The abc conjecture

K≧1, ε>0

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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.

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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


∴