The Legend of Question Six - Numberphile
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- Опубліковано 4 лют 2025
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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...
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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
for real
Tauseef Baggia if you’re so smart then go and become the smartest person in the world
Skip to 5:30... you're welcome.
Lol saw it right away a= 1 b= 0
@@MP-rh2pl prove it, keyboard genius
"Hey, let's give a bunch of teenagers one of the hardest problems ever conceived in mathematics."
too right - The Math Olympiad is for Top Guns!
who are actually kids. Here I thought I was good in maths for my age.
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 ;-)
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.
+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)
Do you know The Legend of Question Six
Me, an intellectual: Why was number six scared of seven?
A: SIX is the more proficient model. (BSG [BATTLESTAR GALACTICA]).
Because seven eight ate (eight) nine.
because seven was a registered six offender
@@lovelypotatoes most underrated comment
Because he had the high ground
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
450/90= 5 so 5 questions. I don't get it
4.5 hours == 270 minutes
3 questions
270/3 = 90
@@halo6mastercheif It's a joke
That's as far as I got. I was proud of myself.
@@wcsxwcsx i understood 19 and was seriously questioning my math
"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
Pls can you see my comment on this video. I proved it.
I THOUGHT I WAS AN IDIOT AND HE SAID 19 AND THERE WAS SOMETHING IM MISSING
@@ezralebowitz3371 omg me too 🤣🤣
@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
😆 🤣
I came up with a solution in 4minutes....it was wrong.
Hum Shak so what’s the point?
@@apacheattackhelicopter5823 he tried
I did in about 10 seconds. It was- it was wrong.
hahahahahahahahhahahhahahahaahahhaahahahahaha
I beat you. I arrived at that solution at 2 minutes.
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
"have you heard of the Legend of question six?"
no, but I've heard of the Emu war 1932
Lan Vu gang gang gang?
Big deal. Have you ever heard of darth Plaguies the wise?
Kiwis
I thought not, it's not a story the schools would tell you
Okay, but have you ever heard of the tragedy of darth plauguas the wise?
I just solved it, but there wasn't enough room here to type it in so I haven't.
Colin Java it’s just 1.
@@jamievlogs7103 you clearly didn't get the joke.
@@somnathdash4428 You're just mad because John Gnash got the fields medal and you didn't.
@@jamievlogs7103 Colin Java´s last theorem ;)
Typical Fermat
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.
JoMomma239 but they patched that in update v1.30
Not gonna lie.. they had us in the first half
And "your" bad at using grammar. YOU'RE welcome.
Seriously, why did more than 430 people actually like your comment? Are they blind?
435 people*, are you blind?
@@nick242 Read it again... "why did *more than* 430 people". I said "more than" for future-proofing.
Are you blind?
"Have you heard of the legend of question six?" No but I have heard of thelegend27.
Yes, but have you heard the legend of the Tragedy of Darth Plaguis the Wise?
Did you heard the tragedy that reach the man?
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?_*
Kortan Shizuka have you heard of The Rise of the Legend of the Tragedy of Dark Plaguis the Wise?
Ah, Man of culture
Why doesn’t anybody give credit to the people who designed these questions. They must be even MORE genius.
Maximoose 2005 there are problem proposers and they are famous among IMO Community
WRONG!! Much easier to pose questions than to solve them.
are you joking?
@@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.
@@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.
Australian maths be like "Oi to the power of mate², carry the roo = shrimps on the barbie."
You forgot to divide by boomerang
And carry the crikey.
💀💀💀💀
Solve for Kano
WHAT 😂
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!
The amazing thing is that A handful of participants were able to do it correctly in stipulated time
it really boggles my mind
for sure
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.
He may as well have first thought of the solution and then formulated the question. LOL!
Or he gave a random question without an answer for it
@@chengkakful if you understood how this question works you would know that's not possible.
@@heartscaless truthfully i dont know what i was aying but it doesnt matter because he was looking for proof and not an answer
@@chengkakful yeah he could possibly find this correlation and tried to prove it but when failed he submitted it just to find out answer.
Why isn't the person who designed this problem revered?
Yeah! just like Goldbach conjecture
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.
Because it was the one user in the video?
Röyal revered is an English word
@@spiderduckpig English has borrowed it from French(Révéré in French).
This took me 8 hours. 7 hours and 55 minutes of thinking, and 5 minutes of smashing my computer.
Ahahahah
In what order?
how can you use computer for this?
@@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.
"Number 6 will shock you!"
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! :)
I'm just referencing those ads for list articles you find at the bottom of other articles on websites like Buzzfeed.
Mathematicians HATE question number six.
LOL
You have 66 currently
Honestly, delivering a problem as a story like this works amazingly. As always, you deliver a prime product :)
cheers
A prime product? No such thing!
No it's not a prime, it's a square.
technically it would still be a product if you multiplied the prime by one
2,5*2 is a product and a prime
*Takes Simon a year to solve
Numberphile: I hope you cracked out your pencils
Me: Nah, I'm good
im kinda want to try...
this made me laugh out loud
HAHAHAHA
I could have googled for the solution though
But why did it take 5 minutes to see the question?
+mcol3 you must hate the movie Jaws!
+Numberphile lololol
"Anticipation"
"Suspense"
"Drama"
"Bad Decisions"
I'll let you pick one yourself...
Because the title isn't just "Question Six"
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?
Did you ever hear the tragedy of Question Six the Impossible? It's not a story Terry Tao would tell you
Fawaz Shah wht z that
Hey you realised, terry and 4 other people on the team got 1 out of 7
It's a Star Wars Reference.
It's a math legend.
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
"[...] one of the hardest problems... EVAH!"
Adam and Evah !
"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.
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!
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.
Christopher Mango wooshhh
@@mangomath2175 I think instead of doing that they should try going to an otologist
@DeVinnie dafuck are you talking about? He clearly said 'second day' like, multiple times.
4:56 I felt so dumb until I realised he said 90 minutes, not 19 xD
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.
+soulcatch looooooool
omg i still thought he said 19 and was so confused until i saw this comment
I had to replay it 3 times and hear him say it another 2 before I realized was actually saying 90 and not 19.
Flocko7x yeah, that's what I did too :'D
every time this video pops up as recommended I think that the thumbnail is a picture of me
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...
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?
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.
It's one. a and b equal one, don't they?
Benjamin Leaber no because then you’d get a fraction as your answer, more specifically 1/2
(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! :)
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.
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
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.
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"...
+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.
0^2 = 0.
4 miniutes in "GET ON WITH IT" stop pandering!
5:28
christ thank you
THANK YOU
Winning comment right there.
thanks man, the video intro is unbearable
When mathematicians desperately try to build suspense.
I really like these kind of videos.
so do we!!!
+Numberphile pls make more👏
I really don't like this kind of videos. I guess it's a matter of taste.
Well, there is 971,151 people do !
I love the passion of this man. Bring him back as often as possible please !
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).
The actual question is discussed starting at 5m10s
that background is the best bit for me - without all that, it is just a hard question!
Maybe.. 5m of repeating how hard it is is a bit repetitive for me at least.
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.
Thanks mate
I agree. Keep up the good work Numberphile. :)
"this is one of the hardest problems [pause] *_EVAH_*"
I was just watching p*rn and accidentally opens UA-cam 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.
🤔
At 0:59, "6 questions worth 7 points each". So the maximum total score is 6 * 7 = 42. I see what they did there.
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.
lewiszim
well is it factorials because day days
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”
@@jaxryz_380 Butt=Blown. Thank you.
I remember Terrance Tao at Flinders Uni
How do you remember everyone
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.
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
Simon is my favourite. Bear in mind that Matt and James have already set the bar astronomically high.
Omg just tell me the problem
You don't need a talent to do these problems, you just need teammates who think with you
@@1mol831 bruh what? It's individual
1 Mol did your head get hit?
@@1mol831 I think you mean that more than one person can solve it faster.
Or maybe you are being... Skeptical or something.
rungratree1 4:48
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.
This is mathematically correct but there are alot u need to consider..... before coming to that level
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!
counterexample: (a,b) = (30,8) lol
I shouldn't have watched this at 5 mins to midnight, I could be up all night.
all *year ;-)
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
Hm checks out.
Have to love the the enthusiasm Simon has for Numbers :)
"It is about being able to solve awesomely hard problems" should have been my life philosophy and goal.
~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!
Robert Chase zero is positive and negative. Or at least it can be.
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.
The integers are
Z = {...-2,-1,0,1,2...}
Positive integers are all integers without a negative sign, so
Z+ = {0,1,2...}
@@chasefuller8496 Nonsense! Z+ = {1, 2, 3, ...}. Did you people attend the University of Contrarian Mathematics?
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.
one of my favorite numberphile videos. the storytelling is best of the best
We need more videos with Simon in them. They're always entertaining to watch.
0:00 Did you ever hear the tradegy of Darth Plagueis the wise?
No? Is it a story the Jedi wouldn't tell me?
Terence Tao has abilities some may consider… unnatural.
Okay, taking a 90 minute break from reading to try. I think I’m going to cry.
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.
This dude cried because he solved s year long math problem… i love this world
He has a certain appealing charm.. but can't figure just what it is
Enthusiasm
minch333
Is it REALLY that simple??
Stephen Mortimer Well it's the occam's razor answer at least!
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)
The charm of danger. With those teeth you expect him to bite somethings head off at any time.
"This question stumped a Fields medalist"
*Random UA-cam Commenters who want to feel special*: "Pathetic."
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
One cool thing about this is that one of the guys that solved the problem during the competition is now running for president in Romania and is currently the mayor of Bucharest!
Lol a and b can't be 0 it says "POSITIVE integers".
+Maxime Couture (Apophyx) Wrong. It is neither.
+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.
+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.
+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.
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.
That competition must have been held on April 1st. What were they thinking?!
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.
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 :)
Guess it shows you how smart the 11 competitors are then if they were able to solve it in such a short time.
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.
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.
I wouldn’t be surprised if they put the frickin Riemann Hypothesis on the Olympiad and someone solved it
Bruh
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.
5:29 for those like me that have no tolerance for long winded intros.
Thank you sir!
Thanks for that. If only your comment was pinned to the top..
Whoever came up with that question, deserves a million dollars.
1988 Question 6 is the hardest question eva!
2020 Question 6: Am I a joke to you?
man our country got a 1/42
1988 Q6 is actually kinda easy by today's standards. Vieta jumping is a standard technique
Crying after solving a problem like this is so understandable. Mathematics is something.
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
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!)
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
Sneaky
@@kaj9947 the sneakiest
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.
@@anthonyruiz8404 A can not be 0 because 0 is not a positive integer
Solved it in under 5 mins. Assume a
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.
This post deserves to be a lot higher.
Wow great imagery thank you, much easier to understand when visually represented
counter example: (a,b)=(30,8). this does not conform to the condition b=a^3
The answer to almost all Numberphile riddles: 'Things you never really need to know the answer to'.
I just come back to this video like once a year, it's that fascinating.
3:16 It actually appears in the Niven's number theory book. It's the last problem of the section 1.2
ALWAYS keep on FIGHTING for ULTIMATE MATHEMATICAL GLORY!!!
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.
Reading that question gives me a lot of anxiety remembering how badly I failed my college Trig and Pre-Calc class 😂
There is a mistake in the vid: a and b are positive integers, none of them can be 0
0 is an integer. All positive whole numbers 1 and up are called counting numbers.
The stated condition holds if a and/or b is 0.
the questions states "positive integer", 0 is neither possitive nor negative
true, i didn't catch that.
how is that a mistake
I remeber Chau Ngo, Vietnam people 7 point of that problem and he also win Field medal
Parker Square demonstration
My old math professor at 7:44. Professor Kung at St. Mary's College of Maryland.
Most if not all of the solutions are where a is the cube of b, or vice versa, The only non-zero solution for 0 < a, b < 20 is a=8 and b=2
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.
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.
*cries in Fermat*
"and if you cant work out that it was 90 minutes, you would struggle with this exam" 😂
"Math is the language in which the secrets of the universe are written."
-Theodore Gray
Hearing about how brilliant these kinds are makes me think of how unhappy i am with my genetics.
Epigeneticism
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.
@@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.
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. ;)
so awesome that 1.5 million people have watched this.
+Numberphile A small error at 6:04 - a and b cannot be zero as they are positive integers. 0 is not a positive integer.
+Max R. yes it is.
+Max R. oh no sorry. i thought you said even.
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)**
"That's numberwang!!!"
This video was 7 minutes longer than it needed to be.
I'm at 6:27 ... but I'm affraid they will explain the solution. My question: *Spoiler alert?*
They dont. Its in another video
The solution is a=2 and b=8 it was really simple took me only 15 minutes tbh
Septic No???
Forget the scores, we should know who answered Question 6 correctly!
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.
This is one of the hottest problems
*_EVAHH_*
42 the answer to everything!
6 questions each giving 7 points = 42
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.
terence tao should def have come with the disclaimer he was 13 at the time :)
**This channel have π million subscribers **
or, π x 10^8
Two solutions: both a and b equal 1. Or one equals 0 and the other equals 1. Both ways give the fraction of 1 which is the square of itself. Took like 5 seconds of thought... now if the solution of higher squares is required, just involves some intuitive reasoning . Set the fraction equal to the square and reverse solve.
“Have you heard of the Legend of Question Six?”
“No.”
“I thought not, it’s not a story Mathemeticians would tell you.”