"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

@blackpenredpen

4 жыл бұрын

Avik Das Lol yea

@Randomstuff-wx2ku

Жыл бұрын

​@@blackpenredpen lots of love from India 💓

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

@blackpenredpen

4 жыл бұрын

Thank you!!

@iridium8562

4 жыл бұрын

blackpenredpen wait what 20mins passed? I was totally shocked when he said 20 minutes,I thought it was 7-8mins. Wow

@LLWN84

4 жыл бұрын

I was also shocked when I saw that the video is 20 minutes long!!!

@lukamiler5824

3 жыл бұрын

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.

@lucyebrada2950

2 жыл бұрын

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"

@blackpenredpen

4 жыл бұрын

Max Haibara he will be super happy!!!

@yoavcarmel1245

4 жыл бұрын

@@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!

@lucaschai5788

4 жыл бұрын

blackpenredpen, had you ever been to IMO?

@SahilGhosh

4 жыл бұрын

@@yoavcarmel1245 how did he know your birthday?

@malawigw

4 жыл бұрын

@@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 :)

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 :)

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

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

@angelmendez-rivera351

4 жыл бұрын

Sebastian Alvarez I mean, that it is the intuitive approach, it's just not very efficient, if I say so myself

@arpitdas4263

4 жыл бұрын

Oh dear, hope you're alright

@AdityaKumar-ij5ok

4 жыл бұрын

Sebastian Alvarez i did that too😅😅

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

@AndyGoth111

4 жыл бұрын

And the rest of the time alternating between considering zero to be positive and forgetting to prove anything at all about squares.

@valeriobertoncello1809

4 жыл бұрын

Then again, the committe members of the competition themselves couldn't solve this problem. So yeah.. it is hard.

@AndyGoth111

4 жыл бұрын

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.

@jatloe

4 жыл бұрын

Nicholas Hlinka ikr

@nicks210684

3 жыл бұрын

@@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.

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.

@sheppsu7353

4 жыл бұрын

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.

@Mutlauch

4 жыл бұрын

@@sheppsu7353 In your induction you assume that both of the k's are equal. What is that assumption based on?

@sheppsu7353

4 жыл бұрын

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.

@tz233

4 жыл бұрын

@@sheppsu7353 Induction probably won't get you to a solution. Still, I think you may be on the right track....

@curtmcd

4 жыл бұрын

Only if a1 = -511 (k = 512).

I’m loving these well explained IMO problems!

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

These videos give me chills please continue the series👍👍

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.

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

THE ENTHUSIASM

@blackpenredpen

4 жыл бұрын

Yes!!

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

@BlokenArrow

4 жыл бұрын

One week later, 3b1b does a whole 20 min animation on it

@ScholarStream_25

4 жыл бұрын

He reads your mind . Lmao

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

@cristaldark4228

4 жыл бұрын

I mean you're not, wrong, but you're not right either XD

@BlaqRaq

4 жыл бұрын

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.

@farhantajwarahmed3340

4 жыл бұрын

Congrats, you have achieved superposition.

@vasundarakrishnan4093

4 жыл бұрын

You have just proved that this cannot be disproved.

@sahilbaori9052

4 жыл бұрын

@@cristaldark4228 Yes, but actually no

謝謝曹老師🙏

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

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

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

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

@euler2149

4 жыл бұрын

xxxprawn this isn’t his solution

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

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

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

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?

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.

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

@johnnath4137

4 жыл бұрын

This is better than the numberphile video, which was heuristic and indicative, but non- rigorous.

Thank you so much! Such a great surprise!

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?

@bengtbengt3850

3 жыл бұрын

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.

@yakobtsv

3 жыл бұрын

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

@cletushumphrey9163

2 жыл бұрын

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

@reeeeeplease1178

2 жыл бұрын

@@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

@Happy_Abe

2 жыл бұрын

@@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.

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

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

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

I love your videos thank you for providing us with this content keep it up

I love these videos ♥️♥️♥️ more imo problems pleaseeee

Thanks for the proof. Love your enthusiasm.

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

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

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

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 ❤️

@bjornfeuerbacher5514

Жыл бұрын

"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. ;)

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

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.

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

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

The legend of question fffffffffffffff 🤣🤣🤣

@aldricbenalan4755

4 жыл бұрын

Lol Dr. Peyam!

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

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.

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 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.

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

You solved the legend question hence you are legendary !

Thank you, i finally understood this

Beautiful!!!, just beautiful ❤

That was amazing!!!!!! 👍👍👍

Wow, Question 6 STRIKES AGAIN!!

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!

@blackpenredpen

4 жыл бұрын

I only teach now. Thank you!

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.

i understood it completely thx bro

I love these IMO problems :)

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

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

We NEED more IMO problems!!

19:14 This enthusiasm The why I love Maths

Beautiful stuff

This was VERY nice!

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

This was uploaded on 4/20/20. Legend

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

Great ❤ Big respect for u Keep it up👍👍👍👍

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

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.

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.

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

Awesome stuff, keep it up

I usually hate proof by well ordering, but I finally understand this problem now so yay.

The sadness in his voice 2:38

Best explanation ever!

Ratio of hypotenuses to the other sides is why different for different values show the graph of this equation

19:20 Nobody: BPRP: gives us a perfect square

Very very nice solution

Good job！ Greeting from Taiwan！

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

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.

A lot of love from Jamaica 🇯🇲

This is such a cool solution, especially after seeing the famous solution.

Wow, it's so nice!!

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 As usual, awesome problem and solution!

@Vung-KTVHCM

2 ай бұрын

Nice question, i think smallest pair happen when their sum is smallest

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.

@felipemallon4101

4 жыл бұрын

If you are interested, Paulo Pereira (the guy of EquacionaMatemática channel) said to me that he has already tried to make contact with you on Instagram but you probably didn't see it. Try to talk to him if you want to :D

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.

@alesgsi3172

5 ай бұрын

you can get k= gcd(a,b)^2

Conplete Generalization update to my proof: Remove the largest shared multiple of A and B, call it G ( A^2 + B^2 ) mod AB = (a^2 + b^2) mod ab • G^2 (a^2 + b • (b mod a) ) mod ab • G^2 (This equation can’t be simplified further, I tried, but I kept looping back to this) Next Let (a^2 + b • (b mod a) ) = (abN + M) N and M are also positive integers The resulting quadratic equation shows that 1 = (aa + bb) / ( ab(N+1) + M) Therefore aa + bb = ab(N+1) + M ( aa + bb - M ) / ab = (N+1) Which is the same as saying M = (aa + bb)mod(ab) Or M = (aa+bb)mod(N+1) Multiply by GG M • GG = (AA+BB)mod(AB) And this mod was already proven to litterally be equal to (AA+BB)/(AB+1) (This also shows that M is an integer from 0 to N) *** Therefore MGG = K, if there ever is a K, Which means MGG is equal to N+1 And that means G^2 = N/M + 1/M Everything here is an integer, therefore M is always equal to 1. *** If K=1: N is 0 and G is 1, and M still is 1. A and B are equal to 1. M, 1, no longer equals the mod functions, since mod1 always outputs 0. Nevertheless, the equation that defined the mod function, AA+BB-M /AB ==> AA+BBmodAB = M would still only make AA+BB= AB+1 true, if M was 1. 1+1-1 / 1 = 0+1 In other words, if K is an integer, it is the square of A and B’s greatest shared multiple.

First understandable solution :)

Excellent upload.

great video I appreciate

Thank you for saving us from Vieta Jumping !

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)

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

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' 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.