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Empty
 one year ago
Let's learn and try to prove the ABC conjecture! (No prior knowledge of InterUniversial Teichmuller theory necessary!)
Empty
 one year ago
Let's learn and try to prove the ABC conjecture! (No prior knowledge of InterUniversial Teichmuller theory necessary!)

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Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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: \[rad(abc)=rad(a)*rad(b)*rad(c)\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Anybody 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
 one year ago
Best ResponseYou've already chosen the best response.0does it help if I say 'everything above makes sense'? xD

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what do u mean its only true for finitely many numbers for some choice k > 1

dan815
 one year ago
Best ResponseYou've already chosen the best response.2arent there an infinite nuumber for c, that we can pick

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Prove it dan, then you'll have found a counter example to the abc conjecture lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2like 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ok now i see why there could be finite okay!

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Haha 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. \[\mu(rad(n)) \ne 0\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I was looking at this \[c=p^{k+1}\] so that we have \[p > rad^k(ab)\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I 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.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2That's only true if a and b are primes

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2\[rad^y(p^x)=p^y\] It's kinda like a prime renamer huh interesting.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2a+b = p^(k+1) p=root_{k+1} (a+b) < this being an integer has some big restrictions

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2wait what do you mean by root_{k+1}

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2``` \[ \sqrt[k+1]{a+b}\] ``` \[ \sqrt[k+1]{a+b}\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Oh it has some big restrictions but it's not a big deal that it does.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Just start here: \[p^{k}=\frac{p^k+r}{2} + \frac{p^kr}{2}\] there are always going to be multiple choices of r that work.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah I'm here I was watching some youtubes what's up

dan815
 one year ago
Best ResponseYou've already chosen the best response.2why are you doing that expression with r

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I 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, \[c=p^k\]\[a=p^k+r\]\[b=p^kr\] as long as r isn't divisible by p, then all 3 numbers are relatively prime.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2cause 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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah ok the divide by 2 has to be in there right I forgot

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay hmm and this is can even be the maximum one to find

dan815
 one year ago
Best ResponseYou've already chosen the best response.2this should produce the largest or close to largest answers from

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I was thinking that but I realized it's flawed

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2cause it's not so much the size of the number it's the amount of unique primes in the number that matters more

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Also 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 \[p^k = (p^kq^r)+(q^r)\] or something like this idk

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Ooooh ok I got an idea \[p^k = q \# + (p^kq \#)\] where q# is the largest primorial not exceeding \(p^k\) this should help us maximize the rad function!

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Actually 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
 one year ago
Best ResponseYou've already chosen the best response.2Oh 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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2but 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
 one year ago
Best ResponseYou've already chosen the best response.2ya ppretty much had an answer without gcd condition because uc an jsut do 2^(A+1) > 2^3K from there

dan815
 one year ago
Best ResponseYou've already chosen the best response.2but i guess thats why this inequality has finite because of that exact gcd reason

dan815
 one year ago
Best ResponseYou've already chosen the best response.2how about something like that though lets take smallest possible primes

dan815
 one year ago
Best ResponseYou've already chosen the best response.2does c also have to be gcd 1 or just a and b

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Ok 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
 one year ago
Best ResponseYou've already chosen the best response.2its easy to see especially if u have p1^a + p2^b, if its the sum of 2 diffrent prime exponents

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2proof, 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*(nm) donezo

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Ok 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
 one year ago
Best ResponseYou've already chosen the best response.2p1^a+p2^b=p3^c where p1 p2 and p3 are all primes

dan815
 one year ago
Best ResponseYou've already chosen the best response.22+3=5 is 1, where abc =1

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2yeah 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
 one year ago
Best ResponseYou've already chosen the best response.2do abc have to be positive

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah abc have to be positive integers but k just has to be a real number greater than 1

dan815
 one year ago
Best ResponseYou've already chosen the best response.2suppose we find a bound

dan815
 one year ago
Best ResponseYou've already chosen the best response.2like c>rac(abc)^1 and c<=rad(abc)^2 then we are done right

dan815
 one year ago
Best ResponseYou've already chosen the best response.2well i guess just finding any possible one is good enough then xD

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ok wait i guess proving there are an inifite right hand sides is not what they want dang it

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2In 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: \[p^n > rad^k(\frac{q\#}{p}(p^n\frac{q\#}{p})p^n)\] \[p^{nk} > rad^k(\frac{q\#}{p}) rad^k(p^n\frac{q\#}{p})\] So to find a contradiction we can show that this is true for an infinite number of cases?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i think procing it is easier then

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah actually I keep getting confused on what the hell it is we're trying to hsow

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2now we gotta show until k k there will be finite solutiond and that is true because we are allowing only integer abcs

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the last statement is like no work and guess, but i think we have to be able to sho that statement

dan815
 one year ago
Best ResponseYou've already chosen the best response.2lol i am back to begining again i see

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2okay lemme see what is the number sign again

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Really like all this means is: \[\frac{p_n \#}{2} = 3*5*7*11*...*p_{n1}*p_n\] Now let's do some abc's

dan815
 one year ago
Best ResponseYou've already chosen the best response.2how come u wanna maximize rad function

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh for contradiction okay

dan815
 one year ago
Best ResponseYou've already chosen the best response.2how are we gonna minimize rad function

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{p_n \#}{2}> rad^k((\frac{p_n \#}{2}2^x)2^x\frac{p_n \#}{2})\] \[\frac{p_n \#}{2}> rad^k(\frac{p_n \#}{2}2^x)2^k \frac{(p_n \#)^k}{2^k}\] Wait... \[1> 2 (p_n \#)^{k1} rad^k(\frac{p_n \#}{2}2^x)\] This is the opposite of what I wanted to do, this is clearly like a bad case to look at rofl

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.2there are only finite solutions for that apprently

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we know that for all other primes where u dont have 2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2theres no way that p1^a+p2^b=p3^c for a,b,c > 1

dan815
 one year ago
Best ResponseYou've already chosen the best response.2since its odd + Odd =/= odd

dan815
 one year ago
Best ResponseYou've already chosen the best response.2now to see how these things change with more primes thrown in to the mix

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we know all odds thrown in nothings gonna come out

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so we know the existanece of 2 is a must

dan815
 one year ago
Best ResponseYou've already chosen the best response.2one has to have a 2 and the other cannot

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Like if we restrict ourselves to primes? Ok I see

dan815
 one year ago
Best ResponseYou've already chosen the best response.2yeah not even with prime products now

dan815
 one year ago
Best ResponseYou've already chosen the best response.2maybe we can get something like this

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2Yeah I think that's the way to go, something like that sounds cool

dan815
 one year ago
Best ResponseYou've already chosen the best response.2it would be kind of interesting to find a bound for like the numver of interger solutions it cannot exceed a b c

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.2i got midterm in 5 hours T_T

dan815
 one year ago
Best ResponseYou've already chosen the best response.2help me with quantum tomorrow after my classes!

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i gotta actually study for that exam

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Haha 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
 one year ago
Best ResponseYou've already chosen the best response.2oh 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
 one year ago
Best ResponseYou've already chosen the best response.2Ok cool I'm gonna go to sleep and dream about the abc since it's like 4 am haha.
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