You, Me and The Legend of Question 6
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- Опубликовано: 19 апр 2020
- 1988 IMO question 6 is usually regarded as the HARDEST question. www.quora.com/What-is-the-tou...
The Legend of Question 6, by Numberphile: • The Legend of Question...
Here I present the solution that I was able to understand from Buzzorange: buzzorange.com/techorange/202... So good!!!
Vieta jumping:
brilliant.org/wiki/vieta-root...
en.wikipedia.org/wiki/Vieta_j...
Proof by contradiction:
sqrt(2) is irrational: • Sqrt(2) is irrational!...
there are infinitely many primes: • Euclid's proof that th...
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blackpenredpen
"Writing down the pf, because sometimes that's all I could do." No truer words can be spoken by a survivor of the proof-based university math curriculum
Bprp: This is my first IMO problem
One week later,
Bprp: The legend of question 6
Avik Das Lol yea
@@blackpenredpen lots of love from India 💓
It's incredible how 20 minutes of math with you feels like only 5min. Great explication.
Thank you!!
blackpenredpen wait what 20mins passed? I was totally shocked when he said 20 minutes,I thought it was 7-8mins.
Wow
I was also shocked when I saw that the video is 20 minutes long!!!
Yeah. He got to the part where he said he's done and drew a square and I was just wondering what's he gonna talk about in the next half of the video.
You guys only felt a few minutes of the time passing because you fell asleep
Imagine going to IMO in 1988, and when you're writing your answer for problem 6, you just write "Happy Bday Yoav Carmel"
Max Haibara he will be super happy!!!
@@blackpenredpen haha I am! i have followed your channel for about 2 years, i learned most of my calculus from you, and i talked about you and recommended you to all my friends in my university program. this birthday surprise was amazing, and i am so thankful for them, and especially for you!
blackpenredpen, had you ever been to IMO?
@@yoavcarmel1245 how did he know your birthday?
@@blackpenredpen "pf: Happy Bday Yoav Carmel"
Is he doing an IMO question using about 1m² of space
I just want to say thanks for keeping up this channel! It is definitely my source of learning advanced mathy-stuff so I can prep for contests like AMC :)
I stopped the video and tried to complete a binomial to find a solution, how innocent I was.
Sebastian Alvarez I mean, that it is the intuitive approach, it's just not very efficient, if I say so myself
Oh dear, hope you're alright
Sebastian Alvarez i did that too😅😅
Bonus problem: let a1, a2,...a1024 be 1024 consecutive integers such that the sum of their cube roots = k, for some positive integer k.
Show that k is a perfect cube.
I don't know if I took the wrong approach or misunderstood the question, but I re-wrote it as the summation from i=1 to 1024 of x_i^(1/3). x = {n, n+1, n+2, ..., n+1023}; n is an element of all integers. So, then it is a summation set equal to k. So I used induction. a^(1/3) + (a+1)^(1/3) + ... + (a+1023)^(1/3) = k. (a+1)^(1/3) + (a+2)^(1/3) + ... + (a+1024)^(1/3) = k. I then ended up with (a+1024)^(1/3) - a(1/3) + k = k. This came out to (a+1024)^(1/3) = a(1/3) which does not have a solution in the real plane, much less an integer. I know I did induction correctly so I assume I made a faulty assumption somewhere along the line.
@@sheppsu7353 In your induction you assume that both of the k's are equal. What is that assumption based on?
Jan Andreas Beecker I’d forgotten about that part. So instead it would be k_1 and k_2 and so the induction would not work out. Thx for correcting me.
@@sheppsu7353 Induction probably won't get you to a solution. Still, I think you may be on the right track....
Only if a1 = -511 (k = 512).
Wow you're the best. I've been trying to understand vieta jumping for a while now, but now I understand it much better now thanks to you. Awesome video!!
Most of the times, I can't relate to the videos you make since I haven't acquired enough mathematical background! But this one felt exactly like the proofs I usually get taught at school. It felt great to watch this and to understand every bit, especially on such a hard problem! Love your videos :)
bruh in numberphiles video he spent half the time talking about how hard it is
And the rest of the time alternating between considering zero to be positive and forgetting to prove anything at all about squares.
Then again, the committe members of the competition themselves couldn't solve this problem. So yeah.. it is hard.
Agree that it is hard, but I feel the Numberphile videos were showmanship and popularization with a minimum of substance, cashing in on how legendarily hard the problem is without meaningfully advancing the scholarship. Talking about Vieta jumping is fun, and the history is interesting, but the crucial link to squares (and exclusively squares) is lacking, as far as I can tell. And where is the rigor? Treating zero as positive is a bad start for constructing a proof! Yes, it helps explore the nature of the problem space, but the videos never come back from that point to address the problem as written.
Nicholas Hlinka ikr
@@AndyGoth111 yeah the numberphile one proved there are infinitely many solutions where k is a square. But that’s not what the question asks. You need to show that there are no values of a,b where k is an integer but not a square.
I learned about this method during preparation for my local math tournament. It's very interesting to learn more about this, especially on your channel. Good luck
THE ENTHUSIASM
Yes!!
Oh my god, this was an amazing solution, and a fantastic problem.
I'm trying to improve my technique by solving these kind of questions, so your videos help me a lot.
Fantastic! I love these videos! The presenter’s enthusiasm is so contagious!
Me: Want a video with th solution of the famous Question Number Six
*blackpenredpen, 5 minutes later* You, me and The Legend of question Number Six
One week later, 3b1b does a whole 20 min animation on it
He reads your mind .
Lmao
Exam paper:...Prove k is a perfect square
Me: ex: a=8 b= 2....k=4...
HENCE PROVED :)
I mean you're not, wrong, but you're not right either XD
The question is, prove that whenever the fraction yields a whole number, that number is a perfect square. So you only cited a case where it is true. You have to show that it follows every time.
Congrats, you have achieved superposition.
You have just proved that this cannot be disproved.
@@cristaldark4228 Yes, but actually no
I’m loving these well explained IMO problems!
Wow increduble. Also today I was revising exercises which imply vieta jumping. HereI put one of them
Find all pairs of positive integers (m, n), so that the following expression: (m^2+mn+n^2)/(mn-1) is also a positive integer.
Thanks a lot! I had watched the Numberphile videos and couldn't wrap my head around the solution.
These videos give me chills please continue the series👍👍
A really brilliant proof as always
I tried to prove this question by contradiction And I supposed that the square root of a^2+b^2 over ab+1 isn't an integer... This proposition led us to the fact that this latter is a number between to successive integers N and N+1 But unfortunately nothing discloses
A great acknolwledgment goes to you ❤️
"And I supposed that the square root of a^2+b^2 over ab+1 isn't an integer.."
Why? It is given in the statement that this _has_ to be an integer. You are contradicting the very thing which is _given_ in the text. _Not_ the statement which you are supposed to prove. That's not a good start. ;)
Hey Bprp, could you make a video about the polylogarithm function just like what you did with the W function? I'm sure many people would be interested!
PS : Love your vids keep it up :D
The legend of question fffffffffffffff 🤣🤣🤣
Lol Dr. Peyam!
I’ve waiting for this video a long time.
I’m very excited
12:22 -- There's space on the wall beside the whiteboard. :D
That's quite an upgrade from that problem about sums of digits :D. Vieta jumping is a beautiful technique
i literally had to take a pill to calm down because of how excited i was to see your solution to the problem
xxxprawn this isn’t his solution
What a coincidence, I was just watching the numberphile video on this when you uploaded this
This is better than the numberphile video, which was heuristic and indicative, but non- rigorous.
2:44. as a college student, I feel you
Thank you for making us excited and happy.
I love your videos
Awesome!!! I have to think it about 4 times but I can't, but your explanation makes it easier than I think about question
Thank you so much! Such a great surprise!
Want tons of more imo's, specially challenging ones. You can present them too easily to understand! Thanks a lot
謝謝曹老師🙏
Interesting to see the legend of question 6. Did not know what it was, but it was clearly interesting to watch!
Hey BPRP, great video!
And happy birthday, Yoav!!!
Very well explained! Understood the whole thing. Great question.
I feel something is incomplete, what is the strict definition of the smallest pair? Do you order on the highest element of the pair? In that case, how do you know that the found a2, is not the smallest element of some other pair?
Indeed the proof is incomplete. I saw a comment just when the video was posted which gave a correction of the proof. I think BPRP pinned it, but now I can’t find it.
The idea is that even if a1 and b1 are the smallest, a2 will be a smaller solution, so then if a2 was used instead of a1, we would get a3 from the quadratic, which will be smaller than a2, and hence, repeating it will give you infinitely smaller numbers which will not work, hence the contradiction
If we define order on the highest element of the pair as you said, gaps aren't too hard to fill: first of all, suppose a_1=b_1. Then 2a_1^2/(a_1^2+1) = k, adding and subtracting 2 from the numerator we have 2 - 2/(a_1^2+1) = k, in other words a_1^2+1 divides 2. Note that a_1^2+1>1, so it must be 2, thus a_1=1. This case is easy to verify. On the other hand, suppose a_1 is strictly greater than b_1. a_2 and b_1 are a pair of positive integers that satisfy the given equation, where a_2=b_1, we are done because a_2 would be the highest element of the pair, and if a_2
@@yakobtsv but that was not was OP meant
The problem (BPRP's sol. has) is:
What do we mean when we say
"The smallest a, b which solve the eq."
Is (1,7) smaller or is (3,4) smaller?
Since we have pairs of numbers, "smaller" needs to be defined first
@@reeeeeplease1178 Yeah that is ambiguous but that can be easily fixed I’d think by letting b_1 be the minimal b for all possible solutions, and then with that value for b, find the minimal a and call that a_1.
From this, it follows a_1 is greater than equal to b_1 and the rest of the proof should follow the same way.
This minimal pair under this definition should be (a_1,b_1)= (1,1)
I agree the proof was missing this clarity, but it seems to be corrected by this and I therefore don’t understand why the IMO problem was as famously difficult as it was so maybe I’m wrong. Please correct me if so.
This is brilliant! I think some clarification is needed when stating (a1, b1) are the smallest solutions. Pairs of numbers are not always comparable, as in, which is smaller between (4, 8) and (5, 7) for example? One could argue we are ordering the pairs (a1, b1) and (a2, b2) by looking at the values a1,2, but for completeness, I think we should also prove a2
I love your videos thank you for providing us with this content keep it up
It could be also solved positively. First we may notice that it works for A=0 and any B and B=0 and any A and in these cases, K is always perfect square. Second notice that for A=B the only working couple is A=B=1 so we don't have to deal with any more A=B. And last, assume we have A>B>0 for which the K is whole number, then we can prove that there is A1
Thanks for the proof. Love your enthusiasm.
I'm very happy about this video, because the only person who solved this problem perfectly back then was 1 point off from a gold medal and was from my country.
yaaaaaaay this was the question that i asked u about a long time ago here in the comments and in the instagram ! !
I love these videos ♥️♥️♥️ more imo problems pleaseeee
Wow, Question 6 STRIKES AGAIN!!
Hey blackpenredpen person, I really enjoyed this video. You are a good pedagogue.
Numberphile has a two part video dealing with this problem also and showed the other way you can prove this. It is also easier to understand if you are more of a visual learner and have a hard time dealing with some of these formal proofs. Also contains some factoids about the problem also.
Oh! Man as you normally do, very good job. In maths there is no difficulty, I think there is complexity which is actually different. Coping with complexity and finding out a way out is a question of a strong and robust background and that´s what you succesfully perform. Go ahead, great. Let me tell you that I enjoy your videos.
You solved the legend question hence you are legendary !
I Love your way of explaining maths....Love you and maths as well...😁
OMG i just watched numberphile's Legend q6 yesterday, and this video got posted! btw, can we continue the serie for IMO qsns pls?
Heyy blackpenredpen! I love your videos and they really help with my revsion. I was wondering if you could do one covereing Fourier Transforms, in a similar style to your laplace ones?
Please do more IMO problems @blackpenredpen, these videos are amazing!!
Interestingly, k not only has to be a perfect square, it also has to be the square of min(a, b) for the "b^2 - k" step to produce a zero a_2.
wow!!😲 that's an incredible way of thinking!!!
Beautiful!!!, just beautiful ❤
We NEED more IMO problems!!
This was uploaded on 4/20/20. Legend
BPRP, I think you decided on a nice and practical board. Now I know to abbreviate BIRTHDAY, thanks
something about solving that quadratic equation. since it is a symmetric polynomial equation, if a1 is a solution, then so is b1. makes it a little quicker to factor out
OHHHHHH ! I DIDN"T EXPECTED THIS QUESTION FROM BPRP !!
For once I am as excited as you about maths thank sir
That was amazing!!!!!! 👍👍👍
19:14 This enthusiasm
The why I love Maths
@blackpenredpen but does the reverse work? Must all perfect squares have an integer solution to [a^2 + b^2]/ab+1?
very well explained! but i guess you deserve a larger whiteboard👍
Ratio of hypotenuses to the other sides is why different for different values show the graph of this equation
Thank you, i finally understood this
the painful silence after "sometimes that was all i could do" 2:45
Such big mathematics such small boards!! Haha great work as usual.
Thank you for saving us from Vieta Jumping !
I usually hate proof by well ordering, but I finally understand this problem now so yay.
I think demonstration of this proof can be made stronger and more friendly by exploring what can be possible if you explained what implications there are if a_2 can equal to zero. Namely, you will see that a_2=0 is a valid solution that will always give you K =(b_1)^2, but it does not violate the basic assumption of the situation which is that a_1 is the SMALLEST POSITIVE Integer that allows K to be an integer. By combining that (1) Quadratic equation made up of integer coefficients must either have two real solution or two imaginary solution, (2) a_1 cannot be the smallest positive integer that satisfy the "*" condition if K isn't a perfect square, and (3) if a_1 is the smallest positive integer solution, then a_2 from "**" equation must be zero and K must equal to (b_1)^2 , I believe the proof would have been made more complete, and also easier to understand for more casual viewers.
you can get k= gcd(a,b)^2
This was VERY nice!
Good job with another IMO problem! Meanwhile I'm here still being unable to make AIME
Isn't problem step by step and all things will happen...
I love these IMO problems :)
Very nice proof!
Should have pointed out that when K is a perfect square, a2=0 and K=b^2 satisfies the criteria but a2 is not a positive integer.
Finally, a solution that I understand!
The sadness in his voice 2:38
Great ❤
Big respect for u
Keep it up👍👍👍👍
For initial solutions, set b = a^3, k=a^2 => ([a]^2 + [a^3]^2)/([a][a^3]+1) = (a^2)(1+a^4)/(a^4+1) = [a^2]
Given a solution (a1^2+b1^2)/(a1 b1+1)=k, you can get more solutions by setting a2=b1, b2=k*b1 - a1. This can be repeated.
For example:
(2^2 + 8^2)/(2*8+1) = 2^2
(8^2 + (4*8-2)^2)/(8*(4*8-2)+1) = (8^2 + 30^2)/(8*30+1) = 964/241 = 2^2
(30^2+(4*30-8)^2)/(30*(4*30-8)+1) = (30^2+112^2)/(30*112+1) = 13444/3361 = 2^2
(3^2 + 27^2)/(3*27+1) = 3^2
(27^2 + (9*27-3)^2)/(27*(9*27-3)+1) = (27^2 + 240^2)/(27*240+1) = 58329/6481 = 3^2
(240^2 + (9*240-27)^2)/(240*(9*240-27)+1) = (240^2 + 2133^2)/(240*2133+1) = 4607289/511921 = 3^2
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.
19:20
Nobody:
BPRP: gives us a perfect square
First understandable solution :)
This is such a cool solution, especially after seeing the famous solution.
Best explanation ever!
Awesome stuff, keep it up
You could’ve used the result that stipulates that a2 is different than zero, and use it in the equation above. That gives you : k isn’t a perfect square => a different than kb, then use a proof by case : let a=b=1, which gives you a contradiction because a=b=k=1 => a=kb
someone get this man a whiteboard....
Bigger whiteboard... 😂
BPRP, I've been wondering for a while now. But do u do any mathematics research? Or are you solely focused on teaching? Love the videos!
I only teach now. Thank you!
IT WAS MY BIRTHDAY AS WELL ON THE 21ST THANK YOU BPRP!!
Nice!!! Happy belated bday Llewelyn!
3 years late bro
But happy birthday :)
A lot of love from Jamaica 🇯🇲
i understood it completely thx bro
Wow, it's so nice!!
But what does it mean to say that a1,b1 is the smallest solution ?
Because I could in theory have another solution a2,b2 with a2 < a1 and b2>b1, right ?
As I understand it, we have a contradiction only if there is such a thing as a smallest solution because otherwise our contradiction might show the non-existence of a not well define "smallest" solution rather that the non existence of such a solution at all.
Since the least number principle is equivalent (classically) to a principle of mathematical induction, I wonder what the direct proof of this would look like - simultaneous induction on a and b?
Hi, I am still waiting for a reply from you about to show US how can we calculating t by hand, Exp: by calculating the first non trivial zero , when zêta(1/2+it) = (1/2-it)?
Good job！
Greeting from Taiwan！