abc Conjecture - Numberphile

If we increase all the numbers from one to infinity to the power of 2, the distance between the obtained numbers is 1/4/9/16/25/36/49/64/81/100/121/144. It used to go like 3/5/7/9/11/13/15/17. Well, now (25x25 becomes 625)---(24x24 becomes 576)---(26x26 becomes 676)..676-625=51.. ....625-576=49.....

In+Bin=Kin=Pin=(9+16)=25=49

For those watching in 2023: This proof has STILL not been accepted by the community. So the A B C conjecture is still unsolved 😮

√abc

So this may be ignorance but with the abc conjecture...

A method of proof is to show that there is a constant k for all positive x

C< k (RaRbRc)^(1+x)
Ra is Rad(a) etc..
Couldn't you get a pretty much trivial or elementary solution from
C =u*Rc
where u is all higher powers of factors in rad(c)

Since x is only positive we may start with the inequality

Rc<(RaRbRc)^(1+x)
u*Rc< u* (RaRbRc)^(1+x)
C<u*(RaRbRc)^(1+x)

I know it can't be that easy and y'all are going to tear me up lol

I know I didn't show the lowest bound for k necessarily but i think that is k> u/(RaRb)


Same kind of trick

Start with inequality valid for all positive x (as long as n is not taken to the limit of oo)

(RaRbRc)/(1+1/n) < (RaRbRc)^(1+x)

Get Rc alone on LHS and multiply both sides by u

uRc<(1+1/n)*u/(RaRb) (RaRbRc)^(1+x)

Where uRc=C

C< (1+1/n)*u/(RaRb) (RaRbRc)^(1+x)

Take n to the limit at oo and you get (u/RaRb) thus k must be greater than u/RaRb

I need that wallpaper

Your presentation is well wicked

Seems to me, not correct?! the conjecture says rad(abc)^k<c has finitely many abc-triples if k>1. Am I correct?

Time to see the math news of the past 10 years!

The chance to win one million dollars has arrived, so please do your best if you are smart.

Who's bed are they writing this on? Lol

Sooo…. how’s the check process going on this. 10 years should be enough time, right? It’s ONLY 500 pages and Mochizuki is notoriously difficult to deal with. I know that Peter Scholze was working with him, but that didn’t pan out. Would be curious for a follow-up (if you haven’t already done it).

1026+81=1107 not 1105

Still unproven

you did not explain how you calculate rad(abc)^k

Some think he is Satoshi Nakamoto.

If you're bored, and you have access to a pen and paper, a great way to pass the time is to do math. Like trying to manually calculate the first 50 digits of pi. Or reinventing all of math. Or proving Riemann's Conjecture. Or finding the first 100 digits of Grahams number.
The possibilities are endless, and you just need some pen and paper, which kind of act like extended memory for your brain. Think of your mind as the RAM, and the paper as the hard disk.

Random numbers are not related except repeatedly. If only one of a,b, c is random then the values are not related, is most likely!

Still being reviewed because he didnt use PEMDAS

I have the proof but I left it in another comment...

That they are all gonna get shot by the national guard?

if = whether

Well wicked

Tell us if he's right, mathematicians- It's been 10 years! Interuniversal geometry doesn't sound that complicated, lol

@ Numberphile You need to make your 3s clearer when you write them. The plural of
"exception" is "exceptions."

it's called the radical because it's well wicked 🤣

you still haven't solved the rubiks cube yet

any updates on this?

Professor James: "I know nothing about it"
Me: ;-(

Love these early numberphile videos done on James’ bedsheets

2+3=5 boom. Conjecture over. (I am kidding of course)

I am going to put in a large amount of effort into incorporating “well wicked” into my daily vocabulary

I've come from the future year of 2021. Wikipedia says it is still not resolved xD

ABC seems to be tough. Just seen Marouane Rhafli attempt to prove a partial solution to beal's conjecture

2+5=7

Nice pigeons.

Sir my name is nitesh
I math research topic
A+B=3
A-B=-1
A×B=2
A÷B=0.5

Y'all
18²+19=7³
I tried everything but I can't build upon that can someone help me out

I cant imagine what proving this involves. I literally can't imagine

a+b=c[]=[[ a+b]=[ a+b)^=O ] a= -b ]= a^o=O ]=[ a=O ]=[ -b=O] b=O ] [[[ 0+o= c ]= O =C ]]]

a+b=c[]=[[ a+b]=[ a+b)^=O ] a= -b ]= a^o=O ]=[ a=O ]=[ -b=O] b=O ] [[[ 0+o= c ]= O =C ]]]

Shinichi Mochizuki created bitcoin

Anyone here from Matt's new video on the ABC conjecture?
I've tried to understand what's going on till radicals and quality, but after that I just lose it.

I hate math😤😤

Still hasn't been proven.

I could finally understand what the meaning of thin conjecture is. As always your explanation is really easy to grasp.

They are rare. They are infinitely many. They exist.