﻿ The Last Fermat's Theorem - Experiments

# Experiments

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## Fermat coefficients

In an effort to move closer to solutions of LF-theorem could therefore make sense to know the rules for sums of Fermat coefficients. Unlike the sequence of numbers np in a sequence of numbers f(n,p) sums exist. For given p we will write f(n,p) = f(n).

### Sums of Fermat coefficients

In case pεP the numbers np indicate total number of instances in systems G(n,p). This number is approximately p times the number of self classes and numbers of self classes correspond to Fermat's coefficients. f(n,p) = (np−n)/p.

For p=2:

```(a,b,c)       f(a)+f(b)= f(c)  (a,b,c)      f(a)+f(b)= f(c)
─────────────────────────────  ───────────────────────────
( 4,  6,  7)    6 + 15 = 21    ( 8, 28, 29)   28 +378 =406
( 5, 10, 11)   10 + 45 = 55    ( 9, 11, 14)   36 + 55 = 91
( 6,  7,  9)   15 + 21 = 36    ( 9, 36, 37)   36 +630 =666
( 6, 15, 16)   15 +105 =120
( 7, 10, 12)   21 + 45 = 66
( 7, 21, 22)   21 +210 =231
```

For p=3: For p=5:

```(a,b,c)     f(a)+f(b)= f(c)    (a,b,c)  f(a)+f(b)= f(c)
───────────────────────────    ────────────────────────────
( 9, 15, 16)  240+1120 =1360   (13,16,17) 74256+209712=283968
(21, 55, 56) 3080+55440=58520
(31, 56, 59) 9920+58520=68440
```

For higher values of Fermat coefficients it shows that to meet the equation f(a) + f(b) = f(c) both addends must be large enough. Values f(a), f(b) and therefore also a,b gradually flatten (to close levels). In the case of p = 2 for the solution (13, 78, 79); ie. 78 + 3003 = 3081, gives the ratio (a + b)/c the value of 1.15190.

Another solution is close to the value (a+b)/c = √2 = 21/2:

```(1728,1768,2472)     1492128+1562028 =3054156  1,41424
(1738,1768,2479)     1509453+1562028 =3071481  1,41428
(1740,1794,2499)     1512930+1608321 =3121251  1,41417
(1751,1765,2486)     1532125+1556730 =3088855  1,41432
```

Similarly, in the case of p=3 ratio (a+b)/c grows towards the value 22/3.

```( 923,1287,1429) 262109848+ 710581872 =972691720  1,54654
( 969,1002,1242) 303284080+ 335337002 =638621082  1,58696
(1352,1479,1787) 823774952+1078407920 =190218287  1,58422
```

Generally, from the relationship 2∙f(a) = f(c) it follows: a/c = 21/p tj.2a/c = 2(p−1)/p.

## The decisive sequences

### Sums in decisive sequences

We will call decisive sequence - the first difference sequence of a given arithmetical sequence. E.g. in the sequence of second powers it is the sequence R2 = {1,3,5,7,9,11,13,...} and in the sequence of third powers R3 = R3 = {1,7,19,37,61,91,127,.....} We are interested in what must be addends (of elements of decisive sequence), so that their sum is the k-th power of a number..

Eg. in sequence R3 (k=3):

```Twoo addends:
-------------
3781+4219=8000 = 20^3
19+17557=17576 = 26^3
7651+25117=32768 = 11347+21421=32768 = 32^3
...
```
Numbers ...,20,26,32,... are of the form 2 + 6n = 2 + 2kn
```Three addends:
-------------
7+91+631=37+61+631=61+271+397=91+169+469=127+271+331=...=729=9^3
7+1261+2107=19+919+2437=37+169+3169=37+547+2791=37+1387+1951=...=3375 = 15^3
19+331+8911=61+3781+5419=91+1519+7651=...=9261 =21^3
```
Numbers ..,9,15,21,... are of the form 3 + 6n = 3 + 2kn

A similar result is obtained for 4, 5 or more addends. Generally, for p addends, the resulting power has the form:

#### (p+2kn)^k

.

This is simply the consequence of the fact that if a^2+b^k=c^k, k ε P then a+b-c = 2kq, q ε N.