The Legend of Question Six - Numberphile

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Simon Pampena discusses the famous Question 6 from the 1988 International Mathematical Olympiad. More links below...
Second part of this video: • The Return of the Lege...
International Mathematical Olympiad (includes links to all previous Olympiad results and papers): www.imo-offici...
Simon Pampena: / mathemaniac
Terence Tao: www.math.ucla.e... (thanks to him for the photos)
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[@SickSkilz]

It took him 1 year to solve the problem. It took him another year to get to the point of telling us what the problem was

[@rboyle6706]

for real

[@brendenlim2158]

Tauseef Baggia if you’re so smart then go and become the smartest person in the world

[@camplethargic8]

Skip to 5:30... you're welcome.

[@MP-rh2pl]

Lol saw it right away a= 1 b= 0

[@camplethargic8]

@@MP-rh2pl prove it, keyboard genius

[@ym-wi3cp]

I am glad I was at least able to calculate how 4.5 hours for three questions mean 90 minutes for each question on average

[@tobyfastjr]

450/90= 5 so 5 questions. I don't get it

[@halo6mastercheif]

4.5 hours == 270 minutes 3 questions 270/3 = 90

[@Blade.5786]

@@halo6mastercheif It's a joke

[@wcsxwcsx]

That's as far as I got. I was proud of myself.

[@mspika]

@@wcsxwcsx i understood 19 and was seriously questioning my math

[@ArkanMatlub03]

Do you know The Legend of Question Six Me, an intellectual: Why was number six scared of seven?

[@Dresden0010]

A: SIX is the more proficient model. (BSG [BATTLESTAR GALACTICA]).

[@andreyaquilon9184]

Because seven eight ate (eight) nine.

[@lovelypotatoes]

because seven was a registered six offender

[@harshavardhanvs3354]

@@lovelypotatoes most underrated comment

[@d3generate804]

Because he had the high ground

[@adamweishaupt3733]

"Hey, let's give a bunch of teenagers one of the hardest problems ever conceived in mathematics."

[@numberphile]

too right - The Math Olympiad is for Top Guns!

[@vinayvyraveraja7102]

who are actually kids. Here I thought I was good in maths for my age.

[@PhilBoswell]

Isn't there a story about Gauss walking into his classroom late for a lesson, and the teacher has written an "impossible problem" on the board, only Gauss doesn't hear that bit and solves it? Maybe it's an Urban Legend but the trope of solving a problem because you didn't know it was supposed to be impossible is a fun one to play with ;-)

[@John_Ridley]

yes, but one of them solved it in the allotted time. This isn't a competency test as you usually get in school, this is an EXCELLENCE test. The questions MUST be difficult enough that the majority of the best competitors won't be able to finish them all. Else how would you score it? I competed in my state's math competition test for several years in high school. It was a lot of fun. I still have the question sheets. I didn't even know how to start a few of the questions but it was fun to think about anyway.

[@metallsnubben]

+Phil Boswell Not impossible, but intended to be very time-consuming so that the teacher could take a break, and it only took him a moment. The problem was supposedly to add together all the integers from 1 to 100 (or was it 1000?) and while the other kids just did that brute force without questioning, he badically came up with the general formula for sums of that type, (n+1)*n/2 (where n is the highest integer)

[@dieke8978]

"And if you can't figure out that's ninety minutes, you're gonna struggle with the whole exam" You didn't have to attack me like that

[@feelthereal5742]

Pls can you see my comment on this video. I proved it.

[@ezralebowitz3371]

I THOUGHT I WAS AN IDIOT AND HE SAID 19 AND THERE WAS SOMETHING IM MISSING

[@zoeeichholzer9782]

@@ezralebowitz3371 omg me too 🤣🤣

[@sandorrclegane2307]

@Jump Jack my guy you apparently solved IMO problems and don't know that 4,5 hours divided by 3 is 90 minutes? There's something wrong here

[@onetwothreefour-s1n]

😆 🤣

[@lanvu9323]

"have you heard of the Legend of question six?" no, but I've heard of the Emu war 1932

[@tylerpirc]

Lan Vu gang gang gang?

[@Elijahluke1995]

Big deal. Have you ever heard of darth Plaguies the wise?

[@suwinkhamchaiwong8382]

Kiwis

[@narwhalestorm9881]

I thought not, it's not a story the schools would tell you

[@thatnerdoverthere2248]

Okay, but have you ever heard of the tragedy of darth plauguas the wise?

[@hamzahk.8721]

I came up with a solution in 4minutes....it was wrong.

[@apacheattackhelicopter5823]

Hum Shak so what’s the point?

[@Flaruwu]

@@apacheattackhelicopter5823 he tried

[@talhatariqyuluqatdis]

I did in about 10 seconds. It was- it was wrong.

[@blackpinktanked6809]

hahahahahahahahhahahhahahahaahahhaahahahahaha

[@pocojoyo]

I beat you. I arrived at that solution at 2 minutes.

[@DavidRussell323]

I think one of the biggest reasons I love this channel is that it's not really a maths channel--it's more of a place that tells stories through difficult questions, and often shows you different ways of thinking about these problems. The stories these professors tell are always super enchanting

[@maximusmawle2904]

Why doesn’t anybody give credit to the people who designed these questions. They must be even MORE genius.

[@ghauramahabaduge9746]

Maximoose 2005 there are problem proposers and they are famous among IMO Community

[@ReasonableForseeability]

WRONG!! Much easier to pose questions than to solve them.

[@dean7763]

are you joking?

[@abogmus8904]

@@AndreFranca99 if they do not how to solve the problem they cannot ask the participants for an answer since there can be no answer at all.

[@yyhra]

@@AndreFranca99 they do, try to design a complicated math problem, that requires specific steps to solve. Could be an equasion or a proof for something. The solutions are usually pretty apparent, if the designer doesn't know how to solve the problem himself, the solutions either turn out to be gibberish which makes them appear false, or if it's a proof of something, it looks and feels really intransparent.

[@bm-br3go]

"Have you heard of the legend of question six?" No but I have heard of thelegend27.

[@kortanshizuka5442]

Yes, but have you heard the legend of the Tragedy of Darth Plaguis the Wise?

[@tomascanevaro4292]

Did you heard the tragedy that reach the man?

[@malikmister]

I've never heard of thelegend27, but I have heard of _The Legend of Zelda._ We all have, but... Have you heard of *_The Legend of Korra?_*

[@greysquirrel404]

Kortan Shizuka have you heard of The Rise of the Legend of the Tragedy of Dark Plaguis the Wise?

[@MrZukokku]

Ah, Man of culture

[@DefeatLust]

Well, it didn't take me a year but it did take me about 10 hours... it is difficult but if you really think about it, all you have to do is press Right button, Right button, B, Right Trigger, Right, Left, Right, Left, Right, Left into your xbox controller and you can beat gta fairly easily. Your welcome.

[@SirSteak1]

JoMomma239 but they patched that in update v1.30

[@jancovanderwesthuizen8070]

Not gonna lie.. they had us in the first half

[@albond]

And "your" bad at using grammar. YOU'RE welcome. Seriously, why did more than 430 people actually like your comment? Are they blind?

[@nick242]

435 people*, are you blind?

[@albond]

@@nick242 Read it again... "why did *more than* 430 people". I said "more than" for future-proofing. Are you blind?

[@colinjava8447]

I just solved it, but there wasn't enough room here to type it in so I haven't.

[@jamievlogs7103]

Colin Java it’s just 1.

[@somnathdash4428]

@@jamievlogs7103 you clearly didn't get the joke.

[@jamievlogs7103]

@@somnathdash4428 You're just mad because John Gnash got the fields medal and you didn't.

[@evaristegutierrez8260]

@@jamievlogs7103 Colin Java´s last theorem ;)

[@kevindehulsters]

Typical Fermat

[@yungmilopkl]

Australian maths be like "Oi to the power of mate², carry the roo = shrimps on the barbie."

[@mambodog5322]

You forgot to divide by boomerang

[@malikmister]

And carry the crikey.

[@Life-er6mq]

💀💀💀💀

[@dekudude8888]

Solve for Kano

[@jeninarvaezmelo656]

WHAT 😂

[@yashrawat9409]

The amazing thing is that A handful of participants were able to do it correctly in stipulated time

[@abhaysharmafitness]

it really boggles my mind

[@abhijeetsingh5681]

for sure

[@t.k.-s.4212]

I love how this all builds up for 5.5 min, just to get more and more admiration and respect for the problem. Very much enjoyed that!

Ye. Terrence may have had 1 out of 7.. But we have to revere the maker of the question, because he made the question AND found the answer for it.

[@xaris5309]

He may as well have first thought of the solution and then formulated the question. LOL!

[@chengkakful]

Or he gave a random question without an answer for it

[@heartscaless]

@@chengkakful if you understood how this question works you would know that's not possible.

[@chengkakful]

@@heartscaless truthfully i dont know what i was aying but it doesnt matter because he was looking for proof and not an answer

[@taintedPot]

@@chengkakful yeah he could possibly find this correlation and tried to prove it but when failed he submitted it just to find out answer.

[@fawazshah]

Did you ever hear the tragedy of Question Six the Impossible? It's not a story Terry Tao would tell you

[@abinashah9085]

Fawaz Shah wht z that

[@tankigamingwithrichardrock9930]

Hey you realised, terry and 4 other people on the team got 1 out of 7

[@rossrandomstuff9365]

It's a Star Wars Reference.

[@fi4re]

It's a math legend.

[@quack3891]

In case anyone who doesn't know what this is This is the tragedy of Darth Plagueis, of course, the Jedi will never tell you about it

[@vikingslayer34]

This took me 8 hours. 7 hours and 55 minutes of thinking, and 5 minutes of smashing my computer.

[@mustafakaratas7326]

Ahahahah

[@leswhynin913]

In what order?

[@NoNameAtAll2]

how can you use computer for this?

[@RomanHold]

@@NoNameAtAll2 you use Google in a brute force and optimal way and are going through forums and ask people how to express certain things eg. how to formulate a proof of contradiction that is valid in a sub set of parameters, which also have an infinite cardinal size. And because it doesn't apply for "this" specific counter set of rules, it necessarily has to apply to the other set of rules, which is what you are meant to show. Which is possible by breaking sth down to the two fundamental principles/theorems of the classic logic.

[@512TheWolf512]

Why isn't the person who designed this problem revered?

[@rickywang3648]

Yeah! just like Goldbach conjecture

[@royal6355]

Why do you use a French word that I have to look up? You could say respected highly. Anyway, Thanks for teaching me a new word.

[@DespOIcito]

Because it was the one user in the video?

[@spiderduckpig]

Röyal revered is an English word

[@royal6355]

@@spiderduckpig English has borrowed it from French(Révéré in French).

[@Intertayne]

"Number 6 will shock you!"

[@numberphile]

ha ha - that is only naughty if you are making people endure 1,2,3,4 and 5 for no good reason - we can't help it that question 6 is called question 6! :)

[@Intertayne]

I'm just referencing those ads for list articles you find at the bottom of other articles on websites like Buzzfeed.

[@Schobbish]

Mathematicians HATE question number six.

[@toss4409]

LOL

[@shikharsrivastava5020]

You have 66 currently

[@mundanest]

*Takes Simon a year to solve Numberphile: I hope you cracked out your pencils Me: Nah, I'm good

[@taintedPot]

im kinda want to try...

[@lemlok]

this made me laugh out loud

[@robbiepanteleev7549]

HAHAHAHA

[@robbiepanteleev7549]

I could have googled for the solution though

[@NemosChannel]

4:56 I felt so dumb until I realised he said 90 minutes, not 19 xD

[@soulcatch]

OMG. I was scrolling through the comments looking for how he got 19 minutes. Thought I was an idiot. Probably still am, but at least I know it wasn't for this.

[@divineinterventionasyouwis5725]

+soulcatch looooooool

[@jadude378]

omg i still thought he said 19 and was so confused until i saw this comment

[@Flocko_]

I had to replay it 3 times and hear him say it another 2 before I realized was actually saying 90 and not 19.

[@NemosChannel]

Flocko7x yeah, that's what I did too :'D

[@Nathar45]

Honestly, delivering a problem as a story like this works amazingly. As always, you deliver a prime product :)

[@numberphile]

cheers

[@DavidVaughan00]

A prime product? No such thing!

[@Djorgal]

No it's not a prime, it's a square.

[@camdenfitzgerald2557]

technically it would still be a product if you multiplied the prime by one

2,5*2 is a product and a prime

[@caciowskij768]

"[...] one of the hardest problems... EVAH!"

[@ThomasJr]

Adam and Evah !

[@jaakkojokelainen5124]

I was there in 1988 and got one point like Terence Tao! This video inspired me to try this again and after a week of solving I am pretty sure I got a proof...

[@chrisc7248]

I came up with a= any positive integer (1,2,3,4 etc..) and b= that numbers cube (1, 8, 27, 64 etc...). Was that the proof?

[@AroundTheBest]

Same, though I made the assumption that the square of A^2+B^2/(AB+1) = A (instead of X). Using A allowed me to reduce to B = A^3, though this is missing the step where X would have to equal A. Only spent 10 minutes on it though. Maybe I'll look into this later.

[@benjaminleaber4691]

It's one. a and b equal one, don't they?

[@sydelixyt]

Benjamin Leaber no because then you’d get a fraction as your answer, more specifically 1/2

[@jwhiteheadcc]

(0+0)/(0+1)=0, and (1+1)/(1+1)=2/2, right? Or maybe I did it wrong. Anyways, that gives 2 solutions and isn't actually proving the claim. Now I want to see strategy/s they used to find the proof! :)

[@lancetschirhart7676]

every time this video pops up as recommended I think that the thumbnail is a picture of me

[@flashrogue4376]

4 miniutes in "GET ON WITH IT" stop pandering! 5:28

[@pattymaboy5425]

christ thank you

[@MrGrga987]

THANK YOU

[@_EvilKam]

Winning comment right there.

[@TempestPhaedra]

thanks man, the video intro is unbearable

[@kd5nrh]

When mathematicians desperately try to build suspense.

[@devinnie7572]

"Three questions per day. It was the third day, and so the third question was question six" Now that is some complicated maths. Let me give my brain some time to process that, I'll get to the rest of the video in a moment.

[@exonizu.1597]

3 per day, 3rd day, the third question was question six. Let me get this straight. 3x2 equals 6, if it were the 3rd day it would be nine. But the third question was six as it was stating, but if were the third question 3 per day, then you would wait 2 days for question 6 therefore it would be impossible. It will only equal 3 - 3 per day, the third was question six. Only three!

[@malikmister]

I know I'm replying to a year old comment, but he clearly said it was two days and it was on the second day.

[@SkKedDy]

Christopher Mango wooshhh

[@nightcoreandanimeislife1888]

@@mangomath2175 I think instead of doing that they should try going to an otologist

[@SpydersByte]

@DeVinnie dafuck are you talking about? He clearly said 'second day' like, multiple times.

[@alcesmir]

There seems to be a worrying amount of people who don't understand the question and think that supplying one example where (a^2+b^2)/(ab+1) is a square solves the question. The question asks you to show that in *all* cases where the fractions turns out to be an integer, that integer is square. All cases. *Not* one case. All cases. And for people spazzing out about 0 being included in the video, the statement to be proven holds for a=0 or b=0 as well.

[@SsJVasto]

a=0, b=0 doesn't work... 0²+0²/0*0+1 ==> 1+1/0+1 ==> 2/1 ==> 2 2 is not a square number

[@SsJVasto]

I see what you meant now, I'm not removing my previous statement (I don't believe in censoring my stupidity), and I totally didn't see the question in that way. I didn't realize that the theorem was "Prove that if the result is an integer, it MUST be a square". I stupidly thought it was just "Find a result that's an integer and a prime"... Sorry.

[@SsJVasto]

The question does state, however, that the *a* and *b* variables must be positive, and I'm not convinced that 0 is a positive number... I don't know if that's debatable or not, I've always seen "positive numbers" as "numbers greater than 0". He also only includes "0" in his list of "integers" not "positive integers"...

[@alcesmir]

+Justin Drobey Including zero is still an error in presenting the problem, but it's an error that doesn't change the nature of the problem and it makes the theorem a tad more general. The theorem to prove is quite a surprising one. For me it's very unexpected and beautiful that this expression can take on only fractions and a very specific subset of integers. I haven't played around with the problem too much yet, but I suspect there might be something special about the possible fractions as well.

[@Pulsar77]

0^2 = 0.

[@henryg.8762]

"this is one of the hardest problems [pause] *_EVAH_*"

[@mcol3]

But why did it take 5 minutes to see the question?

[@numberphile]

+mcol3 you must hate the movie Jaws!

[@noodleb2248]

+Numberphile lololol

[@deamon6681]

"Anticipation" "Suspense" "Drama" "Bad Decisions" I'll let you pick one yourself...

[@blacxthornE]

Because the title isn't just "Question Six"

[@james0xaf]

Would you prefer it if the problem was at the start and the bit explaining it will probably take a lot of hours to crack came after?

[@theworldeatswithyou]

I really like these kind of videos.

[@numberphile]

so do we!!!

[@Excalibur-lw5ie]

+Numberphile pls make more👏

[@Ezullof]

I really don't like this kind of videos. I guess it's a matter of taste.

[@paradox4140]

Well, there is 971,151 people do !

[@martinpaddle]

One of the participants of the 1988 IMO who was able to solve the problem (and win a gold medal with a perfect score) is Ngô Bảo Châu, who would also go on to win the fields medal (in 2010).

[@ilducedimas]

I love the passion of this man. Bring him back as often as possible please !

[@HowRandomIsRandom]

At 0:59, "6 questions worth 7 points each". So the maximum total score is 6 * 7 = 42. I see what they did there.

[@lewiszim]

Douglas Adams wrote in the Hitchiker's Guide to the Galaxy that 42 is the answer to the ultimate question of life the universe and anything. The problem is that we don't know what the question is. We do, however, learn in the third book that the question is not 6*7.

[@want-diversecontent3887]

lewiszim well is it factorials because day days

[@jaxryz_380]

lewiszim did you know that ASCII 42 represented an asterisk, which is basically used as a “whatever you want it to be symbol”. The giant computer was asked “what is the meaning of life” and the computer responded how a computer would. “Whatever you want it to be”

[@ultimaxkom8728]

@@jaxryz_380 Butt=Blown. Thank you.

[@jayantsingh9008]

I was just watching p*rn and accidentally opens KZbin and this was in my recommendation , not gonna lie this has more logic and concept than what i was watching before,and even more interesting.

[@thej3799]

🤔

[@scarcesense6449]

I shouldn't have watched this at 5 mins to midnight, I could be up all night.

[@irrelevant_noob]

all *year ;-)

[@rdbchase]

~6:00: "a and b can be any whole number, including zero ..." -- huh? Not if the problem states that a and b are positive integers, and it does!

[@DepFromDiscord]

Robert Chase zero is positive and negative. Or at least it can be.

[@foxxygearreviews7754]

Robert Chase if they can be any positive integer then a can equal b can equal 1 and the equation seems to work unless I’m missing something obvious.

[@chasefuller8496]

The integers are Z = {...-2,-1,0,1,2...} Positive integers are all integers without a negative sign, so Z+ = {0,1,2...}

[@rdbchase]

@@chasefuller8496 Nonsense! Z+ = {1, 2, 3, ...}. Did you people attend the University of Contrarian Mathematics?

[@borisrodriguez8575]

Matter of convention, it looks like. Here in the States, positive means strictly greater than zero and negative strictly less. Zero is just zero. For the numbers 0, 1, 2, ... we say nonnegative.

[@mathsinmo4372]

please check this solution a²+b² can be written as (a²+b²)(1+ab) - ab(a²+b²) and as (1+ab)|(a²+b²) then ab(a²+b²) should be equal to zero In case 1, when a² + b² = 0, the expression (a² + b²)/(1 + ab) simplifies to 0/(1 + ab) = 0, which is indeed a perfect square. In case 2, when ab = 0, the expression (a² + b²)/(1 + ab) simplifies to (a² + b²)/(1 + 0) = (a² + b²)/1 = a² + b². Since ab = 0, it follows that a² + b² = (a + b)², which is a perfect square. Therefore, based on these two cases, it can be concluded that for any values of a and b, the expression (a² + b²)/(1 + ab) is always a perfect square.

[@sierpinskibrot]

Case 1 only shows it is a square when both a&b are 0. Case 2 only shows it is a square when one or both of a&b is 0. There are solutions like 2,8 where a^2+b^2≠0 and ab≠0

[@Red_wine]

0:00 Did you ever hear the tradegy of Darth Plagueis the wise?

[@masterrecon118]

No? Is it a story the Jedi wouldn't tell me?

[@magicmulder]

Terence Tao has abilities some may consider… unnatural.

[@andyxu9969]

Very cool problem! One of my favorites. In fact, one can prove that all solutions can be generated by (k, k^3) for all integers k>0 .(sans order) The argument is surprisingly simple: FIx (a^2+b^2)/(ab+1)=x, and then see that if (a,b) is a solution with a+b minimized, then (xb-a, b) is also a solution with the same x-value (not too difficult compared to the other problems), and if a>b the a>xb-a as well. The only way out is if a=xb and you get the previous solution. (forces k=b^2) Admittedly, this solution exploits a rather modern technique used as Vieta jumping, which basically solves a quadratic in one of the variables. Tells you how much more difficult the problems have gotten these days!

[@SpeedyMemes]

counterexample: (a,b) = (30,8) lol

[@nqobilemsomi3656]

I have no idea why I'm watching this video, I still count using my hands... it's just giving my flashbacks of high school maths classes

[@fernwehn5925]

Hm checks out.

[@aparthia]

The actual question is discussed starting at 5m10s

[@numberphile]

that background is the best bit for me - without all that, it is just a hard question!

[@aparthia]

Maybe.. 5m of repeating how hard it is is a bit repetitive for me at least.

[@D0mmac]

To be honest, I started to get really annoyed after hearing how hard it is for three minutes and still not knowing what it actually is. It's a bit like clickbaiting pages. "hey, let me tell you about this really cool problem... lol, nope, after peaking your interest I'm just telling you all the background that you don't care about". Yes, the background might be interesting, too. But once you say that there's this really cool problem, the single thing I want to know most right now is what the problem actually is. Just tell it, don't tease people for 5 minutes.

[@oioki87]

Thanks mate

[@gr-gx4zy]

I agree. Keep up the good work Numberphile. :)

[@MrTej780]

Simon is my favourite. Bear in mind that Matt and James have already set the bar astronomically high.

[@santoshkmallick5781]

I tried a basic method:- Say that (ab+1) divides (a^2+b^2) and the divisor is +ve integer "k". This solving shall give us an equation:- That a=b^3 or b=a^3 which when put in the main question gives result as a^2 or b^2 which is the perfect quare of an integer.

[@amosmacdonald9573]

This is mathematically correct but there are alot u need to consider..... before coming to that level

[@gammaknife167]

At the same time I saw the title of this video, I was in the middle of working through my M2 maths book. I was on Q6.

[@WalrusRiderEntertainment]

I remember Terrance Tao at Flinders Uni

[@cyrusbeiryrao2428]

How do you remember everyone

[@moroccangeographer8993]

"It is about being able to solve awesomely hard problems" should have been my life philosophy and goal.

[@tetradb_]

Have to love the the enthusiasm Simon has for Numbers :)

[@JoeHinkle11]

Okay, taking a 90 minute break from reading to try. I think I’m going to cry.

[@tjspeirs75]

one of my favorite numberphile videos. the storytelling is best of the best

[@StephenMortimer]

He has a certain appealing charm.. but can't figure just what it is

[@minch333]

Enthusiasm

[@StephenMortimer]

minch333 Is it REALLY that simple??

[@minch333]

Stephen Mortimer Well it's the occam's razor answer at least!

[@StephenMortimer]

minch333 That's it.. this OldGuy don't shave anymore (once every 2 months..OFF.. with the hair clippers... I go from fuzz ball to billiard ball look)

[@YouHolli]

The charm of danger. With those teeth you expect him to bite somethings head off at any time.

[@BYMYSYD]

Haha. I like how those problems in number theory are so simple to state...even an 8th grader could under stand what the question asks. But to solve them requires maths of a much larger caliber.

[@0vectoria485]

It soo easy even i am not genuise in math but a^2+b^2>=2ab and because a,b are positive entegres so a^2+b^2>=ab If we place a ,b by the smallest value it seems like that a=1. b=0 We find 1>=0 So we can add 1 to the right side a^2+b^2>=ab+1 so a^2+b^2\ab+1>=1 AND 1 is square integre

[@Arkhanno]

We need more videos with Simon in them. They're always entertaining to watch.

[@TheKrazyLobster]

"This question stumped a Fields medalist" *Random KZbin Commenters who want to feel special*: "Pathetic."

[@concernedcommenter8258]

Bruh I thought I solved it but then I realized I had to prove why it worked and then I gave up theres no way I could

[@ericzeigler8669]

I solved this with a great insight and shortcut that took about 45 minutes. I wrote it in the margin of this video, but changed my text view size and lost it.

[@umbreon8527]

Lol a and b can't be 0 it says "POSITIVE integers".

[@Deathranger999]

+Maxime Couture (Apophyx) Wrong. It is neither.

[@Deathranger999]

+Maxime Couture (Apophyx) Oy guys. Simon can make a mistake. A positive integer is defined as any integer n > 0. So obviously 0 doesn't work. Similarly for negative.

[@Deathranger999]

+Maxime Couture (Apophyx) Oy guys. Simon can make a mistake. A positive integer is defined as any integer n > 0. So obviously 0 doesn't work. Similarly for negative.

[@Deathranger999]

+Maxime Couture (Apophyx) Oy guys. Simon can make a mistake. A positive integer is defined as any integer n > 0. So obviously 0 doesn't work. Similarly for negative.

[@mina86]

Ultimately it doesn’t matter in this question since you have to prove for all a, b pairs. The fact that a=0 or b=0 is a trivial solution doesn’t help you with cases where a and b are positive.

[@jorgej4071]

3:16 It actually appears in the Niven's number theory book. It's the last problem of the section 1.2

[@bobingstern4448]

I wouldn’t be surprised if they put the frickin Riemann Hypothesis on the Olympiad and someone solved it

[@agrajyadav2951]

Bruh

[@SelfishNeuron]

I would already cry if I got that question asked in an exam, not only if I were able to solve it. (but I would cry hard if I could solve it!)

[@rungratree1]

Omg just tell me the problem

[@1mol831]

You don't need a talent to do these problems, you just need teammates who think with you

[@missclick7609]

@@1mol831 bruh what? It's individual

[@apacheattackhelicopter5823]

1 Mol did your head get hit?

[@shreerangvaidya9264]

@@1mol831 I think you mean that more than one person can solve it faster. Or maybe you are being... Skeptical or something.

[@JS-iu3ce]

rungratree1 4:48

[@veta557]

It’s not hard to prove if you visualize a*a and b*b as squares on paper and also ab+1 as a rectangle plus one square (the +1 part), where 1 is a fraction of a*a square. This fraction is always a square of an integer because in order to fit (ab + 1) into (a*a + b*b) you need to fit that +1 part into a*a square the whole number of times. That said, the b should be always equal (a*a)*a = a^3 to satisfy the condition of this equation. Try to use b=a^3 to see that it works. It took me about 30-40 minutes to visualize, understand and explain this solution.

[@chrissdehaan]

This post deserves to be a lot higher.

[@KingCat3rax]

Wow great imagery thank you, much easier to understand when visually represented

[@SpeedyMemes]

counter example: (a,b)=(30,8). this does not conform to the condition b=a^3

[@TopFloorEricc]

Reading that question gives me a lot of anxiety remembering how badly I failed my college Trig and Pre-Calc class 😂

[@DavidEriksson372]

That competition must have been held on April 1st. What were they thinking?!

[@John_Ridley]

What do you mean? this is a world-wide competition of the elite of students. One student solved it excellently and several solved it. This isn't a competency quiz, this is a hard test intended to single out the few top guns in the world. it's EXPECTED that most competitors won't be able to solve all the problems, in fact it's necessary.

[@DavidEriksson372]

Thanks for clarifying that, I just thought that a problem that even most math professors couldn't solve, would be used in a test aimed for children and teens. Thanks for showing me it in a different perspective :)

[@11Sparky111]

Guess it shows you how smart the 11 competitors are then if they were able to solve it in such a short time.

[@s0gu0001]

If you were able to solve 3 of the 6 problems, you will probably get a medal, if you solved 4 perfectly, you probably get a gold medal. Problems 3 and 6 are those extra hard ones most people will not solve.

[@alexanderstiefelmann5982]

For very gifted children and teens tough. To get to the International Math Olympiad, you have to qualify via lower-level olympiads (f.x. in Germany it is school, then city, then state, then country). I never got past the state level (once got the second prize and the state level though, and it is still an unsolved mystery to me why I was not invited for the country level olympiad to which I prepared fiercely). And you know what? Those tasks are often kind of - you may struggle to find the solution, but once you see the solution you don't stop wondering how easy it was..... But I also noticed the tasks got harder with time. The International tasks of early 1960s are actually about the same difficulty level as the State tasks of 2005.

[@tasdau]

5:29 for those like me that have no tolerance for long winded intros.

[@98swarup]

Thank you sir!

[@YG-do3dg]

Thanks for that. If only your comment was pinned to the top..

[@septimusheap1778]

5:15 that laugh a professor has when he knows that you will fail at the task. 'well the best mathematicians in the world could not solve this problem in 6hours, well i give you 90 min' you are welcome.

[@FlyingTurtleLP]

I'm at 6:27 ... but I'm affraid they will explain the solution. My question: *Spoiler alert?*

[@FlawzDesigns]

They dont. Its in another video

[@septic1448]

The solution is a=2 and b=8 it was really simple took me only 15 minutes tbh

[@void9720]

Septic No???

[@wengeli4370]

Teacher: All right students you have one question to attempt and you are given 8760 hours. The exam starts now.

[@lucaschai5788]

1988 Question 6 is the hardest question eva! 2020 Question 6: Am I a joke to you?

[@zawadulhoque4511]

man our country got a 1/42

[@shantanunene4389]

1988 Q6 is actually kinda easy by today's standards. Vieta jumping is a standard technique

[@MrMomoro123]

+Numberphile A small error at 6:04 - a and b cannot be zero as they are positive integers. 0 is not a positive integer.

[@wierdalien1]

+Max R. yes it is.

[@wierdalien1]

+Max R. oh no sorry. i thought you said even.

[@NamesAreJustAdvertisingIDs]

Hm, interesting. I basically threw 0 into this and called it a day, but I never drew a distinction between "positive" and "even" until now. I guess a simple reminder which just now occurred to me is that "'Even' cannot necessarily equal 'positive' since 'positive' also includes those 'odd' numbers," or that if we say that 0 = even, then all non-zero numbers = odd, though they would all ("all" as in all numbers greater than 0) still be considered positive.** (o_0)**

[@prakhargodara5898]

Solved it in under 5 mins. Assume a

[@ViKtoReinKILL]

My old math professor at 7:44. Professor Kung at St. Mary's College of Maryland.

[@thgodfry]

I made a comment earlier, saying that i solved it, and im sure i did. its not a simple a=1, b=2. you have to show that you can solve (A^2 + B^2) / (AB + 1) = X^2 where 'X' is an integer greater than or equal to 1. so, A=1 and B=1 works, but so does A=2 and B=8 (X=2 so X^2=4) you're trying to solve for the sequence of answers. I first assumed that i could make B = A+n, where n is an integer > or = 0 so A and B can be the same so i rewrote the function as (A^2 + (A+n)^2) / (A(A+n)+1) = (x^2) / 1. i wrote is as a ratio cause it made it easier in my brain. then i moved AB+1 to the other side of the equation and tried to solve for X^2 = 4. so, (A^2 + (A+n)^2) = (4)(A(A+n)+1), where 'n' is the value that allows the 2 sides to be equal, from here it was a plug and chug in Desmos graphing calculator to find the intercept between the 2 at which all numbers are integers -> try all values of n = 1 - 10 6 is the only one that works when N=6, the point (2,68) is the intersect so A=2, B=8, X^2 = 4 from this i concluded that B = A(X^2). or you could say that i assumed this was relationship between A and B so (A^2 + (A(X^2))^2) / (A(A(X^2))+1) = (x^2) / 1 from here i thought, what if i make A = X? that seems to hold true for 1 and 2 A = X becomes (X^2 + (X(X^2))^2) / (X(X(X^2))+1) = (x^2) / 1 OR ((X^2) + (X^6)) / ((X^4) +1) = X^2 THEREFORE A = X , B = X^3 when 'X' is an integer > or = 1 and that is the solution for the sequence if X=5 -> (25 + 15625) / (625 + 1) = 15650/626 = 25

[@kaj9947]

Sneaky

[@thgodfry]

@@kaj9947 the sneakiest

[@anthonyruiz8404]

He said it can also be 0 so let A=0 and B=1. With that being said the addition of both squares is 1, and the product if the variables plus one is also 1, therefore making the equation 1/1 or the square of 1.

[@thgodfry]

@@anthonyruiz8404 A can not be 0 because 0 is not a positive integer

[@heartscaless]

To the people who keep saying the answer is 1, they're not asking for a solution to the expression, because it already tells you that the possible solutions are all squares. What they're asking is that you prove why that is always the case.

[@Xeverous]

There is a mistake in the vid: a and b are positive integers, none of them can be 0

[@dDoublevisioNn22]

0 is an integer. All positive whole numbers 1 and up are called counting numbers.

[@Arkalius80]

The stated condition holds if a and/or b is 0.

[@Xeverous]

the questions states "positive integer", 0 is neither possitive nor negative

[@dDoublevisioNn22]

true, i didn't catch that.

[@arachnid1483]

how is that a mistake

[@leirgauk]

ALWAYS keep on FIGHTING for ULTIMATE MATHEMATICAL GLORY!!!

[@tinhtranvan2928]

I remeber Chau Ngo, Vietnam people 7 point of that problem and he also win Field medal

[@DataJuggler]

The answer to almost all Numberphile riddles: 'Things you never really need to know the answer to'.

[@MrAskolein]

Crying after solving a problem like this is so understandable. Mathematics is something.

[@eduardciuca217]

This problem is hard! But it doesn't compare to problem 6 at Romanian Masters of Mathematics 2016 ! " A set of n points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets A and B. An ABtree is a configuration of n − 1 segments, each of which has an endpoint in A and the other in B, and such that no segments form a closed polyline. An AB-tree is transformed into another as follows: choose three distinct segments A1B1, B1A2 and A2B2 in the AB-tree such that A1 is in A and A1B1 + A2B2 > A1B2 + A2B1, and remove the segment A1B1 to replace it by the segment A1B2. Given any AB-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps." From 113 participans, only one did it completely and took 7 points, and a few scored 1 or 2 points !

[@jamirimaj6880]

Forget the scores, we should know who answered Question 6 correctly!

[@joester610]

Have you heard of the legend of question 6? It's not a legend any Jedi would tell you

[@garbagedump5705]

The moment I looked on the paper I almost left my iPad on my bed and went to another room

[@doughboi007]

Hearing about how brilliant these kinds are makes me think of how unhappy i am with my genetics.

[@tessarae9127]

Epigeneticism

[@colinjava8447]

Yeah, but they aren't even the top 1%, they are a much smaller group than that. I'm just pleased to be a lot better than average in maths, number theory was never really my thing anyway really.

[@AhnafAbdullah]

@@colinjava8447 I think it's just a matter of education. I'd love to have to have the knowledge and be able to solve problems like this, but we weren't even taught things like this in school... So I didn't even have the tools to make this question accessible.

[@jayluck8047]

Ahnaf Abdullah - In contrast, I had the education - the tools - however it wasn’t until I used raw artistic creativity that the Maths disciplines made sense. Some can draw a stained glass window with a box of crayons, others can manufacture one by incorporating the box itself. ;)

[@humanmold]

"That's numberwang!!!"

[@YashSingh-nx7pb]

Me : I am tired of my math study , let's take a break KZbin : wanna see some brain cells killing math problem ??? Hey KZbin , for what thing you are taking revenge on me ???

[@blackapple2938]

This is one of the hottest problems *_EVAHH_*

[@911gpd]

Parker Square demonstration

[@MrHeroicDemon]

that evil laugh at the end is him gone completely mad over this at one point for a year.

[@audiodood]

"Math is the language in which the secrets of the universe are written." -Theodore Gray

[@nathanielsharabi]

doesnt positive intagers mean the set 1,2,3,4... not 0,1,2,3,4... that would be a non negative set simon wrote down. ????

[@UpstairsPancake]

Yes, but it still works, because if a and b are both zero or if one of a and b is zero, then you have 0 or b^2 respectively, both perfect squares.

[@madduck09]

+SquishyBananaBread but if 0 is included you can just say a=0, and b=2

[@UpstairsPancake]

madduck09 Yep. That gives a perfect square. Doesn't prove the statement though.

[@DIYOrDIE777]

What do you have to prove? I didn´t understand

[@UpstairsPancake]

La Rebel You have to prove that it's true for every pair of natural numbers. Not just one pair.

[@tik5150_]

Someone has probably mentioned this already but the problem said a and b are positive integers. So neither are equal to 0, but 0 was included in this explanation. I suppose it doesn’t ultimately come up in the solution but something to make sure to point out for people giving this problem a try.

[@changchia-huan2177]

0 can be considered both positive and negative in some definitions -Wikipedia

[@thej3799]

@Chang Chia-huan yeah I believe this has been inferred on this channel, too. So,I stand by my solution.

[@rychlost2268]

hold my beer, i can do this

[@kei5985]

I just wanna at least pass my gcses and these teens tryin to figure out what life equals to

[@davidt1152]

(1^2 +1^2)/(1*1+1)=2/2=1, so a solution of the type they request exists. We discard all fraction solutions as contradictory. Alternately, in addition to the solution of (a^2 +b^2)/(a*b+1) being the square of an integer, it could be a prime number or a composite, non-squared number. Starting with a prime number as it is easier, we test (a^2 +b^2)/(a*b+1)=p. (a^2 +b^2)/(a*b+1)=p a^2 +b^2=p(a*b+1) a^2 +b^2=p(a*b)+p a^2 +b^2-p(a*b)=p which implies (a^2 +b^2)/(a*b+1)=a^2 +b^2-p(a*b) a^2 +b^2=[a^2 +b^2-p(a*b)](a*b+1) a^2 +b^2=[a^3*b+a^2] +[a*b^3+b^2]-p(a*b)^2-p 0=a^3*b+a*b^3-p(a*b)^2-p p=a^3*b+a*b^3-p(a*b)^2 = ab [a^2+b^2-p(a*b)] p=a*b*p so p is prime iff a*b = 1, which we already established (within the realm of acceptable solutions) means that p=1 as well. So, now we assume that we have a composite, non squared number, say p is replaced with n*m, where n =/= m. (a^2 +b^2)/(a*b+1)=n*m ... (as above) a^2 +b^2-p(a*b)=n*m and so, (a^2 +b^2)/(a*b+1)=a^2 +b^2-n*m(a*b) ... (as above) n*m=a*b*n*m again requiring a*b to equal 1, and thus a contradiction as n*m = 1 by consequence of our initial test. Therefore, all solutions must be as requested. So what's wrong with this answer?

[@loloolaf6359]

The answer expected to this problem at imo was just : "I have discovered a truly marvelous proof of this, which this margin is too narrow to contain." 90 minutes to find the courage to write this.

[@niranjanaanil3597]

*cries in Fermat*

[@yetanotherhma2]

terence tao should def have come with the disclaimer he was 13 at the time :)

[@iainfulton3781]

The pairs of integers that fit the equation are x^(2n-1) - (n-2)x^(2n-5) + T(n-4)x^(2n-9) - TT(n-6)x^(2n-13) + TTT(n-8)x^(2n-17) - TTTT(n-10)x^(2n-21) + ... where T(n) is the triangle number TT(n) is the triangle number of the triangle numbers and TTT(n) is the triangle number of the triangle numbers of the triangle numbers and so on. If you substitute n = n - 1 you get the other pair and if the power becomes negative you stop the formula. So if n = 11 you get a=(x^21 - 9x^17 + 28x^13 - 35x^9+15x^5- x) b= (x^19 - 8x^15 + 21x^11 - 20x^7 + 5x^3) cause T(11-4)=28 TT(11-6) = 1+3+6+10+15 =35 TTT(11-8) = 1+1+3+1+3+6=15 TTTT(11-10) =1 and T(10-4)=21 TT(10-6)=1+3+6+10=20 TTT(10-8) = 1+1+3=5. All the coefficients add to either (1,1) (1,0) (0,1) (0,-1) (-1,0) or (-1,-1) so that x = 1 will result in 1.

[@johnsmoke1785]

42 the answer to everything! 6 questions each giving 7 points = 42