錣（shi-ko-ro) 60 Stars Astrology Season Holiday Extra 8 of 3 Combinations of Numbers for Measuring Something part 7 " Memories of the Journey with Fermat Numbers "

60 STARS ASTROLOGY

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SEASON HOLIDAYS "An Extra"

錣（shi-ko-ro) 60 Stars Astrology  Season Holiday Extra 8 of 3

Combinations of Numbers for Measuring Something part 7

" Memories of the Journey with Fermat Numbers "

   

1. Well, I've been quietly continuing my journey with Fermat numbers until today, and something got me thinking.......


it's about that so-called "ABC Conjecture".  


Of course, Tanu-chan, is a liberal arts student and a complete amateur!

🌟　 🌟　 🌟

The "ABC Conjecture" states that  for mutually prime natural numbers (a, b, c) that satisfy  "a + b = c",  

when we define "d" = rad ( abc ) ......that is, when we define "d" as the combination of prime factors (without repetition) obtained by prime factorizing "a", "b", and "c" 

then,  c < d^(1+ε) ..... Probably c < d²

This is that conjecture, maybe!


For example, 

If a and b are 5 and 7,  then "c" = a + b = 12.

abc = 5 × 7 × 12 = 2² × 3 × 5 × 7 

So "d" = rad (abc) = 2 × 3 × 5 × 7 = 210.

"d" is clearly larger than "c"; this is the usual case.

....However, occasionally "d" is slightly smaller than "c".  

For example, if a and b are 1 and 8, then "c" is 9.  
abc = 1 × 8 × 9 = 1 × 2³ × 3²,  so  "d" = rad (abc) =2 × 3 = 6.  
......In this case, "d" < "c".




Also, for example, if we calculate with different numbers:
If a and b are 11 and 13, then "c" is 24.

abc = 11 × 13 × 24 = 2³ × 3 × 11 × 13,  so "d" = rad(abc) = 2 × 3 × 11 × 13 = 858.

....Clearly, "d" is larger than "c".


【While there are cases where "c" is larger than "d", they are very few.
And no matter how large "c" is, it remains within "c" < "d²".】

This is the "ABC Conjecture".
......Well, probably.....It's probably something like this.



2. A Japanese professor researching this presented the idea to the academic community that the "ABC Conjecture" requires separating addition and multiplication. 

This has caused major confusion worldwide.


Now, returning to Fermat numbers,

F(n) = 2^{2^n} + 1.

Considering what “power” means here, we see there are different types of powers.

Type A exponentiation is like this: a ⇒ a² ⇒ a³ ⇒ a⁴
This type increases in a summation-like manner.


Type B exponentiation is like this: a ⇒ a² ⇒ (a²)² ⇒ ((a²)²)²
This type multiplies the larger number again and again, making it grow bigger and bigger. It's the multiplication type.

The Fermat numbers F(n) are like the following:
F(0) = 2^(1) + 1 = 3
F(1) = 2^(2) + 1 = 5
F(2) = 2^(4) + 1 = 17
F(3) = 2^(8) + 1 = 257

F(4) = 2^（16）＋1=65537


So, yeah, it's a B-type power.

By the way, well, there's another type, the C-type power, which grows even faster:
a ⇒ a ^a ⇒ a ^a ^a ⇒ a ^ a ^ a ^a

Fermat numbers are 2^{2^n} + 1, so they resemble type C, but they're a bit different. They're of the b-type.




3. So, in terms of primes, powers (squares), and distinguishing between multiplication and addition, I feel like there might be some commonality between the "ABC conjecture" and "Fermat numbers".　

Well, this is just a guess, as a raccoon dog—a small animal....but I think things like twin primes are surely related too.   

The distinction between multiplication and addition... loosening the tight coupling between multiplication and addition...  

So I tried reading a book on IUT theory aimed at the general public. Of course, I couldn't understand it at all from a certain point on.   


...........But it looks big fun ! ! ! 

If a = 5 and b = 27, then   "c" = 5 + 27 = 32   
abc = 5 × 27 × 32 = 2⁵ × 3³ × 5 = 4320
"d" = rad (abc)  = 2 × 3 × 5 = 30

Oh!  "c" is bigger than "d"!  

How strange!     


....And so, my memories of the journey became a beautiful photograph , together with cutting-edge theory, maybe!  



That's all for today.  


Tanu-chan💓  TOKYO-TANUKI💛