I was fascinated by Fermats conjecture that there are no integer numbers x,y,z and x·y·z >0 such than xn+yn=zn for n >2.
Richard Feynman tried to prove this conjecture (now proven): His idea was to use a probabilistic theory approach and show that the probability for a triple x,y,z to satisfy the equation is extremely small. Of course this is not a proof.
When I was around 14 years old a tried a similar approach trying to filter out combinations and hoping that at the end I can get all of them so that finally the conjecture is proved.
I observed that there is a property which I call the decimal sum (or digit sum) which I called L(x) or L-value of an integer x:
For example L(45634) = 4+5+6+3+4=22 and L(22) = 2+2 = 4. So finally L(45634) = 4. The L function is applied until a single digit comes out. I found that numbers with a L value = 1,4 or 7 form a subgroup under multiplication with the following Cayley table:
|
* |
1 |
4 |
7 |
|
1 |
1 |
4 |
7 |
|
4 |
4 |
7 |
1 |
|
7 |
7 |
1 |
4 |
Looking at the properties of this group I could proove that for integers a,b,c,n,m,k> 0
a2n + b2m = c2k can be only true if abc º0(mod3)
and also only true if abc º0(mod5)
The main group is formed under the multiplication * defined as L(a)*L(b) = L(a*b)
Its Cayley table is:
|
* |
1 |
2 |
4 |
5 |
7 |
8 |
|
1 |
1 |
2 |
4 |
5 |
7 |
8 |
|
2 |
2 |
4 |
8 |
1 |
5 |
7 |
|
4 |
4 |
8 |
7 |
2 |
1 |
5 |
|
5 |
5 |
1 |
2 |
7 |
8 |
4 |
|
7 |
7 |
5 |
1 |
8 |
4 |
2 |
|
8 |
8 |
7 |
5 |
4 |
2 |
1 |
Similar from this group it follows that if a º/º0(mod3) then L(a6n) = 1, for n>0
Other results are:
a4k+2 + b4m+2 = c4n+2 is not true if abcº/º0(mod5)
I have found other groups such as the numbers
(50k ± a)2 with a Î {4,6,14,16 ,24}
These numbers form a group under multipilcation. Also the number 4,6,14,16 and 24 form a cyclic abelian group under multiplication *. The unity element is 24.
|
* |
4 |
6 |
14 |
16 |
24 |
|
4 |
16 |
24 |
6 |
14 |
4 |
|
6 |
24 |
14 |
16 |
4 |
6 |
|
14 |
6 |
16 |
4 |
24 |
14 |
|
16 |
14 |
4 |
24 |
6 |
16 |
|
24 |
4 |
6 |
14 |
16 |
24 |
For example 4*14 = 6 since 4·14º6(mod50)
If we define the product a*b = c where abºc(mod50)
It follows that (50k ± a)10nº24(mod50)and L((50k ± a)10n) = 7.
Of course L(a) = b is equivalent to aºb(mod9) if a =/= 0
For all integer numbers a Î N can be written as a= 9n+i. Three sets L0, L1, and L2 can be then defined
L0 = {x | x º i(mod9) and iº 0(mod3) }
L1 = {x | x º i(mod9) and iº 1(mod3) }
L2 = {x | x º i(mod9) and iº 2(mod3) }
It follows that and L1 and L2 form a group and L1 is a subgroup. Therefore if x Î L1 or L2 then
x6n = 1 i.e
x6n º 1(mod3)
x6n º 1(mod9)
L0 forms a group under addition with the following Cayley table:
|
+ |
0 |
3 |
6 |
|
0 |
0 |
3 |
6 |
|
3 |
3 |
6 |
0 |
|
6 |
6 |
0 |
3 |
L(x) = j if x º j(mod9).
The addition is defined by the corresponding L value
Example:
For 12+5 = 17 we have L(12) + L(3) = 3+5 = L(17) = 8. Therefore we write 3+5=8
and multiplication
12·5 = 60
i.e.
L(12) L(5) = 3*5 = L(60) = 6. Therefore we write 3*5 = 6
It follows that
a6k+5 + b6l+5 =/= c6m+5 if a·b Î L2
Combining various such groups more complex properties can be found for multiplication and summation of powers.
Undergraduate Texts in Mathematics
Graduate Studies in Mathematics
