem is established.

19.5. Fermatís last theorem for exponent "4"

Only the case "" from Fermatís last theorem is proved by a primitive method. As we said before, Fermat found this proof by himself. He used formulas of general answer of below equation:


Indian mathematicians also knew these formulas. Here, we start from proving these formulas. We know that for equation (1) it is enough to find its primitive answers. It is clear that if  is an answer of equation (1) and then  is also its answer. In other side, for every answer of (x, y, z), at least one of  or  is even. In fact, if both of  and  are odd, then "" is in "" form that it canít be equal to second power of an integer number (because all of second power, "", are in "" or "" form). Also it is clear that besides answer, () is also an answer of equation.

According to this points it is resulted that it is enough to find a primitive answer from equation (1) so that all of three numbers,,  and  are positive and  be even.

Pretheorem. For every two positive and integer numbers "m" and that are coprime, with this thesis that one of "m" and "n" is even and another one odd, below formulas include three positive integer numbers that is a primitive answer of equation with condition of being even of :


Conversely, every three positive integer numbersare a primitive answer of equation (1) with thesis of being even of number, it can be expressed as form of formulas (2) that "m" and "" are coprime and one of them is even and another one is odd.


Proof. Below identity:

Show that numbers (2) (that are positive numbers clearly) are an answer of equation (1) that more over,  is an even number. If these numbers have a common divisor like , then below numbers must be divisible by :

 It means "" because according to the thesis "m" and "n" are coprime, but if "", then number ""is even. So,  and  are even or both of them are odd that it is impossible, because according to the thesis, one of "m" and "n" must be even and another one must be add, And this proves that answer (2) is a primitive answer.

Conversely, we suppose that  is a primitive answer includes positive numbers and more over, even number"". Since  and  are odd, therefore, "" and "" are even. Suppose:


That  and are positive clearly. Every common divisor of  and  will be also a divisor of "" and ". Therefore "", it means that  and  are coprime. In other side:


Therefore, according to the pretheorem, positive numbers "m" and "n" (that clearly are coprime and one of them is odd and another one is even) exist so that:

Then "" it means "" and so:

    ,           ,      

For completing the proof, it is enough to attend that"".Proving of the pretheorem finished.

Now, we can prove Fermatís last theorem for"". We prove validity of the general proposition.



19.5.1. Theorem.

Below equation hasnít non-zero answer in set of integer numbers.


Proof. We use contradiction for proving so we suppose that for equation (3), a non-zero answer exist in set of integer numbers. It is clear that without any disorder to generality, we can consider this answer as it contains the positive numbers that are coprime two-to-two. Since in every set of positive integer numbers, there is the smallest number, therefore, among these answers for equation (3), answer  exists so that  is the least possible value. We attend to this answer more.

As we proved for answer of equation (1), one of   and  must be even, here, we suppose that  is an even number. It is clear that this assumption makes no disorder in the generality. Because:


Since numbers  and  and  are positive and coprime two-to-two, and  is an even number, therefore, according to the pretheorem two coprime numbers "m" and "" exist that one of them is even and another one is odd so that:

If "" and "", then:

that it is impossible, because, as we know the second power of every odd number must be in "" form. Therefore, "m" is odd and "n" is even. We suppose""and"", so:

 Because numbers "m" and "" are coprime, it results:

in which "" and "t" are positive integer numbers and coprime.


Specially we see:


Since "t" and "" are coprime, again we can use pretheorem for this equality.

Therefore, two coprime numbers "a" and "" exist that one of them is odd and another one is even so:

Because "a" and "b" are coprime. It is resulted from the first equality (according to the pretheorem) two integer numbers "" and "" exit so that for them:

Therefore, the third equality can be written below form:

and it means that  is  primitive answer from equation (3) that includes positive numbers. So, according to the answer, we must have:


that below meaningless inequality is resulted:

Therefore, the assumption of existing answer for equation (3) makes paradox. It means that this equality doesnít have any non- zero answer in set of integer numbers.

19.6. Fermatís last theorem for exponent "3"

We said before that in 1768, Euler proved Fermatís last theorem for case "" for the first time. Here, we reconstruct this proof. Euler expressed this pretheorem at the beginning.

Pretheorem. If two number "a" and "b" that are coprime, have this property that the number "" is the third power of an integer number, then two integer number "s" and "t" will exist so that:

At first we explain that how we can use this pretheorem for proving Fermatís last theorem.

We suppose that Fermatís last theorem doesn't be correct for "", it means integer numbers like, and  exist so that:


We know that numbers, and  can be considered coprime two-to-two. Then, only one of these three numbers can be even. In other side, it is clear that all of three numbers can not be odd (because sum or subtraction of two odd numbers is an even number). Therefore, one and only one of, and  is even. Without any disorder to the generality  can be considered as an even number. In fact, if  be an even number, we can change the role of  with, if   be even we can change role of  with  by changing the sign because:

Between all of integer numbers and triplet  that adapt in equation (1), with assumption  to be even, it is possible to choose it so that  has minimum possible value. Such minimum triplet numbers exist because in every set of positive integer numbers, the smallest number exists.

Since  and  are odd, therefore, below numbers are integer numbers:




Therefore, one of numbers "p" and "q" is even and another one is odd. More over these numbers are coprime clearly.

According to (1) and (2):

If we suppose"":


Since, one of two numbers "p" and "q" is even and another one is odd, "" will be an odd number. So, it is resulted from that "q" is divisible by "4" (it means "q" is an even number and "p" is an odd number).

            According to pretheorem, multiplication of two numbers that are coprime, can be a third power of an integer number if and only if every one of them be third power of an integer number. In other side, numbers "" and"" are coprime only when numbers "q" and "" be coprime and it is possible if and only if every ("p" and "q" are coprime) "q" isnít divisible by "3". So, if we suppose that "q" isnít divisible by "3", then it is resulted from (3) that every one of numbers "" and "" is a third power of an integer number.

But, according to pretheorem, if "" be a third power of an integer number, then:

That "s" and "t" are integer numbers. Because "p" is odd, so, from equality "", it results that "t" is odd and "s" is even, in other side, because "p" and "q" are coprime, "t" and "s" also will be coprime. Since  is a third power of an integer number, therefore:

also must be equal to the third power of an integer number. This proves that below number is also third power of an integer number:

Numbers "", "" and "" are coprime. In fact, if "" and "" have common prime divisor "", then "", because "" is an odd number.

Therefore "" must be common divisor of "s" and "", but since "t" and "s" are coprime, it is possible only for "". So, if numbers "" and "" have common prime factor "", then "", because both of them are odd numbers.

More over, below numbers are divisible by "":

That again it is possible only for"". Therefore, in both of two cases, number "s" and then number "q", will be divisible by "" in spite of the assumption. Since multiplication of numbers "", "" and "" that are coprime,   two-to-two, is the third power of integer number. Therefore, every one of them must be third power of an integer number. It means that three integer numbers "", "" and "" exist so that




So, with start from three numbers, we can obtain three new numbers that adapt in equation (1) and more over have this property that their first number namely "", is even number. Since "" therefore  . Because we have "", so  and then:


So  that it is paradox to the assumption of minimal of triplet numbers .

This paradox proves that number "q" must be divisible by "3":

that "r" is an integer number (and divisible by "4"). So, according to (3):


If integer numbers  and "" be divisible by a prime number like "", then "", because if it isnít so, then number "p" become divisible by "", it means that it has a common divisor with "q". But if "", then below numbers:

Namely "p" and "q" must be divisible by "" that is impossible. Therefore, numbers  and "" are coprime.

So, it is resulted from (4) that both of these two numbers are the third power of an integer number and so, according to the pretheorem (if we use it about ""), below equalities are resulted:


That "s" and "t" are integer and coprime. More over, it is clear that "t" is even (because "r" is even) and then "s" will be odd. In other side, below integer number is equal to the third power of an integer number:

Since numbers "s" and "t" are coprime and more over, one of them is even and another is odd. Therefore, number,  and  are coprime two-to-two. So, every one of them is third power of an integer number so that integer numbers "", "" and "" exist:

But in this situation we will have:

And also:


[1] . Because we have  


[3]. Set of regular numbers are defined by formula of prime numbers. For more information refer to appendixes I (21.3).

[4]. For calculating determinant, refer to "H.M" new method that is written in end of book in appendixes I (21.2)

[5]. For avoiding the answer of general equation, refer to (H.M) table method in appendixes  (21.1) at the end of this book.