- Empty

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

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- Empty

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 \(k>1\)

- Empty

@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

- Empty

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

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## More answers

- Empty

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.

- FireKat97

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

- dan815

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

- dan815

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

- Kainui

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

- Kainui

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

- dan815

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

- Kainui

No because a+b=c

- dan815

ohhhhhhhhhhh

- dan815

ok now i see why there could be finite okay!

- Kainui

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.
\[\mu(rad(n)) \ne 0\]

- dan815

true

- dan815

lol

- Kainui

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

- dan815

oo ok ok

- dan815

a + b = c
and a*b

- Kainui

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.

- dan815

c=p^(k+1)
p > (ab)^k

- Kainui

That's only true if a and b are primes

- Kainui

\[rad^y(p^x)=p^y\]
It's kinda like a prime renamer huh interesting.

- dan815

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

- Kainui

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

- dan815

K+1 th root of A+B

- Kainui

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

- dan815

yea lol

- Kainui

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

- Kainui

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.

- dan815

u there?

- Kainui

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

- dan815

why are you doing that expression with r

- Kainui

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.

- dan815

ohh i see! ok ok

- Kainui

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

- dan815

/2

- Kainui

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

- dan815

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

- dan815

this should produce the largest or close to largest answers from

- dan815

rad(abc) for small rs

- Kainui

I was thinking that but I realized it's flawed

- Kainui

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

- dan815

oh thats ttrue

- Kainui

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

- Kainui

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!

- dan815

|dw:1444290020985:dw|

- Kainui

Actually one small thing to fix, is if that's true then this definitely means q# is divisible by p... Or maybe it means its likely, so I'll just write:
\[\frac{q \#}{p}\]

- Kainui

Oh you can't do that cause we require that \(\gcd(a,b)=1\) and here they're both divisible by 2 so that won't work

- dan815

aww dang

- Kainui

but I like where you're going with that, for a second I was like about to throw my crap down and start working on that cause mixing addition and multiplication like that would be cool. Maybe we can use that idea though somehow.

- dan815

ya ppretty much had an answer without gcd condition because uc an jsut do
2^(A+1) > 2^3K from there

- dan815

but i guess thats why this inequality has finite because of that exact gcd reason

- dan815

how about something like that though lets take smallest possible primes

- dan815

2^a + 3^b= 2^N

- dan815

does c also have to be gcd 1 or just a and b

- Kainui

Ok this looks fun and useful. Also if \(\gcd(a,b)=1\) then this implies \(\gcd(a,c)=\gcd(b,c)=1\) so you're good.

- dan815

oh that makes sense

- dan815

its easy to see especially if u have p1^a + p2^b, if its the sum of 2 diffrent prime exponents

- Kainui

proof, suppose b and c share a factor in common, then you can write
b=f*m
c=f*n
a+b=c
a+f*m=f*n
a=f*(n-m)
donezo

- Kainui

Ok I took your thing and generalized it slightly, plug in 2 and 3 if you want:
\[p^a+q^b=c\]
This will give us
\[c > rad^k (p^aq^bc)\]
\[c > p^kq^k rad^k(c)\]
Maybe moving the c's to the same side is nice:
\[\frac{c}{rad^k(c)} > p^kq^k\]
Or taking the root:
\[\frac{\sqrt[k]{c}}{rad(c)} > pq\]
Just things to look at that are kinda useful maybe.

- dan815

is this ever possible

- dan815

p1^a+p2^b=p3^c
where p1 p2 and p3 are all primes

- dan815

2+3=5 is 1, where abc =1

- Kainui

\[3^2+2^4=5^2\]

- dan815

ok i like this

- Kainui

yeah I was also thinking like maybe we can find some sorta weird contradiction if we look at like
\[x+y+z=c\]
and then we group them into 3 different ways to form a and b and compare them.

- dan815

do abc have to be positive

- Kainui

Yeah abc have to be positive integers but k just has to be a real number greater than 1

- dan815

hmmmmmmmmmm

- dan815

suppose we find a bound

- dan815

like c>rac(abc)^1 and c<=rad(abc)^2
then we are done right

- dan815

well i guess just finding any possible one is good enough then xD

- dan815

ok wait i guess proving there are an inifite right hand sides is not what they want dang it

- Kainui

In an attempt to find a contradiction:
since \(q\#\) maximizes the rad function,
\[max = rad(q \#)\]
I pick a+b=c (dividing out p from the primorial so I don't have factors in common)
\[ \frac{q\#}{p} + (p^n-\frac{q\#}{p})=p^n\]
We get:
\[p^n > rad^k(\frac{q\#}{p}(p^n-\frac{q\#}{p})p^n)\]
\[p^{n-k} > rad^k(\frac{q\#}{p}) rad^k(p^n-\frac{q\#}{p})\]
So to find a contradiction we can show that this is true for an infinite number of cases?

- dan815

i think procing it is easier then

- Kainui

Yeah actually I keep getting confused on what the hell it is we're trying to hsow

- dan815

we know that rad(abc) will have a base > rad(c )
so at some point
rad(abc)^k > c, for some k ,after k is a certain value this will always be true

- dan815

now we gotta show until k k there will be finite solutiond and that is true because we are allowing only integer abcs

- dan815

the last statement is like no work and guess, but i think we have to be able to sho that statement

- dan815

lol i am back to begining again i see

- Kainui

I got it I think i mixed myself up here, this is what I want to do:
\[(\frac{p_n \#}{2}-2^x)+2^x = \frac{p_n \#}{2}\] This will maximize and minimize what we want I think haha.

- dan815

okay lemme see what is the number sign again

- Kainui

Really like all this means is:
\[\frac{p_n \#}{2} = 3*5*7*11*...*p_{n-1}*p_n\]
Now let's do some abc's

- dan815

okk

- dan815

how come u wanna maximize rad function

- dan815

oh for contradiction okay

- dan815

how are we gonna minimize rad function

- Kainui

\[\frac{p_n \#}{2}> rad^k((\frac{p_n \#}{2}-2^x)2^x\frac{p_n \#}{2})\]
\[\frac{p_n \#}{2}> rad^k(\frac{p_n \#}{2}-2^x)2^k \frac{(p_n \#)^k}{2^k}\]
Wait...
\[1> 2 (p_n \#)^{k-1} rad^k(\frac{p_n \#}{2}-2^x)\]
This is the opposite of what I wanted to do, this is clearly like a bad case to look at rofl

- Kainui

Yeah to minimize the rad function we have to use powers of 2 and 3 I think like you were doing earlier haha. In my mind I was trying to maximize c, but the way I did it ended up not working out the way I intended.

- dan815

there are only finite solutions for that apprently

- dan815

how about using that

- dan815

we know that for all other primes where u dont have 2

- dan815

theres no way that p1^a+p2^b=p3^c
for a,b,c > 1

- dan815

since its odd + Odd =/= odd

- Kainui

oh true

- dan815

now to see how these things change with more primes thrown in to the mix

- dan815

we know all odds thrown in nothings gonna come out

- dan815

so we know the existanece of 2 is a must

- dan815

one has to have a 2 and the other cannot

- Kainui

Like if we restrict ourselves to primes? Ok I see

- dan815

yeah not even with prime products now

- dan815

no* even with

- dan815

maybe we can get something like this

- dan815

for
a , b , c
where a b c are some product of primes, if we can say the number of integer solutions is a function of the number of primes u are using or something like that

- Kainui

Yeah I think that's the way to go, something like that sounds cool

- dan815

it would be kind of interesting to find a bound for like the numver of interger solutions it cannot exceed a b c

- dan815

for a given a b c

- Kainui

Yeah definitely, I think looking back at this, it kind of makes sense more to me to look at after this as being a useful form:
\[\frac{c}{rad^k(c)} > rad^k(a)rad^k(b)\]

- dan815

i got midterm in 5 hours T_T

- dan815

help me with quantum tomorrow after my classes!

- dan815

i gotta actually study for that exam

- Kainui

Haha damn I was just about to say the rad^k function is like a projection operator. Check this out:
\[\langle a,b,c \rangle \rightarrow \langle k, k, k \rangle\]
Alright sounds fun what chapters in kaye should I try to read or whatever I wanna be on top of it so we can go through it confidently and fast.

- dan815

oh its stuff u already know the mid term is only up to last weeks stuff, so its the same stuff we were doing for my assignment i just gotta go over that in detail, ill try to go thru the lecture slide problems, and if we are working fast then we can keep going ahead

- Kainui

Ok cool I'm gonna go to sleep and dream about the abc since it's like 4 am haha.

- dan815

ok night dood

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