You, Me and The Legend of Question 6 

"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

It's incredible how 20 minutes of math with you feels like only 5min. Great explication.

Imagine going to IMO in 1988, and when you're writing your answer for problem 6, you just write "Happy Bday Yoav Carmel"

Is he doing an IMO question using about 1m² of space

I stopped the video and tried to complete a binomial to find a solution, how innocent I was.

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 :)

bruh in numberphiles video he spent half the time talking about how hard it is

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 :)

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!!

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 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

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

Fantastic! I love these videos! The presenter’s enthusiasm is so contagious!

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.

I’m loving these well explained IMO problems!

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

Exam paper:...Prove k is a perfect square Me: ex: a=8 b= 2....k=4... HENCE PROVED :)

That's quite an upgrade from that problem about sums of digits :D. Vieta jumping is a beautiful technique

Thanks a lot! I had watched the Numberphile videos and couldn't wrap my head around the solution.

I’ve waiting for this video a long time. I’m very excited

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.

These videos give me chills please continue the series👍👍

i literally had to take a pill to calm down because of how excited i was to see your solution to the problem

What a coincidence, I was just watching the numberphile video on this when you uploaded this

Awesome!!! I have to think it about 4 times but I can't, but your explanation makes it easier than I think about question

12:22 -- There's space on the wall beside the whiteboard. :D

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?

Interesting to see the legend of question 6. Did not know what it was, but it was clearly interesting to watch!

Want tons of more imo's, specially challenging ones. You can present them too easily to understand! Thanks a lot

Thank you for making us excited and happy. I love your videos

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 ❤️

Thank you so much! Such a great surprise!

Very well explained! Understood the whole thing. Great question.

謝謝曹老師🙏

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

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?

2:44. as a college student, I feel you

Hey BPRP, great video! And happy birthday, Yoav!!!

OMG i just watched numberphile's Legend q6 yesterday, and this video got posted! btw, can we continue the serie for IMO qsns pls?

I love these videos ♥️♥️♥️ more imo problems pleaseeee

yaaaaaaay this was the question that i asked u about a long time ago here in the comments and in the instagram ! !

The legend of question fffffffffffffff 🤣🤣🤣

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.

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.

Please do more IMO problems @blackpenredpen, these videos are amazing!!

Thanks for the proof. Love your enthusiasm.

@blackpenredpen but does the reverse work? Must all perfect squares have an integer solution to [a^2 + b^2]/ab+1?

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

Wow, Question 6 STRIKES AGAIN!!

Hey blackpenredpen person, I really enjoyed this video. You are a good pedagogue.

You solved the legend question hence you are legendary !

I Love your way of explaining maths....Love you and maths as well...😁

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!

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?

BPRP, I think you decided on a nice and practical board. Now I know to abbreviate BIRTHDAY, thanks

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)?

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.

We NEED more 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.

wow!!😲 that's an incredible way of thinking!!!

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.

OHHHHHH ! I DIDN"T EXPECTED THIS QUESTION FROM BPRP !!

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.

Good job with another IMO problem! Meanwhile I'm here still being unable to make AIME

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

I know that i might ask something too much, but would be nice if u post most of your soultions in a printable version? (Hope for your reply)

I don't understand how is "smallest" defined for a pair. If for some k, we have two pairs of solutions, (a1, b1) and (a2, b2) and we have a1 > a2 and b1 < b2, which one is smaller? Would it be necessary to prove there are no such solutions (if there aren't)? As usual, awesome problem and solution!

This was uploaded on 4/20/20. Legend

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

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

I wonder about how they approved that question, I mean what was the process and who brought up the question to the committee etc etc

It looks from the proof that k has to be b² (when a≥b), since the only way we used the assumption that k wasn't a perfect square was to say that it wasn't b². Is that true, or did I miss something?

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.

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

Such big mathematics such small boards!! Haha great work as usual.

Awesome stuff, keep it up

Hi blackpenredpen, thank you for your proof, it makes me learn a lot. I have a question to ask. In your proof, you mention that, " Let a, b be the smallest soln to (*)" , a and b is a pair, how do you define the smallest of a pair number (a, b)? Let's say, if a and b is the smallest, it may happen that there is another pair (a', b') such that a' < a , but b' > b. So, how to determine which pair is smaller? On the other hand, the solution a and b are symmetric, which means that if (a, b) is a solution, then (b,a) is also a solution. In your proof, you assume that WLOG, a1>=b1, which then imply that there may be a smaller solution a1 when we interchange a1 and b1. So the conclusion a2 > a1 could not be a contradiction.

For once I am as excited as you about maths thank sir

Thank you, i finally understood this

Thank you for saving us from Vieta Jumping !

the painful silence after "sometimes that was all i could do" 2:45

could someone explain to me where k being a perfect square comes into this? why would the contradiction a2

Hello I think that there is a simlper solution: In order to get a whole number the determinant has to be factored : root of k^2*b^2 - 4(b^2-k) it is only possible if b^2 = k witch means that k is a perfect sq.

Wait, you said at the beginning that you had some trouble understanding this solution? But if you just google vieta jumping, and go to like the wiki page, you can find essentially the same solution.

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.

I love these IMO problems :)

Hey. bprp. Solve this one - it is my problem I have 10 naturals 1,2,3,...10 written down in a blackboard. Now, I select i naturals a_1, a_2 ... a_i among these, i ≥ 2and I erase them and replace it by a single number given by 2(a_1+a_2+a_3....a_i)^2+7 is a_1+a_2+a_3....a_i is a perfect square > 1 or else we replace it by a_1+a_2+....a_i+5. We stop when there's one number left in the board. Find the expected value of the final number on the board. *Note : This is insanely intuitive and difficult*

i took like five ibuprofen so i think they’re all just starting to work it’s so weird it’s like whenever I lift my arm it‘s like a pitching like nerve pain but I’m gonna push through

It looks like you are actually proving a much stronger statement than the original question. Your proof shows that not only does k have to be a perfect square, but that k = b^2.

Love the enthusiasm. Dear @blackpenredpen, something is not perfectly sound to me: you are assuming 1) [ k is a non-square positive integer] but also 2) [there exist positive integers (a, b) such that eq (*)] and 3) [taken two a= a1, and b = b1 you assume that a1 + b1 is minimal] and 4) [ a1 >= b1 ] WLOG. When you arrive to a contradiction the wrong assumption can be one (i.e. any) of the given *four* assumption, given that these condition are in logical AND . Hence ... I am not sure one can jump to the conclusion so easily and fast. Something is missing. Or not?

Many viewers noticed that saying (a1,b1) is minimal does not make any sense, some even proposed fancy corrections which tend to increase the level of difficulty of the proof. However I have a correction which if fairly simple. Indeed by defining a1 as the smallest positive integer such that there exist (b1,k) verifying (*) and a1

This was VERY nice!

Just a general question: the equations dont "know" that we assumed a1 to be the smallest solution, so shouldnt that fact somehow be represented in the equation? I mean, where in the equations did the assumption of a1 being the smallest solution actually have an effect?

Finally, a solution that I understand!

Hey, Blackpenredpen I'm from Brazil and really like your videos. Here in Brazil some youtubers who teach math are starting to propose some math challenges to each other and their videos are really fun, i wanna know if you want to me to ask if they could make contact with you, in english of course, to participate the challenges too. I know that some of them have already met your youtube channel and would be fun to see you included.

Hey just wondering, is it true that for all n natural , √n can only be integer or irrational? If so, could you maybe make a video on that, it could be interesting