Question

Let's learn and try to prove the ABC conjecture! (No prior knowledge of Inter-Universial Teichmuller theory necessary!)

Answer 1

Ok what the hell is the ABC conjecture? First we have to define a function,

\[rad(n)\]

All it does is gives you a number with the exponents on its prime factorization thrown away leaving only 1s left, like this:

\[rad(12)=rad(2^23^1)=2^13^1=6\]

Now we're almost ready to state the conjecture. We start with two numbers \(a\) and \(b\),

\[a+b=c\]

and here's the conjecture!

\[c> rad(abc)^k\]

This statement is true for only finitely many numbers for any choice of \(k>1\)

Answer 2

@FireKat97 Ok I'm not sure I fully understand this I guess I'm going to try throwing in random examples to play with it to try to figure it out. :P

Answer 3

Ok something interesting is if a and b don't share factors, then neither does c, so they're all relatively prime:

\[\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=1\]

Since that's true I think we can separate out the rad function this way since it won't matter:

\[rad(abc)=rad(a)*rad(b)*rad(c)\]

Answer 4

Anybody interested in trying to figure this out or have any ideas of interesting stuff? I don't wanna feel like I'm yelling in an echo chamber lol.

Answer 5

does it help if I say 'everything above makes sense'? xD

Answer 6

what do u mean its only true for finitely many numbers for some choice k > 1

Answer 7

arent there an infinite nuumber for c, that we can pick

Answer 8

Prove it dan, then you'll have found a counter example to the abc conjecture lol

Answer 9

https://en.wikipedia.org/wiki/Abc_conjecture#Formulations

Answer 10

like suppose
i pick
c^n+1, and a=1,b=1,
then rad(abc)= c
and
c^n+1 = c^k
now i can pick n=k
and get an infinite number right

Answer 11

No because a+b=c

Answer 12

ohhhhhhhhhhh

Answer 13

ok now i see why there could be finite okay!

Answer 14

Haha yeah I feel like this is related to the arithmetic derivative, also did you notice the rad(n) function is multiplicative? Another fun fact although probably useless is this identity you could call it with the mobius function.
\[\mu(rad(n)) \ne 0\]

Answer 15

true

Answer 16

lol

Answer 17

I was looking at this \[c=p^{k+1}\] so that we have

\[p > rad^k(ab)\]

Answer 18

oo ok ok

Answer 19

a + b = c
and a*b <c

Answer 20

I guess this probably always false maybe I should have just written this out:

\[c=p^k\]

\[p^k > rad^k(abp^k)=rad^k(ab) p^k\]
\[1 > rad(ab)\]

So like because of stuff like this there are infinitely many cases where I see that it must be false it's just proving that there are finitely many cases where it's true, kinda weird.

Answer 21

c=p^(k+1)

p > (ab)^k

Answer 22

That's only true if a and b are primes

Answer 23

\[rad^y(p^x)=p^y\]

It's kinda like a prime renamer huh interesting.

Answer 24

a+b = p^(k+1)
p=root_{k+1} (a+b) <--- this being an integer has some big restrictions

Answer 25

wait what do you mean by root_{k+1}

Answer 26

K+1 th root of A+B

Answer 27

```
\[ \sqrt[k+1]{a+b}\]
```
\[ \sqrt[k+1]{a+b}\]

Answer 28

yea lol

Answer 29

Oh it has some big restrictions but it's not a big deal that it does.

Answer 30

Just start here:

\[p^{k}=\frac{p^k+r}{2} + \frac{p^k-r}{2}\]
there are always going to be multiple choices of r that work.

Answer 31

u there?

Answer 32

Yeah I'm here I was watching some youtubes what's up

Answer 33

why are you doing that expression with r

Answer 34

I think I kinda made that weirder than it needed to be, I was just showing we can always find a value of a and b that work,

\[c=p^k\]\[a=p^k+r\]\[b=p^k-r\] as long as r isn't divisible by p, then all 3 numbers are relatively prime.

Answer 35

ohh i see! ok ok

Answer 36

cause you were trying to say that there are restrictions or whatever. I guess I was looking for an infinite number of counter examples I think I'm partly confused too haha

Answer 37

/2

Answer 38

Yeah ok the divide by 2 has to be in there right I forgot

Answer 39

okay hmm and this is can even be the maximum one to find

Answer 40

this should produce the largest or close to largest answers from

Answer 41

rad(abc) for small rs

Answer 42

I was thinking that but I realized it's flawed

Answer 43

cause it's not so much the size of the number it's the amount of unique primes in the number that matters more

Answer 44

oh thats ttrue

Answer 45

Also this way isn't really the only way we can pick numbers a and b where they're some distance from the middle, we could pick like

\[p^k = (p^k-q^r)+(q^r)\] or something like this idk

Answer 46

Ooooh ok I got an idea

\[p^k = q \# + (p^k-q \#)\]

where q# is the largest primorial not exceeding \(p^k\) this should help us maximize the rad function!