Usuario:Huseche33/Taller

FERMAT LAST THEOREM DIRECT PROOF

Introduction

Tells the story that in 1637 Pierre de Fermat conjectured:

"It is impossible to break down a cube into two cubes, a fourth power into two fourth powers, and in general, any power, apart from the square, into two powers of the same exponent. I found a really admirable demonstration, but the margin of the book is too small to contain." .

Pierre de Fermat, Diofanto Arithmetic, 1670 edition[1]

In mathematical terms we can express the situation described by Fermat as

(1)$z^{n}\neq x^{n}+y^{n}$

which means that the equality $z^{n}=x^{n}+y^{n}$ cannot exist, for $n\geq 3$. For $n=2$, there are infinite solutions similar to $5^{2}=4^{2}+3^{2}$. However, we will not go deeper into this topic because it is not the objective of this research.

In 1995, the mathematician Andrew Wiles, in a 98-page article published in the Annals of mathematics, demonstrated the semi-stable case of the Taniyama-Shimura Theorem, previously a conjecture, which relates the modular forms and elliptic curves. From this work, combined with Frey's ideas and the Ribet Theorem, the demonstration of Fermat's Last Theorem follows.

However, the evidence provided by Walis, and his collaborators, is an indirect proof because it does not directly demonstrate the Fermat conjecture but instead demonstrates the Taniyama conjecture which, as mentioned above implies the Fermat's last theorem (FLT). And therefore this proof provides very little practical and direct knowledge about the heart of the problem formulated by Fermat.

The proof presented here is direct, directly attacks the meaning of the expression $z^{n}=x^{n}+y^{n}$, is the first and only proof of this class and does not require advanced concepts in theory of numbers, for this reason it can be understood by anyone who has enough interest in the matter to take the job of doing a disciplined reading of this article. This also makes it aspire to have an audience much larger than the one that can be attributed to Wiles' work.

Trivial results

Before dealing fully with the theory of gaps and its relationship with FLT, it is convenient to analyze some trivial results that are necessary to contextualize the situation.

Prime powers

When $n$ is a composite number $n=p\cdot m$, then the FLT statement is transformed into (assuming equality existed):

$z^{n}$	$~=$	$x^{n}+y^{n}$
$z^{n}-(x^{n}+y^{n})$	$~=$	$0$
$z^{mp}-x^{mp}-y^{mp}$	$~=$	$0$
$(z^{m})^{p}-(x^{m})^{p}-(y^{m})^{p}$	$~=$	$0$
$w^{p}-u^{p}-v^{p}$	$~=$	$0$

Where do we change variables $w=z^{m}$, $u=x^{m}$, and $v=y^{m}$, with which finally it is possible to show that:

(2)$z^{p}+x^{p}+y^{p}=0$

where we add the exponentiated variables in attention to the fact that their exponent is an odd prime.

Theorem 1: Assuming that $z^{n}+x^{n}+y^{n}=0$ exists, then it is enough to consider $n$ as an odd prime exponent.

Proof. It is enough with the explanation given earlier in the section Prime powers $\Box$

X,Y,Z Relative primes

Another important case arises when two of the three variables involved share a common factor, that is, $(x,y)=m$. In this case, the variables can be represented by $x=m\cdot u$ and $y=m\cdot v$, so the equation 2 can be written as:

$z^{p}+x^{p}+y^{p}$	$~=$	$0$
$(w\cdot m)^{p}+(u\cdot m)^{p}+(v\cdot m)^{p}$	$~=$	$0$
$m^{p}\cdot w^{p}+m^{p}\cdot u^{p}+m^{p}\cdot v^{p}$	$~=$	$0$
$(m^{p})\cdot (w^{p}+u^{p}+v^{p})$	$~=$	$0$
$w^{p}+u^{p}+v^{p}$	$~=$	$0$

where we have also done $z=w\cdot m$ because it is clear from equation 2 that the third variable involved should also be a multiple of $m$. In this way we have returned, again, to equation 2 but this time it is clear that the variables involved must be relative primes.

Theorem 2: Assuming that $z^{n}+x^{n}+y^{n}=0$ exists, then it is sufficient to consider that the variables involved are relative primes. This is, $(x,y)=1$ and $(z,x)=1$ and similary $(z,y)=1$.

Proof. Again, it is enough with the explanation given earlier in the section X,Y,Z Relative primes $\Box$

Case $z\geq x+y$

Another very trivial case is when one variable is equal to or greater than the sum of the other two variables. For example, suppose that $z=x+y$, then equation FLT becomes:

$z^{p}$	$~=$	$x^{p}+y^{p}$
$(x+y)^{p}$	$~=$	$x^{p}+y^{p}$
${\displaystyle x^{p}+{p \choose 1}x^{p-1}y+{p \choose 2}x^{p-2}y^{2}+\cdots +{p \choose p-1}xy^{p-1}+y^{p}}$	$~=$	$x^{p}+y^{p}$
${\displaystyle {p \choose 1}x^{p-1}y+{p \choose 2}x^{p-2}y^{2}+\cdots +{p \choose p-1}xy^{p-1}}$	$~=$	$x^{p}-x^{p}+y^{p}-y^{p}$
${\displaystyle {p \choose 1}x^{p-1}y+{p \choose 2}x^{p-2}y^{2}+\cdots +{p \choose p-1}xy^{p-1}}$	$~=$	$0$ (contradiction!)

Theorem 3: Assuming that $z^{n}+x^{n}+y^{n}=0$ exists, it is enough to consider that none of the variables can be greater than or equal to the other two, in absolute value.

Proof. It is enough to observe that the last of the equations in the previous system is absurd, that is, we have reached a contradiction. Now, if $z$ is greater than $x+y$, the previous argument is even stronger. $\Box$

Gap succession

Consider the following ordered sequence of numbers representing powers of 3 of a base that grows by one unit in each element of the sequence, that is:

$1^{3}=1$
	7
$2^{3}=8$		12
	19		6
$3^{3}=27$		18
	37		6
$4^{3}=64$		24
	61		6
$5^{3}=125$		30
	91		6
$6^{3}=216$		36
	127		6
$7^{3}=343$		42
	169		6
$8^{3}=512$		48
	217
$9^{3}=729$
	...

In this sequence there are 4 columns of numbers. The first column includes the valuations of the powers themselves, we call it zero order gaps. The second column defines the consecutive difference between neighboring powers, we call them gaps of order 1. The third column is composed of values that are consecutive differences of gaps of order 1 and are called gaps of order 2. Finally there is a fourth column whose elements are all constant and define the consecutive differences between the gaps of order 2, they are called gaps of order 3, but since it is the last column that is also constant for the entire infinite sequence, we call it the arithmetic distance of the sequence. So, for clarity the zero order gaps are 1, 8, 27, 64, etc. The order 1 gaps are 7, 19, 37, etc. and so on until reaching the arithmetic distance, which is constant and its value is nothing other than the factorial of the power 3, this is $3!=6$.

For further illustration let's see another example with a power of order 4:

$1^{4}=1$
	15
$2^{4}=16$		50
	65		60
$3^{4}=81$		110		24
	175		84
$4^{4}=256$		194		24
	369		108
$5^{4}=625$		302		24
	671		132
$6^{4}=1296$		434		24
	1105		156
$7^{4}=2401$		590		24
	1695		180
$8^{4}=4096$		770
	2465
$9^{4}=6561$
	...

In this sequence we have five columns and the arithmetic distance is equal to $4!=24$.

All these sequences are called gap successions and the theory that studies them we call gap theory. These definitions are necessary to be able to refer to them unambiguously and because in conventional mathematics these studies are unknown, except, of course, for the results that this author has recently released, some of which include proof of the conjecture of Goldbach and the proof of the conjecture of the twin primes that readers can consult in Goldbach Theorem

The Successions Theorem

Given a generating numerical sequence $g=(g_{i})_{i\in \mathbb {Z} }$, the infinite matrix is built

$E(g)=\left[e_{ij}\right]_{i\in \mathbb {Z} ,j\in \mathbb {N} }$

of the following way

• For each $i\in \mathbb {Z} :\quad e_{i0}=g_{i}$
• For each $i\in \mathbb {Z} ,j\in \mathbb {N} :\quad e_{i,j+1}=e_{i+1,j}-e_{ij}$

schematically you have

$E(g)={\begin{bmatrix}g_{-2}&&\\&g_{-1}-g_{-2}&\\g_{-1}&&g_{0}-2g_{-1}+g_{-2}\\&g_{0}-g_{-1}&&g_{-1}-3g_{0}+3g_{-1}-g_{-2}\\g_{0}&&g_{-1}-2g_{0}+g_{-1}&&...\\&g_{-1}-g_{0}&&g_{-2}-3g_{-1}+3g_{0}-g_{-1}\\g_{-1}&&g_{-2}-2g_{-1}+g_{0}\\&g_{-2}-g_{-1}&\\g_{-2}&&\\...&&\\\end{bmatrix}}$

the examples placed in the scheme suggest that each entry $e_{ij}$ can be expressed as a combination linear of generators $g_{i},g_{i+1},...,g_{i+k}$.

Theorem 4​. For each $i\in \mathbb {Z} \quad j\in \mathbb {N}$, we have $e_{ij}=\sum _{k=0}^{j}{\binom {j}{k}}(-1)^{j-k}g_{i+k}$

Proof. (Induction on $j$)

• $j=0$ By definition, for any $i\in \mathbb {Z}$ we have $e_{i}=g_{i}$ and trivially

$g_{i}=\sum _{k=0}^{0}{\binom {0}{k}}(-1)^{0-k}g_{i+k}$

• Suppose that the formula is valid for $j$ (and any $i\in \mathbb {Z}$) by definition

$e_{i}^{j+1}=e_{i+1,j}-e_{ij}$; For both $e_{i+1,j}$ and $e_{ij}$ the formula is valid, then it can be replaced:

${\begin{aligned}e_{ij}&=\sum _{l=0}^{j}{\binom {j}{l}}(-1)^{j-l}g_{i+1+l}-\sum _{k=0}^{j}{\binom {j}{k}}(-1)^{j-k}g_{i+k}\\\end{aligned}}$

(changing index: $k=l+1$)

${\begin{aligned}&=\sum _{k=1}^{j+1}{\binom {j}{k-1}}(-1)^{j-k+1}g_{i+k}+\sum _{k=0}^{j}{\binom {j}{k}}(-1)^{j-k+1}g_{i+k}\\&=(\sum _{k=1}^{j}{\binom {j}{k-1}}(-1)^{j-k+1}g_{i+k}+g_{i+j+1}+(-1)^{j+1}g_{i}+\sum _{k=1}^{j}{\binom {j}{k}}(-1)^{j-k+1}g_{i+k})\\&=(-1)^{j+1}g_{i}+\sum _{k=1}^{j}({\binom {j}{k-1}}+{\binom {j}{k}})(-1)^{j-k+1}g_{i+k}+g_{i+j+1}\\\end{aligned}}$

(by a property of the binomial coefficients)

${\begin{aligned}&={\binom {0}{j+1}}(-1)^{j+1-0}g_{i+0}+\sum _{k=1}^{j}{\binom {j+1}{k}}(-1)^{j+1-k}g_{i+k}+{\binom {j+1}{j+1}}(-1)^{j+1-(j+1)}g_{i+(j+1)}\\&=\sum _{k=0}^{j+1}{\binom {j+1}{k}}(-1)^{j+1-k}g_{i+k}\end{aligned}}$

and this is the formula for $j+1\quad \Box$

Special cases

• If $b$ is a non-zero real number, for each $i\in \mathbb {Z}$ let $g_{i}=b^{i}$. Then, according to the Theorem 1, it is received

${\begin{aligned}e_{ij}&=\sum _{k=0}^{j}{\binom {j}{k}}(-1)^{j-k}b^{i+k}\\&=b^{i}\sum _{k=0}^{j}{\binom {j}{k}}b^{k}(-1)^{j-k}\\&=b^{i}(b-1)^{j}\end{aligned}}$

• If $n$ is positive integer, for each $i\in \mathbb {Z}$ let $g_{i}=i^{n}$

The following section is dedicated to proof that in this case for each $i\in \mathbb {Z}$ we have $e_{in}=n!$

combining this fact with the proven statement, we obtain the following 'surprising expressions' of the factorial

• For each whole number $i\in \mathbb {Z}$,

$n!=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}(i+k)^{n}$

• In special,

$n!=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}k^{n}$

$n!=\sum _{k=0}^{n}{\binom {n}{k}}k^{k}(-k)^{n-k}$

$n!=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{k}(i-k)^{n}$

Combined gap successions

It seems natural that a succession gap can be combined to give rise to new successions with the mathematical advantages that this operation entails.

Successions of products (Twin primes)

In the article on the Goldbach conjecture proof, gap successions were used that are not properly pure powers, such as:

$0\cdot 2=0$
	3
$1\cdot 3=3$		2
	5
$2\cdot 4=8$		2
	7
$3\cdot 5=15$		2
	9
$4\cdot 6=24$		2
	11
$5\cdot 7=35$		2
	13
$6\cdot 8=48$		2
	15
$7\cdot 9=63$		2
	17
$8\cdot 10=80$		2
	19
$9\cdot 11=99$		2
	21
$10\cdot 12=120$		2
	23
$11\cdot 13=143$		2
	25
$12\cdot 14=168$		2
	27
$13\cdot 15=195$		2
	29
$14\cdot 16=224$		2
	31
$15\cdot 17=255$		2
	33
$16\cdot 18=288$		2
	35
$17\cdot 19=323$
	...

the previous sequence obtains the infinite series of the twin primes, that is, 3 and 5 (in 15), 5 and 7 (35), 11 and 13 (143), 17 and 19 (323), etc., as can be easily seen by a simple inspection. This result has been proof in Goldbach Theorem as mentioned earlier.

Two variables fermatians

It seems natural to combine two sequences of powers to obtain a new sequence according to theorem 4, as in the following example:

$20^{3}+7^{3}=8343$
	-972
$19^{3}+8^{3}=7371$		$162=27\cdot 3!$
	-810
$18^{3}+9^{3}=6561$		162
	-648
$17^{3}+10^{3}=5913$		162
	-486
$16^{3}+11^{3}=5427$		162
	-324
$15^{3}+12^{3}=5103$		162
	-162
$14^{3}+13^{3}=4941$		162
	0
$13^{3}+14^{3}=4941$		162
	162
$12^{3}+15^{3}=5103$		162
	324
$11^{3}+16^{3}=5427$		162
	486
$10^{3}+17^{3}=5913$
	...

Note that the two variables always add the same value, 27 in this case, and in addition, the arithmetic distance is the product of this constant by the factorial of the power used.

The Fermat's Last Theorem (FLT)

We are now in a position to get into the very heart of Fermat's last theorem, for which, as the good reader will have imagined, we will use a fermatian that combines three variables, as it was logical to wait. Thus, to the gap sequence of the previous example we can add a third variable to obtain a sequence like the one illustrated below:

$11^{3}-(10^{3}+17^{3})=-4582$
	883
$12^{3}-(11^{3}+16^{3})=-3699$		-90
	793		6
$13^{3}-(12^{3}+15^{3})=-2906$		-84
	709		6
$14^{3}-(13^{3}+14^{3})=-2197$		-78
	631		6
$15^{3}-(14^{3}+13^{3})=-1566$		-72
	559		6
$16^{3}-(15^{3}+12^{3})=-1007$		-66
	493		6
$17^{3}-(16^{3}+11^{3})=-514$		-60
	433		6
$18^{3}-(17^{3}+10^{3})=-81$		-54
	379		6
$19^{3}-(18^{3}+9^{3})=298$		-48
	331		6
$20^{3}-(19^{3}+8^{3})=629$		-42
	289		6
$21^{3}-(20^{3}+7^{3})=918$		-36
	253
$22^{3}-(21^{3}+6^{3})=1171$
	...

In the previous section, it is easy to identify two critical values, these are $18^{3}-(17^{3}+10^{3})=-81$ and $19^{3}-(18^{3}+9^{3})=298$, the first one corresponds to the largest value of the negative values and the second to the smallest value of the positive values, in other words, these are the smaller absolute values throughout the sequence, which are located just where zero order gaps change their signature, that is, the place in the sequence where a zero could exist. We call this place the Fermat boundary. This concept is very important because if there is a functional zero associated with a three-variable fermatian (FLT), that zero must be located here, so that any criteria that prevent us from having zeros in this part of the sequence is potentially proof of Fermat's last theorem.

The case $x^{3}+y^{3}+z^{3}=33$

x^3+y^3+z^42=33

The figure above shows the case of the three-variable fermatian (FLT), namely $x^{3}+y^{3}+z^{3}=33$ where the Fermat boundary is shown but unlike of the fermatian $x^{3}+y^{3}+z^{3}=-85$ the arithmetic distance has reversed its signature, which is a consequence of the symmetry that the fermatian presents.

American mathematician Andrew Booker announced that 33 could express the sum of the cubes: (8866128975287528)^3 + (-8778405442862239)^3 + (-2736111468807040)^3. An achievement achieved through a brute force approach or method; that is, with the assistance of a supercomputer, which executed the algorithm designed by the mathematician for three weeks in a row until he found that solution. For more information about item see https://www.bbvaopenmind.com/ciencia/matematicas/el-misterio-del-numero-33-y-las-ecuaciones-diofanticas/

The case $x^{3}+y^{3}+z^{3}=42$

x^3+y^3+z^42=42

The figure above shows the case of the three-variable fermatian (FLT), namely $x^{3}+y^{3}+z^{3}=42$ where, again, the Fermat boundary is shown. For more information about theme see https://www.europapress.es/ciencia/laboratorio/noticia-solucion-final-enigma-matematico-suma-tres-cubos-20190909112613.html

The cases FLT (3) = 33 and FLT (3) = 42 have been calculated by this author using software of my creation written in Google's Go language. We have chosen this option preferably to the use of formatting in LaTeX because this last strategy would be very laborious given the size of the exercise.

Arbitrary fermatians

We currently know that $FLT\neq 0$, but there are infinite fermatian forms of which FLT is only a special case when three varials are involved. A fermatian of an arbitrary number of variables is defined as:

(3)$F(x_{1},x_{2},\dots x_{k},n)=x_{1}^{n}+x_{2}^{n}+\dots +x_{k}^{n}$

And of course, some meet the condition $F(x_{1},x_{2},\dots x_{k},n)=0$. In the future we abbreviate this expression as $F(x_{i},n)=0$ when there is a singularity, this is, $x_{1}^{n}+x_{2}^{n}+x_{k}^{n}=0$. If there is no singularity in the fermatian we will write $F(x_{i},n)\neq 0$ as it is just natural. Some examples of fermatians with singularity are:

$30^{5}-[29^{5}+19^{5}+16^{5}+11^{5}+10^{5}+5^{5}]=0$

$12^{5}-(11^{5}+9^{5}+7^{5}+6^{5}+5^{5}+4^{5})=0$

$144^{5}-(133^{5}+110^{5}+84^{5}+27^{5})=0$

$20615673^{4}-(2682440^{4}+15365639^{4}+18796760^{4})=0$

$6^{3}-(5^{3}+4^{3}+3^{3})=0$

$9^{3}-(8^{3}+6^{3}+1^{3})=0$

$9^{3}-(8^{3}+5^{3}+4^{3}+3^{3}+1^{3})=0$

$19^{3}-(18^{3}+10^{3}+3^{3})=0$

$41^{3}-(40^{3}+17^{3}+2^{3})=0$

All of them are formally gap sequences (theorem 4) as shown below to illustrate a single case:

$140^{5}-(129^{5}+114^{5}+80^{5}+31^{5})=-4500226624$
	1166327131
$141^{5}-(130^{5}+113^{5}+81^{5}+30^{5})=-3333899493$		-27904470
	1138422661		592590
$142^{5}-(131^{5}+112^{5}+82^{5}+29^{5})=-2195476832$		-27311880		-25440
	1111110781		567150		120
$143^{5}-(132^{5}+111^{5}+83^{5}+28^{5})=-1084366051$		-26744730		-25320
	1084366051		541830		120
$144^{5}-(133^{5}+110^{5}+84^{5}+27^{5})=0$		-26202900		-25200
	1058163151		516630		120
$145^{5}-(134^{5}+109^{5}+85^{5}+26^{5})=1058163151$		-25686270		-25080
	1032476881		491550		120
$146^{5}-(135^{5}+108^{5}+86^{5}+25^{5})=2090640032$		-25194720		-24960
	1007282161		466590		120
$147^{5}-(136^{5}+107^{5}+87^{5}+24^{5})=3097922193$		-24728130		-24840
	982554031		441750
$148^{5}-(137^{5}+106^{5}+88^{5}+23^{5})=4080476224$		-24286380
	958267651
$149^{5}-(138^{5}+105^{5}+89^{5}+22^{5})=5038743875$
	...

It is easy to show that when there is a singularity in a fermatian, all zero order gaps are composite numbers, with the only exception, perhaps, of the gaps that are located immediately before and after the singularity. To see it better, look at the factorization of such gaps in the immediately preceding exercise:

$140^{5}-(129^{5}+114^{5}+80^{5}+31^{5})=-4500226624=-4\cdot 1125056656$
$141^{5}-(130^{5}+113^{5}+81^{5}+30^{5})=-3333899493=-3\cdot 1111299831$
$142^{5}-(131^{5}+112^{5}+82^{5}+29^{5})=-2195476832=-2\cdot 1097738416$
$143^{5}-(132^{5}+111^{5}+83^{5}+28^{5})=-1084366051=-1\cdot 1084366051$
$144^{5}-(133^{5}+110^{5}+84^{5}+27^{5})=0$
$145^{5}-(134^{5}+109^{5}+85^{5}+26^{5})=1058163151=1\cdot 1058163151$
$146^{5}-(135^{5}+108^{5}+86^{5}+25^{5})=2090640032=2\cdot 1045320016$
$147^{5}-(136^{5}+107^{5}+87^{5}+24^{5})=3097922193=3\cdot 1032640731$
$148^{5}-(137^{5}+106^{5}+88^{5}+23^{5})=4080476224=4\cdot 1020119056$
$149^{5}-(138^{5}+105^{5}+89^{5}+22^{5})=5038743875=5\cdot 1007748775$
$150^{5}-(139^{5}+104^{5}+90^{5}+21^{5})=5973142176=6\cdot 995523696$
$151^{5}-(140^{5}+103^{5}+91^{5}+20^{5})=6884063557=7\cdot 983437651$
	...

In the previous sequence you have $144^{5}-(133^{5}+110^{5}+84^{5}+27^{5})=0$ and according to the statement that we have made the number $(144+7)^{5}-[(133+7)^{5}+(110-7)^{5}+(84+7)^{5}+(27-7)^{5}]=0$ should be divisible by 7 as in effect it is, $151^{5}-(140^{5}+103^{5}+91^{5}+20^{5})=6884063557=7\cdot 983437651$. This fact allows us to formulate the following

Theorem 5. Given a fermatian with $F(x_{i},n)=0$, then all zero order gaps, with the exception, perhaps, of the elements immediately before and after of the singularity, are composite numbers.

Proof. The demonstration of this fact is trivial and we leave it as an exercise for the reader. Help: Take $F(x_{i},n)$ and make $F((x+k)_{i},n)$ and develop this last expression using Newton's binomial theorem, you will see that all the resulting numbers (zero-order gpas) are divisible by $k$ and therefore they are composite numbers, except the gaps surrounding the singularity, which, being divisible by unity, could be prime numbers.

Theorem 5 is very important because it could help us look for singularities in a fermatian. It can also be used to prove that there are no singularities in a fermatian and in this way it can be useful in the search for proof of Fermat's last theorem.

Modular arithmetic in gap theory

In the theory of gaps we can make use of Modular arithmetic to obtain advantageous results. First we give some definitions to familiarize non-specialized readers with the concepts of modular arithmetic ​.

For a positive integer $n$, two numbers $a$ and $b$ are said to be congruent modulo $n$, if their difference $a-b$ is an integer multiple of $n$. This congruence relation is typically considered when $a$ and $b$ are integers, and is denoted

$a\equiv b\mod n$

Typically we can write $a$ and $b$ as

$a=pn+r,$

$b=qn+r,$

where 0 ≤ r < n is the common remainder. For example, we say that $10\equiv 1\mod 3$ because dividing 10 by 3 the rest is 1, or equivalently if we subtract 1 from 10 the result is divisible by 3, that is, $3\mid (10-1)$, which can also be written as $(10-1)\equiv 0\mod 3$, that is, $9\equiv 0\mod 3$.

Congruence operations are important in gaps theory because they serve to indicate, among other things, when a number is different from zero, that is, when a number is non-zero. This is because if a number delivers a non-zero remainder, it is clear that said number is not zero. For example, if we say $x\equiv 1\mod 3$, we immediately see that $x$ cannot be equal to zero, that is, $x\neq 0$.

Following this reasoning we can link congruences to fermatian forms $F(x_{i},n)$ and especially to FLT, so we can express $z^{p}+x^{p}+y^{p}\equiv r\mod m$ where is clear that $r$ is the rest and $m$ is the module of the previous operation, where, in addition, you have to $0\leq r<p$.

Homogeneous remains system

We define a homogeneous system of remains when we have:

(4)$F_{j}(x_{i},n)\equiv r\mod m$

that is, when all fermatians leave the same rest with respect to the same module $m$. We could have written too:

$F_{1}(x_{1},x_{2},\dots x_{k},n)\equiv r\mod m$
$F_{2}(x_{1},x_{2},\dots x_{k},n)\equiv r\mod m$
$F_{3}(x_{1},x_{2},\dots x_{k},n)\equiv r\mod m$
$\quad \vdots$
$F_{l}(x_{1},x_{2},\dots x_{k},n)\equiv r\mod m$		...

or equivalently

$x_{11}^{n}+x_{12}^{n}+\dots +x_{1k}^{n}\equiv r\mod m$
$x_{21}^{n}+x_{22}^{n}+\dots +x_{2k}^{n}\equiv r\mod m$
$x_{31}^{n}+x_{32}^{n}+\dots +x_{3k}^{n}\equiv r\mod m$
$\quad \vdots$
$x_{l1}^{n}+x_{l2}^{n}+\dots +x_{lk}^{n}\equiv r\mod m$

but we prefer to use $F_{j}(x_{i},n)\equiv r\mod m$ notation because it is obviously more compact. For example, the following fermatians define a homogeneous system of remains:

$(4)^{3}+(3)^{3}=91\equiv 0\mod 7$
$(5)^{3}+(2)^{3}=133\equiv 0\mod 7$
$(6)^{3}+(1)^{3}=217\equiv 0\mod 7$
$(7)^{3}+(0)^{3}=343\equiv 0\mod 7$
$(8)^{3}+(-1)^{3}=511\equiv 0\mod 7$
$(9)^{3}+(-2)^{3}=721\equiv 0\mod 7$
$(10)^{3}+(-3)^{3}=973\equiv 0\mod 7$
$(11)^{3}+(-4)^{3}=1267\equiv 0\mod 7$
$(12)^{3}+(-5)^{3}=1603\equiv 0\mod 7$
$(13)^{3}+(-6)^{3}=1981\equiv 0\mod 7$

No homogeneous remains system

We define a no homogeneous system of remains when we have:

(4)$F_{j}(x_{i},n)\equiv r_{i}\mod m$

that is, when all fermatians leave,ordinarily, different rest with respect to the same module $m$, although some of the remains could be repeated. We could have written too:

$F_{1}(x_{1},x_{2},\dots x_{k},n)\equiv r_{1}\mod m$
$F_{2}(x_{1},x_{2},\dots x_{k},n)\equiv r_{2}\mod m$
$F_{3}(x_{1},x_{2},\dots x_{k},n)\equiv r_{3}\mod m$
$\quad \vdots$
$F_{l}(x_{1},x_{2},\dots x_{k},n)\equiv r_{l}\mod m$

Normally, fermatians have remains systems that are periodic, as shown in the following example:

$(6)^{3}+(5)^{3}=341\equiv 0\mod 11$
$(7)^{3}+(6)^{3}=559\equiv 9\mod 11$
$(8)^{3}+(7)^{3}=855\equiv 8\mod 11$
$(9)^{3}+(8)^{3}=1241\equiv 9\mod 11$
$(10)^{3}+(9)^{3}=1729\equiv 2\mod 11$
$(11)^{3}+(10)^{3}=2331\equiv 10\mod 11$
$(12)^{3}+(11)^{3}=3059\equiv 1\mod 11$
$(13)^{3}+(12)^{3}=3925\equiv 9\mod 11$
$(14)^{3}+(13)^{3}=4941\equiv 2\mod 11$
$(15)^{3}+(14)^{3}=6119\equiv 3\mod 11$
$(16)^{3}+(15)^{3}=7471\equiv 2\mod 11$
$(17)^{3}+(16)^{3}=9009\equiv 0\mod 11$
$(18)^{3}+(17)^{3}=10745\equiv 9\mod 11$
$(19)^{3}+(18)^{3}=12691\equiv 8\mod 11$
$(20)^{3}+(19)^{3}=14859\equiv 9\mod 11$
$(21)^{3}+(20)^{3}=17261\equiv 2\mod 11$
$(22)^{3}+(21)^{3}=19909\equiv 10\mod 11$
$(23)^{3}+(22)^{3}=22815\equiv 1\mod 11$
$(24)^{3}+(23)^{3}=25991\equiv 9\mod 11$
$(25)^{3}+(24)^{3}=29449\equiv 2\mod 11$
$(26)^{3}+(25)^{3}=33201\equiv 3\mod 11$
$(27)^{3}+(26)^{3}=37259\equiv 2\mod 11$
$(28)^{3}+(27)^{3}=41635\equiv 0\mod 11$
$(29)^{3}+(28)^{3}=46341\equiv 9\mod 11$
$(30)^{3}+(29)^{3}=51389\equiv 8\mod 11$
	...

the previous system of remains is non-homogeneous and is also periodic, with a period equal to the module used, in this case 11.

It is easy to demonstrate that in every fermatian, the system of remains with respect to the same module is periodic. Again, we leave this fact as an exercise for the reader.

The direct proof of FLT

Strategy

The evidence presented here is simple, straightforward and blunt. You do not need to spend half of your life to understand how $x^{n}+y^{n}+z^{n}\neq 0$, on the contrary, a judicious reading, of a few hours, will be enough for the reader to fully understand the meaning of Fermat's statement about the year 1637.

As an author I have the conviction that mathematics should not only be formal but also retain a sense of aesthetics that brings to light its beauty and symmetry, modesty apart, as is done in this research.

Throughout this work you will surely have noticed different ways to demonstrate FLT. However, this time we opted for the use of congruences as will be seen in the following section.

Suppose, for example, that you want to know if $569936821221962380720^{3}-569936821113563493509^{3}-472715493453327032^{3}\neq 0$, without having to calculate the fermatian. Honestly, most professional mathematicians in the world could not do it (I can say it by experience) and all because they do not know or cannot program computers. ​ However, all that would be needed is to calculate the remains system associated with this fermatian as shown below:

$(569936821221962380720)^{3}+(-569936821113563493514)^{3}+(-472715493453327027)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493513)^{3}+(-472715493453327028)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493512)^{3}+(-472715493453327029)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493511)^{3}+(-472715493453327030)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493510)^{3}+(-472715493453327031)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493509)^{3}+(-472715493453327032)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493508)^{3}+(-472715493453327033)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493507)^{3}+(-472715493453327034)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493506)^{3}+(-472715493453327035)^{3}\equiv 3\mod 570409536607016820541$
$(569936821221962380720)^{3}+(-569936821113563493505)^{3}+(-472715493453327036)^{3}\equiv 3\mod 570409536607016820541$
$\quad \vdots$

where we have included the fermatian in question as belonging to a homogeneous system of rest as indicated above in the section that dealt with this issue. In this exercise we have used module $m=570409536607016820541=569936821113563493509+472715493453327032$ while the value of 569936821221962380720 is left as constant.

Alternatively we can also do:

$(569936821221962380715)^{3}+(-569936821113563493509)^{3}+(-472715493453327027)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380716)^{3}+(-569936821113563493509)^{3}+(-472715493453327028)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380717)^{3}+(-569936821113563493509)^{3}+(-472715493453327029)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380718)^{3}+(-569936821113563493509)^{3}+(-472715493453327030)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380719)^{3}+(-569936821113563493509)^{3}+(-472715493453327031)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380720)^{3}+(-569936821113563493509)^{3}+(-472715493453327032)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380721)^{3}+(-569936821113563493509)^{3}+(-472715493453327033)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380722)^{3}+(-569936821113563493509)^{3}+(-472715493453327034)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380723)^{3}+(-569936821113563493509)^{3}+(-472715493453327035)^{3}\equiv 3\mod 569464105728509053688$
$(569936821221962380724)^{3}+(-569936821113563493509)^{3}+(-472715493453327036)^{3}\equiv 3\mod 569464105728509053688$
$\quad \vdots$

And again we have achieved that the fermatian in question belongs to a homogeneous system of remains, not null, where on this occasion we make $m=569464105728509053688=569936821221962380720+(-472715493453327032)$, and the value of -569936821113563493509 remains constant. A third option is still possible, which yields the result:

$(569936821221962380715)^{3}+(-569936821113563493504)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380716)^{3}+(-569936821113563493505)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380717)^{3}+(-569936821113563493506)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380718)^{3}+(-569936821113563493507)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380719)^{3}+(-569936821113563493508)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380720)^{3}+(-569936821113563493509)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380721)^{3}+(-569936821113563493510)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380722)^{3}+(-569936821113563493511)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380723)^{3}+(-569936821113563493512)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$(569936821221962380724)^{3}+(-569936821113563493513)^{3}+(-472715493453327032)^{3}\equiv 3\mod 108398887211$
$\quad \vdots$

This time the value of -472715493453327032 has been left as constant while we take the module $m=108398887211=569936821221962380720+(-569936821113563493509)$. In this way in three different forms we have shown that $569936821221962380720^{3}-569936821113563493509^{3}-472715493453327032^{3}\neq 0$.

In all these cases the remains can be calculated with the help of Fermat's little theorem ​ and without having to evaluate the powers.

The strategy followed in the previous case is summarized as:

$x^{p}+y^{p}+z^{p}\equiv r\mod |y+z|$

$x^{p}+y^{p}+z^{p}\equiv r\mod |x+z|$

$x^{p}+y^{p}+z^{p}\equiv r\mod |x+y|$

Which was to be expected for how much $(x,y,z)=1$, that is, the variables involved do not share common factors as shown in Theorem 2. The $(x,y,z)=1$ condition is essential to prevent the homogeneous system of remains from giving us null remains. In the case studied, the three remains are the same but this does not have to be this way. The remains could be different, however, they must be different from zero by virtue of the condition of relative primes imposed on the variables involved in the fermatian.