﻿ The Last Fermat theorem - Introduction

# The Last Fermat theorem - Introduction

You're reading an English translation of the original Czech page.

The following paragraphs contain a short outlook on the Last Fermat theorem with regards to G-systems.

### Instances in segments

Let c(s) be a number of instances in segment s. In case ap +bp =cp (i.e. the Last Fermat theorem) it should hold:

```         c-1
(1)     ∑(c(s)) = ap
s=b
```

In G(p) the values c(s) are:

```p=2: 1, 3, 5, 7, 9, 11 13, 15, 17, 19, 21, 23, 25, 27 ..
p=3: 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, ...
p=5: 1, 31, 211, 781, 2101, 4651, ...
p=7: 1, 127, 2059, 14197, ...
```

Only in case p=2 such sums are known:
7+ 9 = 42; 17+19 = 62; 11+13+15+17+19+21+23+25 = 122.

In case p=3, i.e. G(3) the values c(s) are:

```  |   0   1   2   3   4  ...  i
--+----------------------------
0 |   0
1 |   1,  7, 19, 37, 61, 91
2 |   8, 26, 56, 98,152,
3 |  27, 63,117,189,
4 |  64,124,208,
5 | 125,
6 |
j
```

In the table it holds:

``` (2a)   R[i,j]=R[i,1]+R[i+1,j-1], E.g. R[3,3]=R[3,1]+R[4,2]=19+98=117
```

Written in an other way:

``` (2b)   R[i,j]=R[i-1,j+1]-R[i-1,1],  E.g. R[4,2]=R[3,3]-R[3,1]=117-19=98
```

The equation (1) must hold also for every module m:

```c(s)
mod2 mod3 mod4 mod5 mod6 mod7 mod8 mod9 mod10 mod11 mod12
-----------------------------------------------------------------
1    1    1    1    1    1    1    1    1    1     1     1
7    1    1   -1    2    1    0   -1   -2   -3    -4    -5
19    1    1   -1   -1    1   -2    3    1   -1    -3    -5
37    1    1    1    2    1    2   -3    1   -3     4     1
61    1    1    1    1    1   -2   -3   -2    1    -5     1
91    1    1   -1    1    1    0    3    1    1     3    -5
127    1    1   -1    2    1    1   -1    1   -3    -5    -5
169    1    1    1   -1    1    1    1   -2   -1     4     1
217    1    1    1    2    1    0    1    1   -3    -3     1
271    1    1   -1    1    1   -2   -1    1    1    -4    -5
331    1    1   -1    1    1    2    3   -2    1     1    -5
397    1    1    1    2    1   -2   -3    1   -3     1     1
469    1    1    1   -1    1    0   -3    1   -1    -4     1
547    1    1   -1    2    1    1    3   -2   -3    -3    -5
```
```ap
mod2 mod3 mod4 mod5 mod6 mod7 mod8 mod9 mod10 mod11 mod12
-----------------------------------------------------------------
1    1    1    1    1    1    1    1    1    1     1     1
8    0   -1    0   -2    2    1    0   -1   -2    -3    -4
27    1    0   -1    2    3   -1    3    0   -3     5     3
64    0    1    0   -1   -2    1    0    1    4    -2     4
125    1   -1    1    0   -1   -1   -3   -1    5     4     5
216    0    0    0    1    0   -1    0    0   -4    -4     0
343    1    1   -1   -2    1    0   -1    1    3     2    -5
512    0   -1    0    2    2    1    0   -1    2    -5    -4
729    1    0    1   -1    3    1    1    0   -1     3    -3
1000    0    1    0    0   -2   -1    0    1    0    -1     4
1331    1   -1   -1    1   -1    1    3   -1    1     0    -1
1728    0    0    0   -2    0   -1    0    0   -2     1     0
2197    1    1    1    2    1   -1   -3    1   -3    -3     1
2744    0   -1    0   -1    0    0    0   -1    4     5    -4
```

Therefore, sum of numbers in a block of some adjacent rows from the first table should be equal to the numbers of one row from the second table (for each column).

#### Some minor dependencies

From the expression (b+p)p-bp = 0 (mod p) and from the structure of diferential progressions of self-classes follows:

s[(b+p)p-(b+p)]/p - (bp-b)/p = -1 (mod p )

#### (b+p)p - bp = 0 ( mod p2 )

E.g. (2+3)3 - 23 = 117 = 0 mod 32; or (2+5)5 - 25 = 16775 = 0 mod 52.

For each p ε P the numbers c(s) satisfy the equation:

#### a(0) + a(p) = constant ( mod p )

E.g. p=3: 1 + 37 = 7 + 61 = 19 + 91 = 37 + 127 = 2 mod 3.

## Search for suitable modules

The Last Fermat theorem (LF-theorem) is assertion, that equation ak+bk=ck (LF-equation) have for kεN, k>2 no solution in (positive) integers a,b,c ε N (a,b,c>0). To have the theorem true for all k > 2 it suffices to prove it is true for k = 4 and all k ε P. When there is no solution of ap+bp=cp there can not be any solution of (aq)p+(bq)p=(cq)p i.e. apq+bpq=cpq.

Evidence is sufficient if limited only to the numbers a, b, c relatively prime in pairs, i.e. (a,b)=1,(a,c)=1,(b,c)=1.

### Segments of G-systems

Denote A=ap, B=bp, C=cp for a<b<c; pεP.

```                                           G(c,p)
┌───────────┐
│           │
N(C) ──>  │           │
│           │
G(b,p)            │           │
┌───────────┐ ─ ─ ┼ ─      ─ ─│
│           │     │           │
N(B) ──> │           │     │    C      │
G(a,p)            │           │     │           │
┌───────────┐ ─ ─ ┼─ ─  B  ─ ─│  −> ┼ ─      ─ ─│
│           │     │           │     │           │
│     A     │  −> │           │     │           │
│           │     │           │     │           │
└───────────┘ ─ ─ └───────────┘ ─ ─ └───────────┘
```

System G(c,p) gains segments of the system G(b,p) and it similarly the segments of G(a,p). Let B = A + N(B), C = B + N(C) = A + N(B) + N(C). According to LF-theorem for p>2 such A,B,C does not exists that A+B=C, i.e. N(C) = A.

### Pythagorean triangles

 Pýthagorás ze Samu , c. 575-509 BC, the ancient Greek mathematician, founder of the philosophical school. According to his teachings, harmony rules to music, to the movements of the heavenly bodies, to microworld and to the human soul itself.

For p=2 such G-systems exists, e.g. G(3,2), G(4,2) and G(5,2) because 3² + 4² = 5²:

```                                       G(5,2)
┌───────┐
│ 0     │
│ 1   5 │
│ 2  10 │
│ 3  15 │
G(4,2)          │ 4  20 │
└───────┘
0               6
1   4           7  11
2   8           8  16
G(3,2)            3  12           9  21
┌───────┐
│ 0     │        5              12
│ 1 3   │        6   9          13  17
│ 2 6   │        7  13          14  22
│ 4     │       10              18
│ 5 7   │       11 14           19  23
│ 8     │       15              24
└───────┘
```

Region N(C) contains v(c)−v(b) = (cp−bp)−(c−b) of self and (c−b) nested instances, i.e. total m(c)−m(b) = (cp−bp) instances divided into (c-b) groups. Order of all systems is the same, classes always have a maximum of p transpositions. Length of groups increases with the degree, must therefore be (c−b) < a, i.e. a + b > c.

### Fermat's congruence

If equation ak+bk=ck is true, then the corresponding congruence must be true for any module rεN: ak+bk≡ck (mod r). We will call Fermat's congruence (F-kongruence) the relationship Za+Zb≡Zc (mod r), where Zx=xk mod r. Solution of F-kongruence will be shortly written in the form: [Za,Zb,Zc].

Validity of the the congruence indicates the validity of the initial equation. E.g. function a³−2x never get rest 1 or 4 by module 5:

```   x             │  0   1    2    3    4
───────────────┼─────────────────────────
x3+2x         │  0   3   12   33   72
x3+2x (mod 5) │  0   3    2    3    2
```

Therefore, the equation x³+2x=a, where a=1 or a=4 can not have any solution for xεZ.

The same procedure is suitable even for functions of several variables. E.g. congruence x²+y²+z²≡7 mod 8 has no solutions. Quadratic residues of module 8 are the numbers 0,1 and 4 and no combination of these numbers can be as sum ≡ 7 (mod 8).

### Differential sequences

The first differential sequence of squares shows the number of instances in particular segments.

```0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ...
```

The existence of Pythagorean triangles means that there are some subsets of numbers in the differenctial sequence of odd numbers, whose sum makes a square.

E.g.: 7+ 9 = 4² 17+19 = 6² 11+13+15+17+19+21+23+25 = 12². Validity of LF-theorem means that for a > 2 such subsets does not exist.

 Nikomachos z Gerasy, around y.100, Alexandrian mathematician of Roman period - the most complete extant interpretation of Pythagorean arithmetic is in his work. Discusses the polygonal and pyramidal numbers.

For k=3 differential sequence {1,7,19,37,...} form so called hexagonal numbers (e.g. 7 represents 6 regularly spaced points around one point in the middle, 19 then adds another 12 points around...)

```  0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000
1, 7, 19, 37, 61, 91, 127, 169,  217, 271, ...
```

Shifting over themselves then so-called pyramidal numbers are formed. Number of points in these pyramids are gradually 1,1+7=8,1+7+19=27,..., so n³.

When there are in a differential sequence of k-th powers such a subsets of numbers, that their sum is a k-th power, the congruences must be satistied also for numbers of these subsets. E.g. because 7+ 9 = 4² then also 7 mod r + 9 mod r = 16 mod r for any module r.

Differential sequences of the k-th powers have certain characteristics. E.g. it follows from periodicity of aritmetical sequences for p ε P: a(i)+a(i+p)= −1 (mod p), i.e. in case p=3 we have: 1 + 37 = 7 + 61 = 19 + 91 = 37 + 127 = 2 mod 3 (see segments).
Therefore the question offers: whether some properties for a > 2 can exclude the validity of some congruences.

### Restrictive module

We will try to differentiate modules r depending on the extent to which they restrict the possible solutions. If numbers nk mod r (for given k and all posible values n) does not fill the entire set of numbers 0,1,...r−1, we call the module r restrictive (for order k).

E.g. n³ mod 9 gives only three possible residues 0,1 and 8, n³ mod 10 all possible residues 0,1,2,3,...9. so for order řád 3 is module 9 restrictive, module 10 not.

We are interested solely by restrictive modules, e.g. for k=2 and 3, r=1,2,..16:

``` r h z
──────── k=2
1 1 0
2 2 1, 0
3 2 1, 1, 0
4 2 1, 0, 1, 0
5 3 1, 4, 4, 1, 0
6 4 1, 4, 3, 4, 1, 0
7 4 1, 4, 2, 2, 4, 1, 0
8 3 1, 4, 1, 0, 1, 4, 1, 0
9 4 1, 4, 0, 7, 7, 0, 4, 1, 0
10 6 1, 4, 9, 6, 5, 6, 9, 4, 1, 0
11 6 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0
12 4 1, 4, 9, 4, 1, 0, 1, 4, 9, 4, 1, 0
13 7 1, 4, 9, 3,12,10,10,12, 3, 9, 4, 1, 0
14 8 1, 4, 9, 2,11, 8, 7, 8,11, 2, 9, 4, 1, 0
15 6 1, 4, 9, 1,10, 6, 4, 4, 6,10, 1, 9, 4, 1, 0
16 4 1, 4, 9, 0, 9, 4, 1, 0, 1, 4, 9, 0, 9, 4, 1, 0
```
```  r h z
──────── k=3
1 1 0
2 2 1, 0
3 3 1, 2, 0
4 3 1, 0, 3, 0
5 5 1, 3, 2, 4, 0
6 6 1, 2, 3, 4, 5, 0
7 3 1, 1, 6, 1, 6, 6, 0
8 5 1, 0, 3, 0, 5, 0, 7, 0
9 3 1, 8, 0, 1, 8, 0, 1, 8, 0
10 10 1, 8, 7, 4, 5, 6, 3, 2, 9, 0
11 11 1, 8, 5, 9, 4, 7, 2, 6, 3,10, 0
12 9 1, 8, 3, 4, 5, 0, 7, 8, 9, 4,11, 0
13 5 1, 8, 1,12, 8, 8, 5, 5, 1,12, 5,12, 0
14 6 1, 8,13, 8,13, 6, 7, 8, 1, 6, 1, 6,13, 0
15 15 1, 8,12, 4, 5, 6,13, 2, 9,10,11, 3, 7,14, 0
16 10 1, 8,11, 0,13, 8, 7, 0, 9, 8, 3, 0, 5, 8,15, 0
```

The number h indicates the number of different residues nk mod r. If the number h is smaller, then module r is more effective. The most efficient modules give two residues (0,1), or three residues (0,1 and p-1).

### Trivial solution

If LF-congruence has solution for given module r only when some of numbers Za,Zb,Zc is zero, we speak about trivial solution:

[0,z,z] ──> ak≡0 (mod m), resp. bk≡ck (mod m)
[z,0,z] ──> bk≡0 (mod m), resp. ak≡ck (mod m)

[z,z,0] ──> ck≡0 (mod m), resp. ak≡bk (mod m)

(Solution [0,0,0] is excluded by initial condition, that numbers a,b,c are relatively prime in pairs).
E.g. expression n² mod 16 takes 4 values: 0,1,4,9 and exists only solutions:

0+1 ≡1, 0+4 ≡4 a 0+9 ≡9 (mod 16)
resp. 1+0 ≡1, 4+0 ≡4 a 9+0 ≡9 (mod 16)

Expression n³ mod 9 gives 3 values: 0,1,8 and only solutions are:

0+1≡1, 0+8 ≡8 and 1+8 ≡0 (mod 9)
resp. 1+0≡1, 8+0 ≡8 and 8+1 ≡0 (mod 9)

In the case of trivial solution is either applied m|ak or m|bk or m|ck, i.e. m|(abc)ak. If mεP, then m|abc.

### Modules of Sophie Germain

Proving of the validity of Fermat's theorem would be easier, if there existed for every order k effective modules whose combination would limit or exclude a possibility of solution.
But this is not so, modules, which give a maximum of three residues, disappear with increasing k. For k=2,3,..13 they are just following:

 Germain, Sophie [], 1776-1831, French mathematician. She dealt with solution to the Last Fermat's theorem. Based on her extensive work she have managed to prove the theorem for all k<100, and later also for many other exponents. She worked also on mathematical physics (theory of elasticity). In correspondence with other mathematicians performed under a male pseudonym (Louis Le Blanc).
```  Modules r
─────────────────
k=  2: 2,3,4,5
k=  3: 3,4,7,9
k=  5: 11
k = 6: 7,8,9,13
k= 11: 23
```

In selected pairs (k,r): (2,5), (3,7), (5,11),(6,13), (11,23),... is k = κ(r) a rε P.

This relationship was used by Sophie Germain to prove Fermat's theorem for some selected exponents. She showed e.g. that if a5+b5=c5 then one of numbers a,b,c must be divisible by number 5.

Sophie Germain further dealt with searching for such auxiliary modules, that would eliminate possible solutions (that she managed for each k<100). She has resolve cases where k and also 2k+1 or more generally 2mk+1 is prime.
It means cases when k=κ(p)=κ(p,2), resp. k=κ(p,2m), for k,pεP.
From her relationships follows e.g. that if a7+b7=c7 then one of the numbers a,b,c must be divisible by number 29 (i.e. 4∙7+1),...

### Consequence of binomial sentence

It follows from binomial sentence (see ):

#### M(n,k) ≡ k (mod n−1)

E.g. M(5,3)= 31 ≡ 3 (mod 4), M(5,7)= 19531 ≡ 7 (mod 4).
As a general rule it is: (ak−bk)/(a−b) ≡ k∙ak−1 ≡ k∙bk−1 mod (a−b).
E.g. (7³−2³)/(7−2) ≡ 3∙7² ≡ 3∙2² (mod 5), tj. 67 ≡ 147 ≡ 12 (mod 5).

### Euler's criterium

Euler's criterium k rozlišení kvadratických zbytků a nezbytků využívá vztahu Q(u,r) = uκ(r) mod r, which takes just three values {−1,0,1} (while a value of 0, only when r|n).

If r divides none of the numbers a,b,c then congruence:

aκ(r)+bκ(r) ≡ cκ(r) (mod r)

can be written to form, without a solution:

±1 + ±1 ≡ ±1 (mod r)

### Divisors of numbers a,b,c

Trivial solutions occur in the case of the following modules r:

```k    Modules r
────────────────────────────────────────────────
2    3,4,5,8,9,16,25
3    3,4,7,9,13,27,49,169,343,2187
4    9,13,17,25,27,81,125,169,289,625,1681,2197
5    4,8,11,16,32
```

For k=2 m it follows from modules m=9, 16 and 25, that 3∙4∙5|abc. The Pythagorean triangles have this property, e.g.:{3,4,5},{5,12,13},{7,24,25},{8,15,17},..

Pro další exponenty k>2 plynou z hodnot zbytků obdobné vztahy.

## Power sums

The sequence of Pythagorean triangular numbers Tn {0,1,3,6,10,15,21,28,36,...} appears among the numbers of classes (self, nested or total). Number T5 = 15 is represented by the 15 points of the triangle with the edge of length 5:

```*
* *
* *
*  Tn = = n(n+1)/2
* * * *
* * * * *
```

It applies:

#### T²n−T²n−1 = n³

e.g. T7+T6 = 28+21= 7² = 49 a T²7−T²6 = 7³ = 28²−21²= 7² = 343.

 Bernoulli, Jacob [bernuli], 1654-1705, Swiss mathematician and physicist known for his work on differential calculus and probability. Used polar coordinates, introduced the concept of 'integral'. Along with his younger brother Johann (1667-1748), studied Leibnitz's differential calculus and applied it to a whole range physical problems. From their workshop comes e.g also so called L'Hospital's rule. He contributed to the development of the calculus of variations. Johann's son Daniel (1700-1782) is known for works on hydrodynamics (Bernoulli equation).

Sums of squares ∑sk (for various k) are called Newton totals. After Newton - Jacob Bernoulli dealt with them in detail.

Let's calculate sequentially how much is n x n squares on a chess board and much is there rectangles in general.

E.g. for n=2 we find one larger square, 4 smaller squares and 4 rectangles.

```     n   squares  rectangles
──────────────────────────
1       1       1
2       5       9
3      14      36
```
```    ┼───┼───┼───┼
│   │   │   │
┼───┼───┼───┼
│   │   │   │
┼───┼───┼───┼
│   │   │   │
┼───┼───┼───┼
```

The numbers of squares appear to be the sums of the serie: 1² + 2² + 3² ... The numbers of rectangles are the sums of 1³ + 2³ + 3³ ... and they simultaneously forms powers (squares): 1², 3², 6²,.. i.e. (1)², (1+2)², (1+2+3)²,...

For every n εN, s=1,..,n it applies:

### Power sequences

We are interested in what are generating functions for sequences {0k,1k,2k,3k,.....} for given number k. Generating function f(x)=1/(1−x)² belongs to sequences {1,2,3,4,5...}. But we could start also from the number 0, i.e. search the function for sequence {0,1,2,3,4,5...}.
It is obtained from the original sequence by multiplying x∙f(x) = x/(1−x)².
By multiplying x we gain sequence that is interchangeable with the original sequence (see interchangeable sequences).

Generating functions fk (for particular k):

```Function                    Sequence
───────────────────────────────────────────────────────
f0=x/(1−x)     0, 1, 1,  1,   1,  1,   1,…
f1=x/(1−x)2      0, 1, 2,  3,  4,  5,   6,…
f2=x(1+x)/(1−x)³       0, 1, 4,  9, 16, 25,  36,…
f3=x(1+4x+x²)/(1−x)4   0, 1, 8, 27, 64,125, 216,…
f4=x(1+11x+11x²+x³)/(1−x)5    0, 1,16,  81,256,625,1296,…
f5=x(1+26x+66x²+26x³+x4)/(1−x)6 0, 1,32,243,512,…
```

Next function is always derivation of the previous function, multiplied by x:

#### f[n](x) = x∙ (f[n−1](x))'

Tedy f1 = x∙f0', f1 = x∙f1'= x²∙f0'', ...

E.g.
```f1' =(x/(1−x)²)' = 1/(1−x)²−x(−2)/(1−x)³ = 1/(1−x)²+2x/(1−x)³
f1' =(1−x+2x)/(1−x)³ = (x+1)/(1−x)³
f2  = x∙f1' = x(x+1)/(1−x)³
```

### Euler triangle

Coefficients of polynomials of generating functions for power sequences form the so-called Euler triangle:

``` 1
1 1
1 4 1
1 11 11 1
1 26 66 26 1
1 57 302 302 57 1
```