You, Me and The Legend of Question 6

[@jameschen2308]

"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

[@avikdas4055]

Bprp: This is my first IMO problem One week later, Bprp: The legend of question 6

[@blackpenredpen]

Avik Das Lol yea

[@Randomstuff-wx2ku]

​@@blackpenredpen lots of love from India 💓

[@osiel_ac]

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

[@blackpenredpen]

Thank you!!

[@iridium8562]

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

[@LLWN84]

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

[@lukamiler5824]

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]

You guys only felt a few minutes of the time passing because you fell asleep

[@maxhaibara8828]

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

[@blackpenredpen]

Max Haibara he will be super happy!!!

[@yoavcarmel1245]

@@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]

blackpenredpen, had you ever been to IMO?

[@SahilGhosh]

@@yoavcarmel1245 how did he know your birthday?

[@malawigw]

@@blackpenredpen "pf: Happy Bday Yoav Carmel"

[@peeper2070]

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

[@joaquinbravo6778]

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

[@angelmendez-rivera351]

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

[@arpitdas4263]

Oh dear, hope you're alright

[@AdityaKumar-ij5ok]

Sebastian Alvarez i did that too😅😅

[@cuboid2630]

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 :)

[@tz233]

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]

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]

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

[@sheppsu7353]

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]

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

[@curtmcd]

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

[@giacomofeltrin7271]

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]

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

[@ScholarStream_25]

He reads your mind . Lmao

[@CDChester]

THE ENTHUSIASM

[@blackpenredpen]

Yes!!

[@ickywitchy4667]

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

[@cristaldark4228]

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

[@Roq-stone]

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]

Congrats, you have achieved superposition.

[@vasundarakrishnan4093]

You have just proved that this cannot be disproved.

[@sahilbaori9052]

@@cristaldark4228 Yes, but actually no

[@MathAdam]

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

[@nicholashlinka7550]

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

[@AndyGoth111]

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

[@valeriobertoncello1809]

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

[@AndyGoth111]

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]

Nicholas Hlinka ikr

[@nicks210684]

@@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.

[@CrazyPlayerFR]

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 :)

[@haztepolvo5809]

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

[@FedericoRulli]

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

[@drpeyam]

The legend of question fffffffffffffff 🤣🤣🤣

[@aldricbenalan4755]

Lol Dr. Peyam!

[@SJ-uy9vk]

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]

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]

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]

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]

@@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]

@@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.

[@philippenachtergal6077]

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.

[@philipwong9557]

a and b are positive integer. Given that a k exists st (expression is true) and k is not a perfect square, then there must be a solution pair with some a. For every "a" that solves the conditions, there must be a smaller "a" that also fits the conditions. And for that smaller "a" to exist an even smaller "a" must exist. etc etc. This is the contradiction. Given any finite integer "a" > 0 , there cannot be infinitely smaller integer values less than "a". So BPRP's condition of "smallest" is not really necessary.

[@ligda_rudna_8133]

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.

[@frozenmoon998]

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.

[@samganzfried4859]

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.

[@SwistakMiecio]

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

[@think_logically_]

"Let (a₁, b₁) be the smallest solution". Generally speaking two pairs cannot be compared to each other. as one member can be bigger, another smaller. Let's try to "save" the solution and rephrase it as the following: "Consider all pairs (a, b) corresponding to (*), where a>=b, pick up the ones with smallest b, call it b₁. If there are several pairs with b₁, take the one with smallest a and call it (a₁, b₁)." Then, the quadratic equation gives another number a₂, so that pair (a₂, b₁) or (b₁,a₂) (depending on which of a₂, b₁ is bigger) also satisfies (*) and it is proven that a₁ > a₂. Then there are two cases: 1) a₂ >= b, which contradicts to the choice of a₁ as the smallest for b₁, or a₂ < b₁ in which case the pair is actually (b₁, a₂) and this contradicts to the choice of b₁ as the smallest in the pair. Sorry for being so pedantic :)

[@jiezhao]

Michael Glickman that part is indeed bothering. Instead, he could define (a1,b1) to be such that a1+b1 is the smallest.

[@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.

[@user-tn2dk2pg2p]

There is an error in the solution, but it is non-central. Taking a minimal pair (a_1,b_1) isn't actually defined well; for example, (4,8) and (5,6) can't be precisely distinguished because there is no clear definition. Arguably the easiest way to get around that is to say that (a_1,b_1) is a pair with minimal sum; at the end, you can conclude the same contradiction, since a_1+b_1>a_1+b_2. Otherwise, there isn't anything major that needs to be changed in the solution. This solution is interestingly subtle in how it works, which is why the idea of Vieta Jumping is so insightful in olympiad problems like this one.

[@blackpenredpen]

Yes I agree. My first assumption on a1 and b1 were not rigorously defined. It reminds me my good old days where I could only get like 7/10 or even less because my first line.

[@aldricbenalan4755]

Wow, Question 6 STRIKES AGAIN!!

[@rockapedra1130]

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

[@renzo711]

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!!

[@robomaglor]

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]

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

[@eeromakinen4222]

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

[@nasrullahhusnan2289]

I work this way: a and b are two positive integers. I mean it as a and b are natural numbers: a,b={1,2,3,...,9}. a,b≠0 Let (a²+b²)/(ab+1)=r (1) As numerator and denominator are both positive, k is positive. (1) can be written as a²-rab+b²-r=0 (2) It is a positive definite quadratic equation in both a and b . Assume that r is not square of an integer. Considering (2) as a quadratic equation in a, it implies that b²-r≠0 as if b²-r=0, (2) becomes a(a-rb)=0 one of the root a=0 contradicting with a,b are natural numbers. We thus reject that b²-r≠0. It means that b²-r=0 --> r=b². (2) is cyclical quadratic equation. The above result can be obtained a is intercanged b, and vice versa. Thus r=a². In either case it yield r as a quadrat of an integer. Is my work correct sir?

[@greece8785]

19:14 This enthusiasm The why I love Maths

[@darkvoqe2054]

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

[@deph__]

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

[@allessioyassine3515]

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]

"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. ;)

[@MizardXYT]

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

[@maximilianthegreatest]

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

[@yoavcarmel1245]

Thank you so much! Such a great surprise!

[@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.

[@徐瑞斌-i8o]

The solution is superb! There is a tiny flaw in the explanation: Not a big deal, but to be more strictly exact, have to say "assume a1 is the smallest solution" rather than say "assume (a1,b1) be the smallest": Otherwise, for example, if both (2,5) and (3,4) are two solutions, which is smaller? From a's perspective, the first one is smaller, but from b's perspective, the 2nd one is smaller.

[@mooshiros7053]

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

[@emanuel3568]

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.

[@lietpi]

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

[@RickofUniverseC-137]

2:17 Idk why but it hit me at such an unexpected moment that I burst into laughter instantly.

[@dantesque17]

I just found out that if (a,b,k) is a proper solution to the equation, then so is (a, a*k-b, k), assuming (a*k-b) is also a positive integer. Example: For (8,2,4) a=8, b=2, k=4. a*k-b = (8*4-2) = 30 (also a positive integer) Therefore, (8,30,4) is also a valid solution. Check: (a^2+b^2)/(a*b+1) = (8^2+30^2)/(8*30+1) = (64+900)/(240+1) = 964/241 = 4 = k. You can make an infinite ladder of valid solutions by constantly replacing the smaller value with the larger value. (8,2,4) (8,30,4) (112,30,4) (112,418,4) (1560,418,4) (1560,5822,4), etc.

[@flameaddressof565]

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.

[@yeast4529]

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

[@johnnath4137]

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

[@Nightmare-ol5ze]

The Mayor of Bucharest Nicușor Dan Got pure medal (42 Points) at the IMO 1988 . He had a special solution for this problem

[@dylandejonge5069]

I’m not sure but I might have found a proof myself. You set the equation (a^2+b^2)/(ab+1)=k where k is a natural number. Solving for a using the quadratic formula you get a nice expansion with a square root. If the thing inside this square root is not a perfect square, for example you end up with sqrt(7), then the result is not a natural number. Since a is a natural number you know the expression inside the square root has to be a perfect square. The expression inside the square root is b^2*k^2 - 4(b^2-k). We set this equal to n^2 where n is a natural number. You ca quickly notice that if k=b^2 the expression will come down to b^2*k^2, that’s always a perfect square so we found some solutions. Now we will look at the situations where k≠b^2. Setting b^2*k^2 -4(b^2-k)=n^2 and solving for b^2, it gives us that b^2=(n^2-4k)/(k^2-4) where all variables are natural numbers. The denominator can be expanded to (k+2)(k-2) but the numerator can’t be expanded like that. Now we set m^2=k so we can rewrite the fraction as (n+2m)(n-2m)/((m^2-2)(m^2+2)). Knowing this equals b^2 and b is a natural umber we know the numerator has to divide the denominator. We can split this into 2 cases. Either (case 1) n-2m divides (m^2-2)(m^2+2) or (case 2) n+2m divides (m^2-2)(m^2+2). We first look at case 1. Since n-2m divides (m^2-2)(m^2+2) we can set n=u(m^2-2)(m^2+2)+2m where u is a natural number. Plugging this in leaves u with just the variable u. Now we stil had an (n+2m) term in de numerator so we are left with u(n+2m)=b^2 notice that k=m^2 so m=sqrt(k). If we use this in the expression we get that u(n+2*sqrt(k))=b^2 which is a natural number. The variables u and n are both natural numbers, so is the number 2 but if sqrt(k) is not a natural number the expression u(n+2*sqrt(k)) will never be a natural number. We have to take the fact that k is a natural number here in consideration but we already stated that in the beginning. We just concluded that sqrt(k) is a natural number so therefore we can see that k has to be a perfect square. A similar thing can be done with the second case and there we also get to the conclusion that sqrt(k) is a natural number. Hereby we’ve proven k is a perfect square so we’re done. Does any one know if my proof is correct and if not, why

[@alvinpalmgren3442]

There are a few issues with this proof: 1. When setting m^2=k, we cannot assume that m is a natural number - this is exactly what we are trying to prove. If m is not a natural number, the statement that "n-2m divides (m^2-2)(m^2+2)" (I think you mean that (m^2-2)(m^2+2) divides n-2m) has no meaning. Consequently, u doesn't have to be a natural number and the proof falls apart. 2. Even if m were a natural number, we wouldn't be able to split the proof into cases like you did unless n+2m and n-2m were coprime. For example, if n+2m=6, n-2m=2 and (m^2-2)(m^2+2)=12 then (m^2-2)(m^2+2) would not divide n+2m nor n-2m but still divide (n+2m)(n-2m). (Just to be clear, these specific numbers wouldn't work but they demonstrate why splitting it into cases isn't always valid.) Don't be too discouraged though, this problem is REALLY difficult. Also, please correct me if I got something wrong myself.

[@yoyokojo651]

I’m loving these well explained IMO problems!

[@pkvlogs5078]

better to use PMI or k= a^2 +b^2 /(1+ab) to get an integral soln. a^2+b^2 > (1+ ab) and (1+ab) must be one of the factor of a^2+b^2 which however converges to conditional soln. means it can only be possible if a= b^3 or b= a^3 K= b^2 or k= a^2 soln.set (b^3,b) and (a,a^3)🙂 e,g; (1000,10) , k=100 or (10,1000), k=100 its always being an interesting thing to solve yewr riddle problems Thank u

[@mileslong7103]

Only being able to write pf is depressingly relatable

[@hichamelyassami1718]

pf=power factor

[@deanociohornocio9149]

This was uploaded on 4/20/20. Legend

[@leswhynin913]

I tried to solve it by looking at squares using pythagorean theorem and some algebraic manipulation. Glad I ran into the video to save hours of wasted time

[@ANUJ.7]

Literally great explanation , I also wanna become physicist and mathematician but Stanford rejected me😭😭😭

[@user-en7dx1qp3k]

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.

[@sakshamsingh4378]

These videos give me chills please continue the series👍👍

[@rodwayworkor9202]

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*

[@thechemtrailkid]

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

[@anilkumarsharma1205]

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

[@philippenachtergal6077]

Toying with this, I found a way to find all the solution for any given k perfect square > 1. Or at least an infinite number of related solutions, I can't prove those are the only ones. From the equation E1) x² - kb x + b² - k = 0 Pose b = 0 Solve that simple quadratic equation to get two possible values of a. Forget the negative one (a0-) which is (kb - sqrt( k²b² - 4 * (b²-k)) ) / 2 , but remember the bigger one (a0+) which is equal to (kb + sqrt( k²b² - 4 * (b²-k)) ) / 2 Now, the problem doesn't allow for 0 so plug the a+ you obtained as b into the equation E1 Solve E1 with the new parameters, your a1- will be 0 but your a1+ gives you the first solution which is (a0+, a1+) (and vice versa, with a1+ always bigger than a0+) Now, if you plug back a1+ as b into E1, and solve E1 you get a2- equal to a1+ (we already have that solution) and a2+ gives us a new, bigger solution (a1+, a2+) And you can continue forever : there is an infinite amount of solutions. ------------- Now, obviously, when we pose b=0, E1 simplifies into x²=k which show that this infinite serie of solutions cannot be started if k is not a perfect square. This, of course, does not prove that there are no solutions outside of the "0 sequence".

[@JohnSmith-vq8ho]

The question is not challenging for IMO participants who are familiar with the technique of Vieta Jumping. I think if it were put on one of the more recent IMO’s, it would be around a question 4, because almost every IMO participant knows the trick now. The difficulty of it in 1988 was that Vieta Jumping was not a standardly taught trick. Those contestant who solved the problem either knew about the trick or came up with the trick on the spot. The latter is amazingly impressive.

[@johnnath4137]

I hadn’t come across the term Vieta jumping until a few days ago. I fail to see any new trick here - there is the Well Ordering Principle, which is taught at the beginning of every number theory course; and there is the method of infinite descent which was invented by Fermat 400 years ago. But I concede that the term “Vieta jumping” seems to be of more recent provenance.

[@clem418]

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

[@sebastiancrone5650]

The sadness in his voice 2:38

[@rushikeshbhor6069]

someone get this man a whiteboard....

[@mohanayare]

Bigger whiteboard... 😂

[@moos.2004]

this is how i did it but i’m not sure if i approached it the right way (btw i just finished high school and just someone who has an interest in math and never really tried a proof before) so lets put the perfect squares in chronological order: 1^2 = 1 2^2 = 4 3^2 = 9 4^2 = 16 5^2 = 25 6^2 = 36 the difference of the difference between them is 2, (like what im saying is that the difference between 1 and 4 is 3, and the difference between 4 and 9 is 5, and the difference between 3 and 5 is 2) and then i proved that (a^2+b^2 | ab+1) has that same characteristic.

[@moos.2004]

and it did

[@moos.2004]

but idk it cant be that simple💀

[@aliyubello15]

Beautiful!!!, just beautiful ❤

[@pauselab5569]

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

[@citronnade1506]

Fun Fact: Mayor of Bucharest (Romania) is one of the guys solving this problem back then and he showed his proof on Tv and it's actually really easy .

[@drpkmath12345]

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

[@nielsstruye5254]

We NEED more IMO problems!!

[@brilliantlyinnovate6039]

You solved the legend question hence you are legendary !

[@moshadj]

When picking the smallest (a,b) in the set S = {(a,b) in z_+ x z_+ : (a,b) satisfies * and a >=b} how should we define (a1,b1) < (a2,b2). My thought is lexigraphical ordering (?) where we look at values of a first and if they are equal then we look at the values of b . This means because the positive integers are well ordered we can pick a subset of S such that all the a's are smallest and equal and then we can use the well ordering of the positive integers again to pick the smallest such b. In this way we have (a1,b1) that is the least element of the set. We can now see from the proof that a2

[@sundeep0207]

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.

[@Stefan96ns]

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!

[@Vung-KTVHCM]

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

[@zohramartini9425]

The person who made this question is a legend too :)

[@flameaddressof565]

My solution is this: AB+1, may only perfectly divide A^2 + B^2, If (A^2 + B^2) / AB ‘s whole number quotient, C, is equal to the remainder, R This can be easily proven geometrically, by combining the area of the squares into one rectangle, then attempting to divide by AB. Therefore, [ A^2+B^2 ]mod[AB] = R = C = (A^2 + B^2) / (AB+1) And then, using the properties of the mod function It can be shown that, if we pick A to be less than B [ A^2+B^2 ]mod[AB] = A • [A]mod[B] And [A]mod[B] is just A Therefore (A^2 + B^2) / (AB+1) is equal to A^2 Anything can be A, and B is the cube of A 0,0 1,1 2,8 3,27, ••• Edit: unforunately, this proof also assumes A is a multiple of B^2. Other solutions do indeed exist, like 30 and 8.

[@flameaddressof565]

How that modular function worked was A • [A+BB/A]mod[B] A•[AmodB + BB/A mod B]modB A is less than B, so AmodB is A BB/A is a multiple of B, so BB/A mod B is zero But this assumes that B can be divided by A into a whole number; that BmodA= 0; B = A•n+0

[@NavneetKapur]

@blackpenredpen Aside from what @VeryEvilPettingZoo stated, I would recommend being clearer in the video that (a1, b1) are the smallest solutions for k. While watching the video today, I assumed you meant that (a1, b1) was the smallest such solution for any k - which makes the solution incomplete.

[@munna_bhai_bsms]

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

[@Joy-be3gk]

謝謝曹老師🙏

[@Blabla0124]

"Let a1, b1 be the smallest solution" doesn't make sense. I think you mean "Let a1, b1 be the solution with b1

[@kshitij7b286]

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

[@Jack_Callcott_AU]

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

[@wanshitong8451]

how about this: (a^2+b^2)/(ab+1) = n, so a^2+b^2-nab=n. let the greater or equal integer be a. a ≥ b let a new integer c = bn-a, c,a,b,n are all integers since a^2+b^2-nab=n and bn = a+c, n=a^2+b^2-a(a+c)=n, n = b^2 - ac since n = b^2 - ac and n > 0 and and a ≥ b it follows c ac (n = b^2 - ac and a,b,n are all positive and c is negative) since |c| > 0 it follows n > a which contradicts a > bn when b ≠ 0 (in the case b = 0 it can be easily observed that n = a square integer, in this case a^2) so c ≥ 0 we are almost done now just substitute a = bn - c (from c = bn-a) into n = b^2 -ac we get: n = b^2 - c(bn-c), **n = b^2 + c^2 - nbc** notice we have the same exact equation we began with except a has become b and b has become c! since the same conditions are present we can rewrite n again with c_2 = cn-b and as we proved b>c and so b>c>c_2>c_3>c_4.......>c_[k-1] > c_k=0 as we keep rewriting with smaller and smaller integers it follows that eventually at some integer in this sequence (c_n) we will arrive at zero. and at this point n = (c_[k-1])^2 + 0^2 - 0 and since all numbers in this sequence are integers n is an integer squared

[@zmaj12321]

That numberphile video mentioned wormholes or something so when i first learned the solution i was very confused

[@GruntDestroyarChannel]

IT WAS MY BIRTHDAY AS WELL ON THE 21ST THANK YOU BPRP!!

[@blackpenredpen]

Nice!!! Happy belated bday Llewelyn!

[@Maths_3.1415]

3 years late bro But happy birthday :)

[@eitanhaim3972]

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.

[@laurisuoranta5512]

Watched only the problem part, at first, and solved for a using quadratic equation. Positive integer constraints for k and the solution (a) already imply that k = b^2 (as turned out to be in the example) and hence a perfect square. So the hardest problem seems to be a trivial problem. However, if somebody gave an example where k is not equal to either a^2 or b^2, I'd admit that there's more to this than what I did.

[@laurisuoranta5512]

Yup, this is "almost trivial". The part which seemed quite evident but that I didn't formally check is that (bk)^2 + 4(k-b^2) is a perfect square IIF k=b^2. To prove this, consider mutually exclusive cases k>b^2 and k

[@manuelpanganoron4529]

try a = 30, and b = 8 the resulting k is still 4 and makes your claim k = b^2 or a^2 false

[@laurisuoranta5512]

​@@manuelpanganoron4529 Thanks, I lost factor 4 from k which made me erroneously conclude that if k

[@laurisuoranta5512]

Looks like writing defining k=b^2 - r and rewriting (bk)^2 + 4(k-b^2) as k^2 = ... (implicit solution) does the trick, but I'll think it over a few times. Anyhow, I'm pretty sure that this problem is just a special case of some general theory in mathematics, one that can be proven using elementary tools. Guys who come up with these don't invent new theorems just for IMOs, it would be too hard. That's why I don't think that truly "legendary" problems are part of the competition: with proper knowledge, they can be reverse engineered and at least one elementary proof constructed systematically.

[@BlokenArrow]

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

[@jaypatel-vd7km]

First understandable solution :)

[@abhishekanand7376]

Thank you for saving us from Vieta Jumping !

[@aminesoukane8961]

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

[@diatom1991]

well, I find a general solution like the follow: let k= r^2, than a = r b = r^3 is the only answer ( here b>a)

[@diatom1991]

I don't expect anyone notic this

[@sebryxs]

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

[@kasuha]

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

[@tropicomango]

2:46 Sometimes that was the only thing I was able to do XD

[@ninthcrusader2355]

Charles Ran Same

[@190santhoshraj5]

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

[@MadScientyst]

This is a great & lucid proof! IMO problems literally fall at the doorstep of Analysis (Real, Complex & Functional). U gotta have a solid background in Discreet, Logic & Proof abilities to tackle them (in particular problems from China & Russia). That's why the 20 mins flew by so quickly...it's the intensity of working out a possible solution! 😂😂

[@kiarash7604]

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