Catalan's Conjecture - Numberphile

[@johnayres2303]

I have purchased lots of brown paper and magic markers but I am still useless at Maths.

[@johnayres2303]

Marcus Campbell Yes I know it is corny

[@dangernoodle8376]

You need sharpies

[@briandeschene8424]

John Ayres ...but you’re very fashionable while being useless. A Kardashian of maths.

[@niemandniemand2178]

dumbass

[@Q_QQ_Q]

*Dr Holly Krieger is so white and redhead that i need my dark glasses to even see* . 😂😂😂😂😂😂😂😂😂😂😂😂

[@johndue2366]

For an elder (+60), average electronic engineer with a major interest in math, this channel is awesome.

[@marios1861]

I'm currently studying electrical and computer engineering. Has your job been fulfilling?

[@jonnamechange6854]

Yeah, of course it has. He's so full he can't even move.

[@amanjain1701]

I know its you electroboom @electroboom

[@YorukaValorant]

I'm 21, Been Watching since 17 :D

[@RetroGamingClashOfClans]

uh, you make me feel old.. im just 17

[@ernestboston7707]

It is more fun to write the equation as 3^2-2^3=1^23

[@petterhouting7484]

Or 3^2-2^3=3-2

[@ernestboston7707]

Dale Kerr Quite eXcellent !!!

[@pablozumaran3997]

Do you mean 3²-2³=1²³? :D

[@MatBaconMC]

@@pablozumaran3997 HOW

[@pablozumaran3997]

@@MatBaconMC Key combinations: AltGr+2, AlgGr+3.

[@feuernarr8330]

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!

[@SciencewithKatie]

That's awesome!

[@Wearefree83]

My conational😎

[@TheMrbaummann]

That course is still infamous at Göttingen uni as the "linear algebra course which almost nobody passed" :D

[@Brien831]

@@TheMrbaummann I passed it in 2019! Preda is totally awesome

[@wolframhuttermann7519]

Göttingen needed such a mathematician after Hilbert, Dirichlez and Gauss.

[@sebastianelytron8450]

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.

[@50mt]

[Catalonia joke]

[@user-wu7ug4ly3v]

There is nothing in the mathematics constitution that allows this conjecture to separate itself.

[@bitterlemonboy]

lol catalonia

[@livedandletdie]

Asturias> Catalonia > rest of Spain > rest of Arab blood filled nations.

[@grecolonsvila]

Damn your racism is over 9000. Get back to your mine.

[@eduardomuller9973]

OMG a new conjecture of math! "This conjecture was already proven" WHY DON'T CHANGE IT TO A THEOREM????

[@jchry3688]

Eduardo Muller It's been proven by Mihailescu You can legally call it Mihailescu's Theorem

[@abdulmuhaimin9780]

Alliteration. The only reason

[@peterjohannsen2183]

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?

[@clickaccept]

It was Catalan's Conjecture. There is no inconsistency in this terminology.

[@smitashripad9757]

Just to sound very very very very very very tough

[@johnchessant3012]

26 is the only number that simultaneously is one more than a square and one less than a cube.

[@maxilexow]

Can you prove it?

[@johnchessant3012]

Maxi Lexow Yes, it uses unique factorization in Z[sqrt(-2)].

[@nazishahmad1337]

now its known as john cessant conjecture

[@sieevansetiawan4792]

This problem currently appears in brilliant advanced weekly problem.

[@caffreys1979]

Yes so is a unique soln to x^2 + 1 = y^3 - 1. (x,y) = (5,3)

[@nymalous3428]

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!

[@guitarraccoon1541]

when you're sitting alone on Valentine's day and numberphile makes a new video. Thank you numberphile, atleast you give me math.

[@SciencewithKatie]

😂👏🏼

[@superchet4026]

And Holly.

[@aviraljanveja5155]

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

[@goldminer754]

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.

[@WolfgangGalilei]

so how did it go? hopefully you learned a thing or two!

[@hexisplus9104]

I love paderborn

[@Brien831]

@@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!

[@liviu445]

Let's go romania.

[@Adraria8]

My 9th grade math teacher called perfect powers “sexy numbers”

[@imagination7710]

Sounds inappropriate tbh

[@thesmart4128]

@Diego Maradonna although those would be called sexy primes

[@becamicusack3598]

@@nexusclarum8000 You sound pointless tbh

[@iamnickyj]

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

[@jubtinn]

As they should be

[@liweicai2796]

2:56 I'm kinda surprised that this was proved algebraically. Most difficult number theory problems seem to be tackled analytically nowadays.

[@gytoser801]

If you look closer it's about groups, space and abstract algebra. How would you prove otherwise

[@Ragwar]

Preda Mihāilescu.........what!?I can't believe that a romanian made it to numberphile I am so proud 🇷🇴🇲🇩🇷🇴

[@ffggddss]

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."

[@JMUDoc]

"For instance, are 25 and 27 the last pair of powers that differ by 2?" Fermat (_that_ Fermat) did prove that 25 and 27 are the only _square and cube_ that differ by two, but that's a very small bite out of the question!

[@24ever66]

Bringing Holly in was the best thing numberphile has ever done

[@harryhirsch3637]

That's not fair to Hannah Fry!

[@kirkland5674]

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.

[@lb5sh]

She's intelligent and incredibly charming. What a perfect combo.

[@UnimatrixOne]

😍

[@paologalli5803]

beautiful beautiful

[@vinayvardhanyt2415]

Beauty with brains

[@double124]

Mihailescu is currently my professor for linear algebra. A very kind and jolly man.

[@noohairdontcare]

I don’t “crush on” KZbin celebs but omg I think I’m in love.

[@BobStein]

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.

[@pumajlr]

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.

[@PS3RatBag98]

Bob Stein What about Trump supporters?

[@Marco9603]

I have a thing for women that show a passion and enthusiasm for something!

[@itsJPhere]

I could listen to Holly explaining anything all day and not mind at all, some people have that special something.

[@lookbacktime9422]

Thumbs up for the Romanian mathematician !!!

[@sharoneisenberg2274]

I thought he was Catalan...

[@instakilogram3230]

A noastră!!

[@borgirvspitsah7329]

@@sharoneisenberg2274 the mathematician who proved the conjecture is Romanian

[@PercivalBlakeney]

@LookBackTime Nice to have a famous Romanian other than Count Dracula. I'll get my coat.

[@stv3qbhxjnmmqbw835]

Theoretically, vampires are intelligent beings

[@andrewcgs]

I love Numberphile videos featuring Dr Krieger! ...Happy Valentine's Day y'all, I guess?

[@alucard87pl]

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.

[@AlabasterClay]

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!

[@wolframhuttermann7519]

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.

[@lftkryo]

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.

[@dneary]

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!

[@pepegapig1489]

Dr Holly Krieger😍😍😍😍

[@jackthmp]

claiming Hannah Fry

[@calebhein8043]

RIGHT?!?

[@Q_QQ_Q]

*Dr Holly Krieger is so white and redhead that i need my dark glasses to even see* . 😂😂😂😂😂😂😂😂😂😂😂😂

[@redaabakhti768]

This boss level beauty

[@margothutton]

Dr. Krieger seems like she'd be just so much fun to hang out with!

[@Yetiforce]

Dr. Holly Krieger is perfect for a Valentine's Day Numberphile!

[@lincolnsand5127]

You're a weirdo.

[@Vextrove]

What

[@V-for-Vendetta01]

@@lincolnsand5127 lol

[@nathanjxaxson]

Numberphile needs some t-shirts and other merch, man. So many cool things you guys cover.

[@sharkinahat]

Of course you got Dr Krieger for valentines... I ain't even mad tho.

[@frankcavallo]

Quarantined sitting on toilet and watching Dr Holly’s videos. This could last for months.

[@ffggddss]

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

[@dlevi67]

Pillai's conjecture says that there are only finitely many misses for any integer value of the miss.

[@timgillam7964]

And 2209 = 47^2 comes shortly after 2187 and 2197 too, so there's another bunched up trio of powers

[@TheLeobc]

Holly Krieger is back 💙💙💙

[@JonSebastianF]

Isn't that the *CGP Grey logo* standing in the corner? :D

[@daggawagga]

that's a funny way to call it!

[@MrDannyDetail]

I think it is. Perhaps it's subliminal cross-promotion.

[@danthebat666]

It's the Nail and Gear! Flag of the Hello Internet podcast.

[@DoctorWhom]

"CGP Grey" is a funny way to spell "hello internet"

[@trevortucker9586]

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"

[@UnSimpluJucator]

OMG a fellow romanian demonstrated this? Nice one.

[@blueghost3649]

Yes, a Romanian

[@SirNobleIZH]

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.

[@ErdTirdMans]

Please do more with Holly!

[@madlad255]

2:50 Wait, that's a French poem, not math!

[@abcdefvxyz4324]

thumbs up for the romanian mathematician

[@hexagon-77]

Are you Romanian?

[@abcdefvxyz4324]

yes

[@mashmax98]

Preda Mihăilescu for President, he's currently hibilitated in göttingen, Germany which is where i am studying

[@lorinczadrienjeno4839]

Sal fra

[@tianyima8642]

My fav math prof is Romanian! And one of my fav ow players. Start to develop a strange fondness for Romanian people :p

[@bombadilo4276]

Fun fact, I used to study at the university where Prof Mihailescou teaches nowadays and actually know him from some lectures. I kind of liked him, je was on the more funny side as far as my professors went. I came to this video fully expecting to see him mentioned here^^

[@georgeice4389]

PLEASE CAN YOU PROVIDE THE EMAIL CONTACT OF THE PROFESSOR MIHAILESCU?THANK YOU

[@phscience797]

What I think is even more interesting about those numbers is: Is every natural number a difference between two of those Catalan numbers?

[@jpdemer5]

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)

[@yourobsession9588]

I MISSED DR KRIEGER SO MUCH

[@thomasgarrett1828]

Very interesting problem. Happy Valentine's Day!

[@justpaulo]

Perfect guest for Valentine's day video.

[@steveyankou4144]

Have you considered changing the name of the channel to "numberwang?"

[@zyaicob]

m a t h p e n i s

[@Danmagnet]

Great video as always. Icing on the cake was the Public Service Broadcasting LP in the background. Excellent choice!

[@ThePharphis]

Great video. Hopefully you'll do more proofs with Holly

[@karlamanea6638]

Preda Mihăilescu is Romanian. I was so happy to see his name in this video.

[@mathsmoica]

*Studying conjectures is my passion.*

[@BobStein]

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.)

[@tungstentoaster]

In my mind, the most natural way to show that x^2-y^3=1 only has the one known solution would be putting x^2 and y^3 on the same axis of a graph and showing that they diverge after x=3; y=2.

[@smurfyday]

If you figure out a way to graph to (not towards) infinity, let us know.

[@OlafDoschke]

There are infinite powers with difference 0, though.

[@OlafDoschke]

Yes, with whole numbers, too.

[@tonydai782]

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

[@mashmax98]

2^2n - 4^n

[@dlevi67]

Even simpler: n = n (each number, no matter if a perfect power or not, has difference 0 to itself)

[@OlafDoschke]

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.

[@juancarlosortiz6756]

I remember hearing one time a conjecture similar to Catalan's; namely, the only (positive Diophantine) solution to x^a-y^b=2 is 3^3-5^2. In other words, 26 is the only positive integer that is "sandwiched" between two perfect powers (25 and 27). Does anyone know if this conjecture has a name, or if it has been proven?

[@CaptainSpock1701]

3:09 - "We don't have time for the next 'couple of years'..." - *Looks at watch* - I thought that was very funny.

[@Mrpallekuling]

For the case y=even: x=odd => (x+1) and (x-1) are both even and have the factor 2^n => x+1=2^n * (some odd factors) and x-1=2^n * (some odd factors). Addition gives 2x=2^n * (some odd factors) => x=2^n-1 (some odd factors) => n=1 since x=odd => x+1=2N+2 and x-1=2N => (x+1)(x-1)=4(N+1)=y^3 => N+1 must be 2 since 4(N+1) is a cube => N=1 => x=3, which is the only solution.

[@PureExile]

I think you've defined n to be the largest integer such that 2^n divides both x-1 and x+1. However, x-1 and x+1 may be divisible by different powers of 2 (e.g. take x=15 or x=23 or...) so dividing them by 2^n does not necessarily give an odd result. And I'm not sure what N is but I don't see how 4(N+1)=y^3 implies N=1 (e.g. 4(15+1)=64=4^3).

[@zeezozeezee7482]

when you’re single and have to watch math videos

[@LuckyTondi]

WANT and CHOOSE, not "have to", pffff

[@kranklg2s]

Dont worry. I'm engaged and I'm still watching math (and some history) videos. ^^ Math loves you!

[@sharofs.6576]

and you don't even study mathematics

[@ilprincipe8094]

XSimoniX so true bro

[@alphascooper7797]

😂

[@rmsgrey]

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.

[@Afdch]

Oooh, a public service broadcasting's race for space! A great album.

[@summertilling]

I like how the thumbnail suggests that it's conjectured that 9-8=1, but that no-one can quite prove it.

[@sam_kant]

Came for the mathematics, stayed for the mathematician.

[@J.P.Nery.N.]

Well, it's nice to see that frame with Ron Graham's handwriting in the background.

[@druid_zephyrus]

Why not x=1 and y=0....1^2-0^3=1 does it not?

[@richardfarrer5616]

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.

[@connecticutaggie]

The general solution to x^2 - y^3 = 1 for integers x and y seems pretty easy. When you factor it as described you get (x+1)*(x-1)=y^3 yields only 3 possibilities: 1) (x+1) and (x-1) are both cubes. As you mention there can be no cubes that meet this - NO SOLUTION 2) y^3 is all contained in either (x+1) or (x-1) and the other = 1 - since y^3 >1 then (x-1)=1 => x=2 => y^3=2^2-1=3 which is not an integer - NO SOLUTION 3) The cube has to be split over the two (x-1) = y and (x+1) =y^2 => (x+1) = (x-1)^2 => x^2=3x =>x=3 => y=(3-1)=2 => 3^2 - 2^3 = 1 = YES 3^2 - 2^3 = 1 is the ONLY solution to x^2 - y^3 = 1 Does the general solution follow a similar path of elimination? Since x^n - y^m = 1 => y=(x^n-1) = (x-1)*(poly of x) => x-1 has to fit one of the same three options described above

[@walterwatson120]

Nail and Gear picture in the background! Nice crossover!

[@rhythmgoel4245]

They grey right?

[@MrAlRats]

I'm interested to know where the difficulty arises from when there are equations involving both addition and multiplication.

[@1234s6]

(x+1)(x-1)=y^3 Hey can anyone explain why BOTH have to be cubes and not just one?

[@jungunddumm8023]

Get a book on number theory

[@adamhrankowski1298]

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.

[@ralfoide]

@@jungunddumm8023 That was rude and unhelpful. Always encourage someone who wants to learn.

[@dbliss314]

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)

[@pythontron8710]

Jung und Dumm last name checks out

[@TheReaverOfDarkness]

I feel like you could make a proof that's easy to understand. Firstly, if you're looking at only powers of two, you will find that if you compare any X^2 with (X+1)*(X-1) you will find that they are always 1 apart. The two numbers must be right next to each other for this to happen. In any instance of X*Y, the distance from X to Y directly correlates with the distance between the product of X*Y versus the product of (X+1)*(Y-1). So we already have a proof that in all cases in which X*Y and (X+1)*(Y-1) are 1 apart, X=Y (or X+1 = Y-1). In similar fashion, we are trying to interpret the change in the minimum amount of separation between two perfect powers as those perfect powers themselves get larger. Now the separation varies back and forth, but there is most likely a pattern, something which the closest partners have in common. What can we notice about some of the first examples of close approaches? 25 (5^2) and 27 (3^3) 125 (5^3) and 128 (2^7) 4900 (70^2) and 4913 (17^3) 32761 (181^2) and 32768 (2^15 or 8^5 or 32^3) Some things I notice immediately: * The numbers on the right are often boolean * The numbers on the right possibly alternate between X^3 and 2^X * The numbers on the left have small exponents I am seeing a similarity between the square roots and the perfect powers. It seems that in either case, you can have a close approach in products when the inputs have the right kind of closeness as well. In the case of perfect powers, it might have something to do with the "most powerful" perfect powers (those with multiple factorizations) also having a perfect power near to and slightly lower than them. You may think this is mere coincidence, but in toying with calculators over the years, I have seen many number patterns emerge, and a lot of them look something like this. One very common feature of comparative number patterns is that when you change a number system slightly to achieve similar output, you seem to always either get the same every time, or not the same every time.

[@pedroespino6988]

I've got the feeling that, all of a sudden, a lot of people are going to become very interested in maths.

[@saparchitekt]

I posit that Dr Holly Krieger is a counterexample of a widely held conjecture, thus proving it wrong.

[@jibran8410]

YASSSSSSSSSSSSSS I LOVE CATALAN'S CONJECTURE. Something about it is just so amazing.

[@gutstallion]

ha, nurd

[@manudehanoi]

it's the green eyes and the smile of the conjecture

[@svendsoemod]

Made a little python program to calculate perfect powers [1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484]

[@habibikante9472]

hey thats amazing ! can you send me the program?

[@svendsoemod]

@@habibikante9472 Thanks, I think i just deleted it when it was done

[@javiercorral7820]

I've been puzzled by 2 cubes in geometry in recent time, would you provide me with your interpretation, please?

[@RohitKumar-lw1mt]

Funny how the most complex of problems could be turned to something so simple , with just a little bit of creativity :)

[@JMUDoc]

If "for any n, there exist perfect powers differing by n" hasn't already been conjectured, I demand that it be named after me.

[@ferociousfeind8538]

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

[@zacharychase7652]

I think it's conjectured that there are no two perfect powers differing by 6.

[@venkatbabu186]

Powers start with higher power of 2. Then power ordered. A square minus cube is one. Folding geometry is unit. 2 and 3 are the closest in switch over jumping into negative dimensions.

[@mrmaxi118]

Where is numberphile live from maths fest?

[@venkateshbabu5623]

How do you develop new maths. Zero is an identity of circular set. The set with elements -1 and + 1 is an other circular set. And -2 +2 is an other. -3 -2 -1 0 1 2 3 is an other. These are the definitions. Why we need them. All equations in physics deals with field. A field is something equivalent to imaginary numbers in maths. All matter particles are derived from fields.

[@mackycabangon8945]

EVERY FREAKING TIME I look at the title I think of Catalonia

[@richardfarrer5616]

Given the Brilliant problem in the end, I suggest taking your Pick of the answers.

[@MultiHanspeterwurst]

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 😂

[@michaelbauers8800]

I found this a bit more intuitive by writing a Python program to iterate through a limited set of numbers, showing their differences. I will post it, in case someone wants to play with it. ***** ShowDifferencesLessThan = 20 GenerateNNumbers = 5001 GeneratePowersUpTo = 50 s = set() for x in range( 1, GenerateNNumbers): for a in range( 2, GeneratePowersUpTo): s.add( x ** a) ss = sorted( s) last=-1 for x in ss: if last != -1 and x - last < ShowDifferencesLessThan: print( "%d\t%d" % (x, x - last)) last=x

[@meirihagever9132]

*"Yeah that's exactly right"*

[@dpie4859]

Dear Dr Holly Krieger: Is there a Holly fan club I can join? I am a big fan of yours.

[@waynewelshans1172]

Dr. Krieger, you are a true unicorn :)

[@nGAhGENVH0Ul]

Why did KZbin recommend this? I hate math and numbers. ...yet for some reason I couldn't stop watching the whole video.

[@JanKowalski-zz8ef]

What a beautiful mathematician!

[@scprivat9519]

I am lucky enough to have Preda Mihailescu as a professor (at Uni Göttingen). You will not find a more happy, funnier (,smaller) or more motivated professor, than Preda in front of a bunch of mathematics, physics and computer science students, in their first semester.

[@TheMrbaummann]

Göttingen really has great professors like Preda and Jörg Brüdern

[@Brien831]

@@TheMrbaummann and the most difficult maths lectures in all of Germany. I had Preda and Bahns in my first semester. Preda taught us the fundamental theorem of algebra by Gauss, constructed the reals using dedekind cuts and did Möbius transformation as well.

[@funkygecko]

Guys anyone got any quick proof on this? My professor said he would pass anyone who proved it with the highest possible mark. I failed when I tried, obviously. ps: I think he was personal friends with Mihailescu, the one who proved it.

[@r_se]

search "an elementary proof of Catalan-Mihailescu Theorem" on google

[@techwithwhiteboard3483]

i have a simple proof for x2 - y3 = 1 but not for y3-x2=1 🙂and its mathematical with symbols and concrete logic unlike here which says 2 cubes cant be close ofcourse they can more cubes are multiplied into one it may be possible to bring them closer may be 10 or 20 anyway if it works for u let me know afterall its a year or maybe 2 now cheers

[@quintopia]

If we extend the statement to include powers of integers, we have two more solutions: 1^2-0^3=1, and 0^2-(-1)^3=1. In the proof given at the end, we have 1 and -1 as perfect cubes differing by 2, and 0 and 2 being an exception to that final equation because the claim that both factors have to be cubes fails if they multiply to zero.

[@dlevi67]

Except that the conjecture states that all bases and exponents must be >1 (otherwise every natural number is a perfect power with exponent 1, and clearly there are infinite number pairs that are separated by 1 unit)

[@quintopia]

dlevi67 which is why I said first I was extending the theorem so that bases could be any integer.

[@dlevi67]

But if you extend the theorem so that bases could be any integer it's no longer true, since there are infinite cases... which makes it far less interesting. Note that the conjecture (or theorem) is NOT only about squares and cubes.

[@quintopia]

dlevi67 nope. The specific instance with a square minus a cube only has three solutions when so extended. I don't know, but I would think that even in the general case, there's still only a finite number of choices for the bases. Perhaps only these three.

[@dlevi67]

quintopia well, if you allow +/- 1 and 0 to be bases there are clearly infinite trivial solution of the form 1^n - 0^m for any n,m... and 0^n +/- (-1)^m with +/- depending on m parity. On the other hand, from the proven conjecture, *as long as it does not depend on the bases being > 0*, it immediately descends that it should also be true for negative bases... as long as the ¦bases¦ > 1. Consider the cases (for bases < -1) and exponents n, m > 1 n, m even: all perfect powers are positive, same as Mihailescu's proof n, m odd: all perfect powers are negative, same as Mihailescu's proof in absolute value n,m even/odd or odd/even: one perfect power is positive, the other is negative => difference between any two opposite sign ¦numbers¦ > 1 is at least 4. The only potential problem is that I don't know if Mihailescu's proof depends in any way on the bases being greater than zero.

[@almanahulzilnicdesuceava5379]

It.s ROUMANIAN and we did not know about him!

[@UnSimpluJucator]

Ikr

[@duartesilva7907]

You could also solve this by transforming y^3 in y*(y^2). That would mean x=y+1 or y=(x+1)^0.5, thus giving x=3 and y=2. Of course, there was another possible disjunction, x=y-1 or y=(x-1)^0.5 but it is impossible for any integer x, y.

[@Myrus_MBG]

I literally typed in “that conjecture where there are only two palindromic powers and they do some things”.

[@gphx]

The specification of 'natural numbers' is the only thing keeping -3^2 and -2^3 from disproving Catalan's Conjecture.

[@smurfyday]

Otherwise it's meaningless.

[@aakksshhaayy]

Two plus two is four minus one that's three. Quick maffs

[@jameshall5171]

Congrats

[@hectooooor]

Old but kinda gold

[@forthrightgambitia1032]

5:17 so as I understand here, as it wasn't that clear, if a factor of y is a cube of one of the two before, then the other must be a cube. Reading up on this, to prove in the case of even numbers you have to convert the equation to x^2 = y^3 +1 and turn it into 3 brackets with the cube root of unity, which can then be whittled down to prove only the case explained.

[@yurisims1798]

3:05 "The proof of this is really advanced." Me: "Ha, hold my beer!" Also Me: *crying* "This is really hard."

[@arturslunga3415]

Why is the fourth cube afraid of 1? Because 1, 8, 27.

[@radaroreilly9502]

If there were more teachers like you, I’d have paid attention in mathematics class.

[@michaelosborne9279]

These videos are like watching Derren Brown. He explains the main part of the trick and then there's some additional flourish which indicates the trick is much bigger and invalidates everything he just told you.