CHAPTER 19
On the history of attempts for proving Fermat’s 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 Fermat’s last theorem:
(1)
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 Fermat’s last theorem.
19.2. An introduction to the chronology of Fermat’s last theorem
When Pierre Fermat propounded his famous theorem in margin
of Diophantus’s book in 350 years ago that equation hasn’t 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 Fermat’s 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, Fermat’s 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 Fermat’s last theorem, because, Fermat’s last theorem was a motive for discovering a lot of properties of numbers. But we must wait for an easier solution for Fermat’s last theorem and from this aspect; mathematician's work is not finished.
We want to clarify the method in which the solution of
Fermat’s 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 Wilson’s theorem, Fermat’s 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:
(2)
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 Fermat’s last theorem.
19.3. Chronology of Fermat’s 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 Fermat’s 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 Fermat’s 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, Fermat’s 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 Fermat’s proposition is "Fermat's great theorem" or "Fermat’s 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 Fermat’s 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 Fermat’s last theorem has not been proved by primary methods.
Because of this, many of mathematicians attempted to prove that Fermat’s 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 Fermat’s 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 Fermat’s last theorem. Just in Guttingen of Germany, more than one thousand solutions were sent to mathematics society in Guttingen, "3" years after propounding Wolfskehl’s 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, Wolfskehl’s prize lost its value and "FERMATISTS" lost their motivation of profit (Mathematicians had given this name to some people who attempted to solve the Fermat’s theorem with weak scientific possibility and useless method).
Because of this and naturally, the flood of "Fermatist’s 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 didn’t 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.
Fermat’s 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 Fermat’s 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 Fermat’s theorem (it means that Fermat’s theorem can not be proved with known methods), But attempts to find the new methods for proving the Fermat’s theorem was valuable and useful for mathematics.
Apparently it is necessary to abandon searching for the primary solution for Fermat’s last theorem. Any way ether it is true or false, entering to primary solution of Fermat’s last theorem isn’t advisable specially for young and inexperienced mathematicians. One of the purposes of this chapter of the book is showing this point that Fermat’s 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 didn’t 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 Fermat’s 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 Fermat’s last theorem doesn’t
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:
(1)
("a" and "b" are integer numbers)
Fermat was unfamiliar with such numbers and it is clear that he couldn’t use them.
In fact, Euler’s 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 Euler’s reasoning was Gauess apparently. In 1825,
Dirichlet and Legendre proved Fermat’s theorem for "" almost
contemporarily. Dirichlet’s proof was printed in 1828 that is very complicated.
In 1921 Pelemelle simplified the proof to some extent.
For the next prime number, namely
, Fermat’s 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 Fermat’s theorem
for all of prime numbers 
. Lame’s method was different from
Euler’s cognition and it was based on arithmetic properties of numbers in below
form:
(2)
That 
are integer numbers and "
" is n-th
root of number "1":
But Liouville found a serious deficit in Lame’s 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 aren’t 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 Fermat’s theorem. At first he thought that he could find the perfect proof of Fermat’s theorem and in 1843, send his letter to Dirichlet. This proof was based on such numbers in form (2) and Kummer repeated Lame’s 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 Kummer’s letter to him.
Kummer found out soon that theorem about unique factorization to prime factors isn’t correct for numbers in form (2) and stopped his attempts to prove it. Kummer didn’t 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:
(3)
That "a" and "b" are integer numbers, number "21" is decomposed to multiple of prime factors by two methods:
Although numbers in form (3) aren’t numbers in form (2), for
every value of "n" (for numbers (2), the same example is
possible only for
).
But we can use Kummer’s 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 Kummer’s 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 Kummer’s proof
in 1894.
But this conclusion of Kummer wasn’t favor, because he
didn’t find even one prime number "
" yet so that Fermat’s
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 "Kummer’s 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 "" isn’t 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
""
doesn’t 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 aren’t reducible) . Kummer called such numbers as
"regular"[3].
Bernoulli’s 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
aren’t 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 can’t mention it here.
It is possible to show that why Bernoulli’s 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 aren’t regular and tried to prove Fermat’s 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 Kummer’s
reasoning but all of them can be reform easily. In 1893, Mirimanoff could prove
Fermat’s 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, Kummer’s name wasn’t even among the candidates).
After Kummer, no serious work had been done for proving
Fermat’s theorem till in 1929, Vandiver proved that Fermat's theorem is correct
when prime powers of "":
I. Second factor "" from "h"
isn’t divisible by "".
II. forms of Bernoulli’s term:
aren’t divisible by "
" .
Testing the condition (2) isn’t 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 doesn’t adapt with
this condition.
The condition (2) is also established for numbers. Therefore,
Fermat’s theorem is correct for all prime numbers .
Euler knew that for verification below equation:
(4)
("" is a prime number greater
than "3")
It is necessary to distinguish the case that none of numbers
"
",
"y" or "z" aren’t 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 non–divisible
numbers by "" as first case of
Fermat’s 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 Fermat’s theorem.
Besides the general case of Fermat’s 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 Fermat’s theorem is
correct for prime numbers "" if number "
" be also
a prime number.
In spite of these, Germain’s wish became hope less and the method which she selected to prove wasn’t useful even for the first case of Fermat’s theorem.
Germain didn’t print her results but wrote them in a letter
to Lagendre, the French mathematician. In 1828 Lagendre published a detailed
thesis about Fermat’s 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 Fermat’s theorem is correct for prime powers of "" when at
least one of these five numbers is prime:
Besides it, the first case of Fermat’s theorem is correct for all of prime numbers smaller than 197.Lagendre’s theorem doesn’t 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 Fermat’s theorem
is correct for every prime power of "" if 
exists then:
1. Number 
be a prime number and not
divisible by ""
2. Number 
isn’t 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 Fermat’s theorem or at least for its first case.
(In spite of this, still Germain’s results aren't clear up
to end. For example, still we don’t 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 Fermat’s theorem
was propounded by Kummer. He proved in 1858 that the first case of Fermat’s
theorem for prime powers of "" is correct when at least
one of these two Bernoulli’s number (
and 
) isn’t divisible by "".
In 1905, Manov extended this result and proved that it is
enough that one of the forms of four Bernoulli’s numbers 
isn’t divisible by
""
and this includes all values of 
.
In 1909, Wieferich proved with use of Mirrimanov’s method
that the first case of Fermat’s theorem is correct for all of prime powers of
""
if 
isn’t
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 Wieferich’s
proof that–Wieferich’s condition can be changed from base "2" to
"3", so that the first case of Fermat’s theorem is correct for every
prime power of "" that at least one of these
two numbers or
isn’t
divisible by "".
In 1912, Furtwangler proved Wieferich and Mirrimanov- Frobenius’s 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 Wieferich’s criterion with every prime number not greater than
"43". This result could prove validity of first case of Fermat’s
theorem for every powers of "" smaller than
"253747889".
In 1934, Wandiver proved that the first case of Fermat’s
theorem is correct for a prime power of "" that its second factor
"",
isn’t divisible by "". This theorem is
interesting because as we said before, no prime power has been known till now
that doesn’t adapt with this condition. But since "" isn’t divisible
by "",
researches are done only for with this method.
* Here this question propound that how we can search about proof of Fermat’s last theorem?
At first, we remind, if three numbers (x, y and z) be integer numbers and adapt in below equation:
(1)
(we don’t 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 don’t have divisor except unit) and for
proving that equation (1) hasn’t 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 Fermat’s 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:
(2)
Has answer in set of integer numbers. Then,
answer 
is
obtained for equation (1). Therefore, it is enough to prove Fermat’s 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 Fermat’s 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 don’t take in account the changing of factor’s 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:
And:
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 isn’t divisible by 
, then 
isn’t also
divisible by ,
it means isn’t
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:
(5)
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):
(6)
Pairs of integer numbers 
and 
exist so that every two numbers
of a pair are coprime and we have:
(7)
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 – 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"":
(8)
This proposition is called Fermat’s small theorem.
If we multiply congruence (8) in "a":
(9)
It is clear that this congruence is also established for 
. Therefore,
congruence (9) is established for every integer number "a" and
Euler arranged Fermat’s 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 
doesn’t
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:
(10)
If in this formula 
, formula (9) is obtained (for 
) .
Now, we can propound Germain’s 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 isn’t
divisible by "p" then:
In other word, it is resulted from formulas (7) of Able:
Therefore:
;
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 
isn’t 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 Germain’s theorem
Suppose that an integer number like "m"
exists for prime number 
so that
1. is a prime number
2. There isn’t 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 Fermat’s
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
"
":
(11)
In fact, if 
that "" is not congruence
with "0" to module "p", then according to Fermat’s
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:
So:
"1" and "
"
It is clear that "1" and "
" adapt in this
congruence, but this point that this congruence hasn’t 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 "" aren’t neighbor
of -th
degree in relative to module 
, clearly, it is resulted that
condition (2) from Germain’s 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 Fermat’s theorem is correct for "".
We know Germain obtained this result by her self (A lot of writers called it as Germain’s theorem).
Therefore, Germain’s general theorem resulted in Lagendre’s
theorem. Remind that Lagendre’s 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 Fermat’s theorem is correct.
Here, we prove it only for "
" because with
growing "m", induction becomes difficult.
Suppose that condition (1) from Germain’s 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 isn’t any neighborhood between roots of below congruence:
(12)
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 hasn’t root or has two roots exactly:
and
In the first case, congruence (12) has two roots
"1" and "" that they aren’t neighbor
clearly. Then condition (2) from Germain’s theorem is established. In second
case, congruence (12) has "4" roots:
Two roots of them can be neighbor only for:
or 
If , then 
is divisible by that it is
impossible for 
. And if 
, then:
Is divisible by, it means 
is divisible by 
that again it
isn’t possible for . Therefore, condition (2) from
Germain’s 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
isn’t a prime number and then this case is also impossible. It means that the
condition (3) of Germain’s theorem is also established.
19.4.2. Remark
Vent’s theorem led to Germain’s theorem. We must only prove
that if prime number in "
" form doesn’t aliquot Vent’s
number namely"". Then condition (3) from
Germain’s theorem is established.
19.5. Fermat’s last theorem for exponent "4"
Only the case "" from
Fermat’s 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:
(1)
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 can’t 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 :
(2)
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 Fermat’s last theorem for"". We
prove validity of the general proposition.
19.5.1. Theorem.
Below equation hasn’t non-zero answer in set of integer numbers.
(3)
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:
(3)
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:
So:
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:
So:
that below meaningless inequality is resulted:
Therefore, the assumption of existing answer for equation (3) makes paradox. It means that this equality doesn’t have any non- zero answer in set of integer numbers.
19.6. Fermat’s last theorem for exponent "3"
We said before that in 1768, Euler proved Fermat’s 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 Fermat’s last theorem.
We suppose that Fermat’s last theorem doesn't be correct for
"",
it means integer numbers like, and exist so that:
(1)
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:
Since:
(2)
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"
":
(3)
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" isn’t divisible by "3". So, if we
suppose that "q" isn’t 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
Then:
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):
(4)
If integer numbers 
and "
" be divisible by a
prime number like "", then "
", because
if it isn’t 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:
(5)
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:
, refer to (H.M) table method in
appendixes 
(21.1)
at the end of this book.