The abc conjecture

K≧1, ε>0

---

The abc conjecture is also based on prime numbers.
It must be abc triples which are  (a,b,c)=a*b*c, a<b, and a+b=c.
You can ignore numbers multiplied by the same.
For example, when there are a=1, b=80, c=81, you see 80=2^4*5, and 81=3^4.
The radical of (1,80,81) is 1*80*81=1*2*5*3=30
It must be less than c.
rad(1,80,81)=30<81.
This is the rule of abc triples.
It is said that you can find abc triples infinitely.
When you see a=1, b=9^n -1, c=9^n, you can't break the rule.
1+9^n -1=9^n. 1<9^n -1
Therefore, (1, 9^n -1, 9^n) is abc triples, if it is less than 9^n.
n=1, rad(1,8,9)=6
n=2, rad(1,80.81)=30
n=3, rad(1,728,729)=546
・
・
・
n would be infinite.

Then you can see quality of the triple.
rad(a,b,c)^q=c
q=log(c)/log(rad)
It is said that quality of abc triple is greater than 1.
q>1

The quality of abc triples is not stable.
It keeps waving, but each dot is finite.

This is the abc conjecture.

You can see the same pattern from prime numbers.

This is the sum of prime numbers in each 1000.

You can find the prime numbers infinitely, but the part is chaotic.

log(rad) must be in the fractal, because fractal multiplied by any numbers is still fractal.
c=9^n is expansion, so log(c)/log(rad) is in fractal.
1 is just the bottom line.

---

Finally, K is arbitrary, so you can adjust c＜X.

ε is a small number which is almost zero.

rad(a,b,c)^q=c
q=log(c)/log(rad)
q>1

c>rad(abc)
This is infinite.

∴

q > 1+ε, and q is not stable.

rad(1,8,9)=6 is q≒1.226294386. It is apparent that 9〈 6^1.3≒10.27061916

rad(1,80.81)=30　⇒　q≒1.29203003　⇒　81〈 30^1.3≒83.22573344

rad(1,728,729)=524　⇒　q≒1.045862642　⇒　729〈 524^1.1≒ 980.0865193

You exclude q < 1+ε.

rad(2,3,5)=30　⇒　q≒0.473197445

rad(2,5,7)=70　⇒　q≒0.45802338

rad(7,11,18)=462　⇒　q≒0.471084865


∴

Chaos and Fractal

There is an impossible triangle.

a+a=2a, so you may say that it is possible. However, it is impossible.

a^2+b^2=a^2, so b^2=0. The height must be zero. It is impossible.

---

a+b=c, and b must be a. As I mentioned, it is ridiculous. However, if a+a=c=2a were possible, you would need to put fractal triangles in it.

a*=b*=a/2. c*=c/2=a

These are fractal triangles.
a* would be a/2,a/4,a/8,・・・a/2^x.
c* would be a,a/2,a/4,・・・a/2^x-1.
2a*=(a/2^x)×2=a/2^x-1=c*=a*+a*.
a*=a/2,so a/2+a/2=a.
∴a+a=2a=c
Therefore, an impossible triangle is possible. When x is ∞, a*=a/2^∞ would be close to zero. If you saw the zero point, you could see the line which includes infinite triangles. It must be fractal.

Square

Prime numbers are also expressed as squares like ancient Babylonian. I found the pattern of prime numbers which were based on fractal. At first, any prime numbers are squared to find none prime numbers.

3^2=9

Then you pile a rectangle. 3*2=6. 9+6=15
You continue to pile rectangles. 15+6=21,21+6=27,27+6=33

You can expand prime numbers.

Then I put numbers in each squares and rectangles.

Any prime numbers which are squared are the sum of numbers in rectangles and squares.
For example, 7^2=9+6+6+10+4+14=49.
11^2=7^2+6+6+4+4+4+4+44=121.
Moreover, you can put the Pythagorean Theorem in it.
7^2=5^2+(√24)^2=25+24=49
11^2=7^2+(√72)^2=49+72=121

Ancient Babylonian mathematics

I have read Hidden Harmonies: The Lives and Times of the Pythagorean Theorem , and I'm interested in Ancient Babylonian mathematics. Ancient Babylonians seem not to have the concept of algebra, but they know how to use geometry. Now we have studied the Pythagorean Theorem, and we use it as it is. However, ancient Babylonians seem to have the almost same stem of it. These are their geometries.

They look like piling squares. You see two of them which are white and blue.

Then, you put dots in each squares.

There are two squares which are purple and blue. You see a dot in the purple square. You also see four dots in blue. However, you distinguish the colors, so there are three dots in blue. It looks like L shape.

Then you see three squares. Therefore, you can add 5 dots in green.

This is the infinite pattern.You would add odd numbers in infinite shape of L. Ancient Babylonians have counted dots.
1+3+5=9=3^2
1+3+5+7=16=4^2
1+3+5+7+9=25=5^2
1+3+5+7+9+11=36=6^2
1+3+5+7+9+11+13=49=7^2
1+3+5+7+9+11+13+15=64=8^2
1+3+5+7+9+11+13+15+17=81=9^2
∞

Now we know the Pythagorean Theorem which is a^2+b^2=c^2. See Ancient Babylonian mathematics.

3^2+4^2=5^2

You can find the same. Of course, you have to search all numbers in Ancient Babylonian mathematics to fit the Pythagorean Theorem because there is no concept about square root in it.



Now we can put algebra into Ancient Babylonian mathematics.

There are three squares which are white, blue, and green. You see also two rectangles.
Therefore, (a+1)^2=a^2+2a+1.
when you put square root in it, you would see the Pythagorean Theorem.

(a+1)^2=a^2+(√2a+1)^2

a must be dots, and 2a+1 is like L shape. You count dots.


In this case, a+1=3, a=2, 2a+1=5.
Therefore, 3^2=2^2+(√5)^2
9=4+5

π(x)

In order to know Riemann's zeta function, you need to understand how to count prime numbers. Basically, there is no pattern to find prime numbers, but when you decide the realm of integer such as 100 or 10000, you can pick up prime numbers and count them. Visual basic is useful. The program is based on sieve of Eratosthenes.π(x) means the sum of prime numbers until x. When x is 5, π(x)=3. The graph is like this.

π(10a) means that x are 10,20,30,40・・・・. You count prime numbers infinitely with the same pattern.

If you practiced π(x), you could understand J(x) and the golden key.

J(x)=π(x)+1/2π(√x)+1/3π*root(3, x)+1/4π*root(4, x)+1/5π*root(5, x)+・・・

1/s*logζ(s)=∫[0→∞]J(x)*x^(-s-1)dx

I also count prime numbers which are the same with π(x).