Catalan's Conjecture - Numberphile
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- Опубліковано 13 лют 2018
- With Dr Holly Krieger from Murray Edwards College, University of Cambridge.
Have a look at Brilliant (and get 20% off) here: brilliant.org/Numberphile
More links & stuff in full description below ↓↓↓
More Numberphile videos with Dr Krieger: bit.ly/HollyKrieger
Her Twitter: / hollykrieger
Some more reading on the topic: www.ams.org/journals/bull/2004...
Open Problems Group on Brilliant: brilliant.org/groups/open-pro...
Editing and animation by Pete McPartlan
Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumberphile
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I have purchased lots of brown paper and magic markers but I am still useless at Maths.
Marcus Campbell Yes I know it is corny
You need sharpies
John Ayres
...but you’re very fashionable while being useless. A Kardashian of maths.
dumbass
*Dr Holly Krieger is so white and redhead that i need my dark glasses to even see* . 😂😂😂😂😂😂😂😂😂😂😂😂
For an elder (+60), average electronic engineer with a major interest in math, this channel is awesome.
I'm currently studying electrical and computer engineering. Has your job been fulfilling?
Yeah, of course it has. He's so full he can't even move.
I know its you electroboom @electroboom
I'm 21, Been Watching since 17 :D
uh, you make me feel old.. im just 17
It is more fun to write the equation as 3^2-2^3=1^23
Or 3^2-2^3=3-2
Dale Kerr Quite eXcellent !!!
Do you mean 3²-2³=1²³?
:D
@@pablozumaran3997 HOW
@@MatBaconMC Key combinations: AltGr+2, AlgGr+3.
My 9th grade math teacher called perfect powers “sexy numbers”
Sounds inappropriate tbh
@Diego Maradonna although those would be called sexy primes
@@nexusclarum8000 You sound pointless tbh
My 9th grade math teacher called me Nick-mobile, then I found out he called Steve, Steve-mobile. I was devastated, though I was special, guess not
As they should be
In my first semester at the Georg-August university in Göttingen (Germany) the linear algebra lecture was given by Preda Mihailescu. Nice to hear his name in one of our videos!
That's awesome!
My conational😎
That course is still infamous at Göttingen uni as the "linear algebra course which almost nobody passed" :D
@@TheMrbaummann I passed it in 2019! Preda is totally awesome
Göttingen needed such a mathematician after Hilbert, Dirichlez and Gauss.
Catalan's Conjecture is too strong a theory and wants to separate from the rest of mathematics. It wants to be in its own independent set. Can't blame it.
[Catalonia joke]
There is nothing in the mathematics constitution that allows this conjecture to separate itself.
lol catalonia
Asturias> Catalonia > rest of Spain > rest of Arab blood filled nations.
Damn your racism is over 9000.
Get back to your mine.
when you're sitting alone on Valentine's day and numberphile makes a new video.
Thank you numberphile, atleast you give me math.
😂👏🏼
And Holly.
26 is the only number that simultaneously is one more than a square and one less than a cube.
Can you prove it?
Maxi Lexow Yes, it uses unique factorization in Z[sqrt(-2)].
now its known as john cessant conjecture
This problem currently appears in brilliant advanced weekly problem.
Yes so is a unique soln to x^2 + 1 = y^3 - 1. (x,y) = (5,3)
OMG a new conjecture of math!
"This conjecture was already proven"
WHY DON'T CHANGE IT TO A THEOREM????
Eduardo Muller
It's been proven by Mihailescu
You can legally call it Mihailescu's Theorem
Alliteration. The only reason
I would think that it is because it's probably an old conjecture, so people are just used to calling it and referring to it as a conjecture?
It was Catalan's Conjecture. There is no inconsistency in this terminology.
Just to sound very very very very very very tough
2:56 I'm kinda surprised that this was proved algebraically. Most difficult number theory problems seem to be tackled analytically nowadays.
If you look closer it's about groups, space and abstract algebra. How would you prove otherwise
I do love that moment of clarity and understanding when learning something new (...I also enjoy watching someone else experience it when I am the one teaching). Most of the math in these videos goes over my head, but I always seem to get just enough to get a brief moment of learning. Thanks again for doing these videos!
when you’re single and have to watch math videos
WANT and CHOOSE, not "have to", pffff
Dont worry. I'm engaged and I'm still watching math (and some history) videos. ^^ Math loves you!
and you don't even study mathematics
XSimoniX so true bro
😂
This Conjecture was proven by Preda Mihăilescu, at the University of Paderborn ! Honored to be able study at the university in 1 month's time ! XD
He is teaching now in Göttingen, you would have even had him in linalg 1&2 and algebra if you started 2 years ago in Göttingen.
so how did it go? hopefully you learned a thing or two!
I love paderborn
@@goldminer754 I had him in my AGLA1 course. He is a really cool guy, his lecture is a bit all over the place though. Proving the fundamental theorem of Algebra to first semesters the Gauss way isnt really cool. Möbius transformations arent nice either for 1st semester students!
Let's go romania.
Of course, there's always a next question(s), once something like this gets settled. Like, is there a point beyond which there are no more differences as small as d, where d is 2 or 3 or ...
For instance, are 25 and 27 the last pair of powers that differ by 2?
Are 125 and 128 the last pair that differ by 3?
Are 2187 and 2197 the last pair of powers that differ by 10?
Etc.
Thanks! This was fun!!
Fred
PS. A reply 2 years ago, by dlevi67, to a similar comment of mine, points out that, "Pillai's conjecture says that there are only finitely many misses for any integer value of the miss."
Bringing Holly in was the best thing numberphile has ever done
That's not fair to Hannah Fry!
That was really fun. I think it is nice to go ahead and start down the right path....even if we can't follow the whole big proof.
Thanks!
I love Numberphile videos featuring Dr Krieger!
...Happy Valentine's Day y'all, I guess?
Thumbs up for the Romanian mathematician !!!
I thought he was Catalan...
A noastră!!
@@sharoneisenberg2274 the mathematician who proved the conjecture is Romanian
@LookBackTime
Nice to have a famous Romanian other than Count Dracula.
I'll get my coat.
Theoretically, vampires are intelligent beings
What I love about this channel is shows that math can be super hard but at the same time doesn’t require you to be from a different planet to understand it.
I can't shake the impression that Dr Krieger reminds me of Jewel Staite (Kaylee in Firefly, Dr Keller in Stargate Atlantis). There's something about the voice that rings the same bells, as well as the way she looks when she smiles.
I've been puzzled by 2 cubes in geometry in recent time, would you provide me with your interpretation, please?
Great video. Hopefully you'll do more proofs with Holly
Great video as always. Icing on the cake was the Public Service Broadcasting LP in the background. Excellent choice!
This channel is great! Thank you for another great video.
Please do more with Holly!
Numberphile needs some t-shirts and other merch, man. So many cool things you guys cover.
Love the mighty Nail and Gear in the background!
Nicely done! Making math entertaining as always!
Nail and Gear picture in the background! Nice crossover!
They grey right?
Preda Mihāilescu.........what!?I can't believe that a romanian made it to numberphile I am so proud 🇷🇴🇲🇩🇷🇴
I'm interested to know where the difficulty arises from when there are equations involving both addition and multiplication.
Mihailescu is currently my professor for linear algebra. A very kind and jolly man.
Isn't that the *CGP Grey logo* standing in the corner? :D
that's a funny way to call it!
I think it is. Perhaps it's subliminal cross-promotion.
It's the Nail and Gear! Flag of the Hello Internet podcast.
"CGP Grey" is a funny way to spell "hello internet"
Very interesting problem. Happy Valentine's Day!
Quarantined sitting on toilet and watching Dr Holly’s videos. This could last for months.
you make mathematical concepts intresting. thank you
Of course you got Dr Krieger for valentines...
I ain't even mad tho.
Came for the mathematics, stayed for the mathematician.
Well, it's nice to see that frame with Ron Graham's handwriting in the background.
I just started learning about elliptic curves, and the curve y^2=x^3+1 is an elliptic curve of rank 0 with a torsion group of order 6. Not only are there no integer solutions (other than (-1,0), (0,+/-1), (2,+/-3)) but there are also no other rational solutions!
I don’t “crush on” UA-cam celebs but omg I think I’m in love.
I've never found any category of people to be categorically excluded from crush potential. Patterns of people are about as useful as patterns in clouds.
Bob Stein
He's propably saying that he's not a tennager who loves somebody just because he like the videos they make. The "pattern" can inform about the people he like or why he likes them. And patterns of people are totally useful. We classify people all the time because of that.
Bob Stein What about Trump supporters?
I have a thing for women that show a passion and enthusiasm for something!
I could listen to Holly explaining anything all day and not mind at all, some people have that special something.
What I think is even more interesting about those numbers is: Is every natural number a difference between two of those Catalan numbers?
Nobody's proved it for the number 6, let alone "every natural number." Or for 14, or 32, or 42, or 50... there's an apparently infinite number of (conjectured) counter-examples (A074981 in the OEIS)
I know the mathematician who proved Catalan conjecture. Prof Mihailescu did it in 2004 and gives lectures in Göttingen where I used to study math.
Dr. Krieger seems like she'd be just so much fun to hang out with!
so really charmming!!
2:50 Wait, that's a French poem, not math!
Perfect guest for Valentine's day video.
I've read that 26 is the only integer that falls directly between a square (25) and a cube (27), and that Fermat proved it? Is this right and is the proof similar to this?
Why not x=1 and y=0....1^2-0^3=1
does it not?
Typically the answers are limited to natural numbers starting from 1. In integers, you are correct, however, and there's also (-1, 0) and (-3,2) as answers.
*"Yeah that's exactly right"*
Excelente video. Los felicito!!!
One step of the proof reminded me of a PBS video about Sohr's algorithm. Intriguing
There seems to be an unstated assumption here. At around 5:30 we're told there are no two cubes that differ by two. But there is one such pair: 1 and -1. Thus, we can expect a solution if x = 0. And putting that into the original equation gives us y = -1.
I've got the feeling that, all of a sudden, a lot of people are going to become very interested in maths.
I'd be interested to know if there is always at least one difference of all the natural numbers; i, 2, 3, 4... If not, which is the first natural number that doesn't work? :-)
Holly Krieger is back 💙💙💙
Dr. Holly Krieger is perfect for a Valentine's Day Numberphile!
You're a weirdo.
What
@@lincolnsand5127 lol
3:09 - "We don't have time for the next 'couple of years'..." - *Looks at watch* - I thought that was very funny.
I MISSED DR KRIEGER SO MUCH
Best kind of valentine's gift! Niiicee
(x+1)(x-1)=y^3
Hey can anyone explain why BOTH have to be cubes and not just one?
Get a book on number theory
The two factors differ by exactly two. We are assuming that y is odd. Since the differ by 2, they can have no common factor. We need 3 copies of each of the factors of y, because y is cubed. These three factors are distributed between the two factors x-1 and x+1. But since those two factors don't share any factors between then, a given triplet of y factors has to be assigned to either of x-1 or x+1. Hence, they are each cubes.
@@jungunddumm8023 That was rude and unhelpful. Always encourage someone who wants to learn.
The factors are the same on both sides of the equation. If (x+1) had a non-cubed factor, and (x-1) did not have that same factor at all, then y^3 must also have that non-cubed factor. y^3 cannot have any non-cubed factors, because y^3 is a perfect cube. Therefore (x+1) must only have cubed factors. Therefore, (x+1) must be a perfect cube. Ditto for (x-1)
Jung und Dumm last name checks out
*Studying conjectures is my passion.*
I conjecture you have yet to find your life's most interesting conjecture.
(Unless that was it. But then it was still true when I conjectured it.)
May I propose the Schultz Conjecture: looking at the list of all integer powers, a separation of every integer value is found eventually, like there is *somewhere* a separation of 1, 2, 3, etc.
Into to Algebra 1. Best explanation I’ve ever heard. 👍
I am a complete amateur in this field but -1 and 1 are cubes and they differ by 2, so why not use them. Or is it just that we are only thinking about positive integers only and not the negative one?
the conjecture says : x^a-y^b=1 with x,a,y,b > 1
Where is numberphile live from maths fest?
I like how I failed every single aspect of math throughout many years of schooling and yet somehow by watching this video I naively thought "oh hey, you're older now Trevor, you'll probably understand what's being said"
Preda Mihăilescu is Romanian. I was so happy to see his name in this video.
here I am, laying in my bed watching youtube videos, not learning for my exam tomorrow and there is a video about one of my professors at the university of Göttingen 😂
EVERY FREAKING TIME I look at the title I think of Catalonia
Brilliantly explained
Given the Brilliant problem in the end, I suggest taking your Pick of the answers.
thumbs up for the romanian mathematician
Are you Romanian?
yes
Preda Mihăilescu for President, he's currently hibilitated in göttingen, Germany which is where i am studying
Sal fra
My fav math prof is Romanian! And one of my fav ow players. Start to develop a strange fondness for Romanian people :p
Dr. Krieger, you are a true unicorn :)
can someone please tell me how she came to the conclusion that (x-1) and (x+1) both have to be cubes?
Thanks for the link to complete proof.
Oooh, a public service broadcasting's race for space! A great album.
An interesting observation I've often wondered about, but had no idea was actually being tackled by mathematicians!
There are a number of other "pretty-close" cases.
5³ - 11² = 125 - 121 = 4 · · · ↓
2⁷ - 5³ = 128 - 125 = 3 . . → these two examples are all the more interesting, because there are *three* powers within a short span (7)
13³ - 3⁷ = 2197 - 2187 = 10
Pillai's conjecture says that there are only finitely many misses for any integer value of the miss.
And 2209 = 47^2 comes shortly after 2187 and 2197 too, so there's another bunched up trio of powers
I love the nails & gears flag in the background in the beginning ;)
Is 1 the only difference that only occurs once? Or the only number that occurs a finite amount of times for that matter?
Have you considered changing the name of the channel to "numberwang?"
m a t h p e n i s
Super awesome video. Question, are there not two cubes that are two integers apart? That being -1^3 and 1^3, and their results would be -1 and 1, respectively. Need to rewatch video as often this stuff goes over my head the first time.
Why does the video's description does not tell anything about the video subject? Not even the name of Mihăilescu?
So we now know that there are finitely many (i.e. 1) that differ by one, what about by 2, 3, 4... etc. Are there finitely many that differ by any finite number? If not, are there any numbers other than one that only finitely many differ by?
There are infinite powers with difference 0, though.
Yes, with whole numbers, too.
Because all powers of powers have a difference of 0 to the base of the power of the power, to the power of the product of the exponents so you can have infinite examples of this
2^2n - 4^n
Even simpler: n = n (each number, no matter if a perfect power or not, has difference 0 to itself)
n^1 is excluded, as far as I see. If these numbers are in, well, you had infinitely many gaps of difference 1...
So that's too simple, I am talking of 2or more representations of the same number. Not all numbers have that, eg 2^2 is the only perfect power (of integers) resulting in 4, but still there are infinite numbers with 2 (or more) perfect powers and thus a difference of 0 between them. That's just not asked for.
OMG a fellow romanian demonstrated this? Nice one.
Yes, a Romanian
Greetings from Romania! Cheers to Preda Mihăilescu😙
so does it work for all separations? Is each separation unique?
Are there an infinity of perfects powers separated by 2? and 3?
Pillai's conjecture is that there are only finite numbers of perfect powers separated by any integer value. Still not proven rigorously, as far as I know.
dlevi67 thanks that's fascinating, i'll check this out
Pillai's conjecture states that each positive integer occurs only finitely many times as a difference of perfect powers. That's a less ambiguous way of stating it.
I fail to see the ambiguity (or the difference - finite or otherwise), especially considering the way in which the question was formulated, but if it makes you happy... ;-)
Dr Holly Krieger😍😍😍😍
claiming Hannah Fry
RIGHT?!?
*Dr Holly Krieger is so white and redhead that i need my dark glasses to even see* . 😂😂😂😂😂😂😂😂😂😂😂😂
This boss level beauty
Dr Holy Krieger ❤️❤️ Long time no see ❤️❤️
How about the generalization? Given a positive integer n, is the number of pairs of perfect powers that differ exactly n always finite?
I literally typed in “that conjecture where there are only two palindromic powers and they do some things”.
If
"for any n, there exist perfect powers differing by n"
hasn't already been conjectured, I demand that it be named after me.
Ah, the Jay-Mu-Doc conjecture
I feel like the larger n is, the more examples of perfect powers differ exactly by n. The problem, in the form of "a^b - c^d = n" is far too free to have any integer n that doesn't also have at least one set of integers a, b, c, and d
I think it's conjectured that there are no two perfect powers differing by 6.
Possibly worth mentioning that if you extend to non-positive integers, you also get (-)1^n and 0^n (but no other pairs separated by 1 - negative integers are only powers if they're odd powers of negative integers). Possibly not worth mentioning it either.
Just wondering, since 1^3 and -1^3 are only 2 apart, could the solution of x also be 0?
Yes, but that's a trivial solution. Same way there's infinite solutions if you set a and b to one.
She's intelligent and incredibly charming. What a perfect combo.
😍
beautiful beautiful
Beauty with brains
I spy a mighty Nail and Gear
No Saturn rocket box in background :(
I'm a year late but I needed to jump in the comments to say this.
Is there any case of n^m = m^n other than 2 and 4? Been curious for a while
The proof at the end can be generalized to even powers of x, right? Because they will also be difference of squares?