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Let's learn and try to prove the ABC conjecture! (No prior knowledge of Inter-Universial Teichmuller theory necessary!)
Mathematics
chestercat
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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 \(k>1\)
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@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
  • 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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Empty
  • 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
  • FireKat97
does it help if I say 'everything above makes sense'? xD
dan815
  • dan815
what do u mean its only true for finitely many numbers for some choice k > 1
dan815
  • dan815
arent there an infinite nuumber for c, that we can pick
Kainui
  • Kainui
Prove it dan, then you'll have found a counter example to the abc conjecture lol
Kainui
  • Kainui
https://en.wikipedia.org/wiki/Abc_conjecture#Formulations
dan815
  • 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
  • Kainui
No because a+b=c
dan815
  • dan815
ohhhhhhhhhhh
dan815
  • dan815
ok now i see why there could be finite okay!
Kainui
  • 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
  • dan815
true
dan815
  • dan815
lol
Kainui
  • Kainui
I was looking at this \[c=p^{k+1}\] so that we have \[p > rad^k(ab)\]
dan815
  • dan815
oo ok ok
dan815
  • dan815
a + b = c and a*b
Kainui
  • 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
  • dan815
c=p^(k+1) p > (ab)^k
Kainui
  • Kainui
That's only true if a and b are primes
Kainui
  • Kainui
\[rad^y(p^x)=p^y\] It's kinda like a prime renamer huh interesting.
dan815
  • dan815
a+b = p^(k+1) p=root_{k+1} (a+b) <--- this being an integer has some big restrictions
Kainui
  • Kainui
wait what do you mean by root_{k+1}
dan815
  • dan815
K+1 th root of A+B
Kainui
  • Kainui
``` \[ \sqrt[k+1]{a+b}\] ``` \[ \sqrt[k+1]{a+b}\]
dan815
  • dan815
yea lol
Kainui
  • Kainui
Oh it has some big restrictions but it's not a big deal that it does.
Kainui
  • 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
  • dan815
u there?
Kainui
  • Kainui
Yeah I'm here I was watching some youtubes what's up
dan815
  • dan815
why are you doing that expression with r
Kainui
  • 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
  • dan815
ohh i see! ok ok
Kainui
  • 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
  • dan815
/2
Kainui
  • Kainui
Yeah ok the divide by 2 has to be in there right I forgot
dan815
  • dan815
okay hmm and this is can even be the maximum one to find
dan815
  • dan815
this should produce the largest or close to largest answers from
dan815
  • dan815
rad(abc) for small rs
Kainui
  • Kainui
I was thinking that but I realized it's flawed
Kainui
  • 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
  • dan815
oh thats ttrue
Kainui
  • 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
  • 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
  • dan815
|dw:1444290020985:dw|
Kainui
  • 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
  • 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
  • dan815
aww dang
Kainui
  • 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
  • dan815
ya ppretty much had an answer without gcd condition because uc an jsut do 2^(A+1) > 2^3K from there
dan815
  • dan815
but i guess thats why this inequality has finite because of that exact gcd reason
dan815
  • dan815
how about something like that though lets take smallest possible primes
dan815
  • dan815
2^a + 3^b= 2^N
dan815
  • dan815
does c also have to be gcd 1 or just a and b
Kainui
  • 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
  • dan815
oh that makes sense
dan815
  • dan815
its easy to see especially if u have p1^a + p2^b, if its the sum of 2 diffrent prime exponents
Kainui
  • 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
  • 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
  • dan815
is this ever possible
dan815
  • dan815
p1^a+p2^b=p3^c where p1 p2 and p3 are all primes
dan815
  • dan815
2+3=5 is 1, where abc =1
Kainui
  • Kainui
\[3^2+2^4=5^2\]
dan815
  • dan815
ok i like this
Kainui
  • 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
  • dan815
do abc have to be positive
Kainui
  • Kainui
Yeah abc have to be positive integers but k just has to be a real number greater than 1
dan815
  • dan815
hmmmmmmmmmm
dan815
  • dan815
suppose we find a bound
dan815
  • dan815
like c>rac(abc)^1 and c<=rad(abc)^2 then we are done right
dan815
  • dan815
well i guess just finding any possible one is good enough then xD
dan815
  • dan815
ok wait i guess proving there are an inifite right hand sides is not what they want dang it
Kainui
  • 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
  • dan815
i think procing it is easier then
Kainui
  • Kainui
Yeah actually I keep getting confused on what the hell it is we're trying to hsow
dan815
  • 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
  • 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
  • dan815
the last statement is like no work and guess, but i think we have to be able to sho that statement
dan815
  • dan815
lol i am back to begining again i see
Kainui
  • 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
  • dan815
okay lemme see what is the number sign again
Kainui
  • 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
  • dan815
okk
dan815
  • dan815
how come u wanna maximize rad function
dan815
  • dan815
oh for contradiction okay
dan815
  • dan815
how are we gonna minimize rad function
Kainui
  • 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
  • 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
  • dan815
there are only finite solutions for that apprently
dan815
  • dan815
how about using that
dan815
  • dan815
we know that for all other primes where u dont have 2
dan815
  • dan815
theres no way that p1^a+p2^b=p3^c for a,b,c > 1
dan815
  • dan815
since its odd + Odd =/= odd
Kainui
  • Kainui
oh true
dan815
  • dan815
now to see how these things change with more primes thrown in to the mix
dan815
  • dan815
we know all odds thrown in nothings gonna come out
dan815
  • dan815
so we know the existanece of 2 is a must
dan815
  • dan815
one has to have a 2 and the other cannot
Kainui
  • Kainui
Like if we restrict ourselves to primes? Ok I see
dan815
  • dan815
yeah not even with prime products now
dan815
  • dan815
no* even with
dan815
  • dan815
maybe we can get something like this
dan815
  • 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
  • Kainui
Yeah I think that's the way to go, something like that sounds cool
dan815
  • 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
  • dan815
for a given a b c
Kainui
  • 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
  • dan815
i got midterm in 5 hours T_T
dan815
  • dan815
help me with quantum tomorrow after my classes!
dan815
  • dan815
i gotta actually study for that exam
Kainui
  • 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
  • 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
  • Kainui
Ok cool I'm gonna go to sleep and dream about the abc since it's like 4 am haha.
dan815
  • dan815
ok night dood

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