Some conclusions

1. The numbers which are multiples of 3 are “invisibility cloaks“ of the true value of numbers because in any non-first power only they transform any number in “packed form“ into 9.

Table 1.3

If we remove the lines that are multiples of 3 from the Table 1.3, we will get a square Table 2 in which the rows and columns of the “archived“ values use the same series of numbers:

а) a series of number 1 corresponds to the column a6

b) a series of numbers 2 and 5 corresponds to the columns  а and a5

c) a series of numbers 4 and 7 corresponds to the columns  a2 and a4

d) a series of number 8 corresponds to the column  a3

Table  2

2.   Let us consider the disposition of “packed“ numbers in the equation аn + вn = cn  , where:

1.1   а , в and c and n  are any finite numbers        

1.2   аn , вn and cn  take on one of the values: 1, 2, 4, 5, 7, 8, 9

1.3  In accordance with the assertion 5., we shall consider the values аn , вn  and cn   for n from 2 to 7.

А.     Let us analyze the column a2 in the table 1.3 for the equation а2 + в2 = с2            

To do this, we shall make a table:

In accordance with the column а2 of the table 1.3, а2 and  в2 have to take on one of the possible values:  1, 4, 7, 9;  besides, their sum c2 also has to take on only these same values.

Table 3.0

In table 3.0, bold font is applied to combination а2 + в2  which allows solution in integers.

Meanwhile, for every finite value of c (from 1 to ∞ ) а and b have a solution in finite numbers and sometimes in different combinations.

Example 10:        

     162 + 632  =  652   =  562 + 332 

Feature of the numbers which are multiple of 3:

1.  Combinations а=3 and в=3, а=3 and в=6,  а=6 and в=6;  

Their ”true” value is hidden behind the “invisibility cloaks“ because in both natural and “packed“ state they are can be divided by 3 only once and without the remainder which is multiple 3. While solving а2 + в2 = с2  , we shall factor out the number 32 = 9:

                9(1+1) = 9х2

                9(1+4) = 9х5      

                9(4+4) = 9х8

These examples can be extended, taking into account the assertion 9, but in these combinations, values inside the brackets will never be equal to 1, 4, 7 or 9.

Removing  “invisibility cloaks”,  which are equal to 9, in the obtained values of с2, we shall obtain values 2, 5, 8 which are not present in the column a2 of the table 1.3 ,  which means that these combinations do not allow solutions in integers.

2.  Combinations  а=3 and в=9, а=6 and в=9;

Their “true” value is also hidden behind “invisibility cloaks” because in both natural and “packed” state they are divisible by 3 with the remainder which is multiple of 3 in one of the component. While solving а2 + в2 = с2,  we shall factor out the number 32 = 9:

                9(1+9) = 9х1

                9(4+9) = 9х4      

These examples can be extended, taking into account the assertion 9, but in these combinations, values inside the brackets will be equal to 1, 4 or 7.

Removing “invisibility cloaks“, which are equal to 9, in the obtained values of с2, we shall obtain values 1, 4 or 7, which are present in the column a2 of the table 1.3 , which means that these combinations allow solutions in integers.

3. Combination а=9 and в=9

Allows multiple division by 9 according to the assertion 9. Solution is allowed only if their multiplicity in initial numbers do not match, and this automatically transfers it to the existing variants, i.e.:

              а)  9(а2+ в2 )    – no solution  

              в)  9(9а2 + в2 ) – possible solution where

            в1)  9а2 + в2  Ξ ( 3а )2 + в2  Ξ   ( 6а )2 + в2  Ξ ( 9а )2 + в2   - possible solution

Summary table 3.0 for с2 is as follows:

Table 3.1

Assertion 10.0.

For a possible solution of the equation а2 + в2 = с2  in integers, it is necessary that:

а)  one of the components (а2) is equal to 9

b)  the other component (в2) is equal to one of the numbers: 1, 4, 7, 9;  

In case both components are multiple of 9, we shall separate this multiplicity by factoring out 9к  until such multiplicity is separated. Then, the assertion 7.0 is simplified into:

Assertion 10.1.

For a possible solution of the equation а2 + в2 = с2  in integers, it is necessary that:

а)  one of the components (а2) is equal to 9

b)  the other component (в2) is equal to one of the numbers: 1, 4, 7;  

Example 11.1:

(12345)2 + (6789)2  =  32 x (4115)2 +  32 x (2263)2 Ξ  32 x (2)2 +  32 x (4)2 Ξ  9 x (4 + 7)  Ξ 9 х 2   –

solutions in integers are not allowed.

Example 11.2:

(264)2+ (6768)2  =  32 x (88)2 + 32 x (2256)2 Ξ  32 x (7)2 + 32 x (6)2 Ξ  9 x (4 + 9) Ξ 9 х 4  

solutions in integers are allowed.

В. Having analyzed, by analogy with а2 + в2 = с2 , all subsequent powers up to the 7th, we obtain values of components which allow solutions in a finite value. Data can be taken from the Table 1.3. To prove FLT (Fermat’s Last Theorem), we need to prove that such solutions do not exist.

For  a3 + в3 = с3 :

( 1 + 8 = 9 ),  ( 1 + 9 Ξ 1 ),  ( 8 + 9  Ξ  8 ),  ( 9 + 9  Ξ  9 )

For    a4 + в4 = с4: 

( 1 + 9 Ξ 1 ),  ( 4 + 9 Ξ 4 ),  ( 7 + 9 Ξ 7 ),  ( 9 + 9 Ξ 9 )

For    a5 + в5 = с5:      a7 + в7 = с7:

( 1 + 1 = 2 ), ( 1 + 4 = 5 ),  ( 1 + 7 = 8 ), ( 1 + 8 = 9 ), ( 1 + 9 Ξ 1 ),  ( 2 + 2 = 4 ),  ( 2 + 5 = 7 ),  ( 2 + 7 = 9 ),  ( 2 + 8  Ξ 1 ), ( 2 + 9 Ξ 2 ),  ( 4 + 4 = 8),  ( 4 + 5 Ξ 9 ), ( 4 + 7 Ξ 2 ), ( 4 + 9 Ξ 4 ),  ( 5 + 5 Ξ 1 ),  ( 5 + 8 Ξ 4 ),  ( 5 + 9 Ξ 5 ), ( 7 + 7 Ξ 5 ),  ( 7 + 9 Ξ 7 ), ( 8 + 8 Ξ 7 ), ( 8 + 9 Ξ 8 ),  ( 9 + 9 Ξ 9 ),  

For    a6 + в6 = с6 :

( 1 + 9 Ξ 1 ),    ( 9 + 9 Ξ 9 )

С.   Values of  а2 + в2 = с2   and   a4 + в4 = с4  match and, upon subsequent squaring, loop: (1, 4, 7, 9;  1, 7, 4, 9; 1, 4, 7, 9;……)  

( 1 + 9 Ξ 1 ),  ( 4 + 9 Ξ 4 ),  ( 7 + 9 Ξ 7 ),  ( 9 + 9 Ξ 9 )

and this means that one of the variants of the proof of   а4 + в4 = с4  

can be impossibility of extraction of the root from biquadrates in finite numbers:

      ( а2 )2+ ( в2 )2= ( с2 )2  =  а4 + в4 = с4    

 D. This analysis takes into account the multiplicity of powers, and since а2 + в2 = с2   is solvable, variants in which сn takes on a value of с2 

                с26 Ξ с8  Ξ с2          

shall be considered as the extraction of the root from cube:

  ( а )26+ ( в )26= ( с )26  Ξ  а8 + в8 = с8  Ξ ( а2 )3+  ( в2 )3= ( с2 )3   Ξ  х3 + у3 = z3        

Е. One of the answers to the question regarding the proof (FLT) may be an answer to the question why while solution for   а2 + в2 = с2   exists, not all initial numbers from ”packed” series 1, 4, 7, 9 provide these solutions.

Regards

Anatoli

anatoli@numbers.ee