Solving the Legendary IMO Problem 6 in 8 minutes | International Mathematical Olympiad 1988

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  • Опубліковано 14 січ 2021
  • #IMO #IMO1988 #MathOlympiad
    Here is the solution to the Legendary Problem 6 of IMO 1988!!
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КОМЕНТАРІ • 364

  • @GothicKin
    @GothicKin 3 роки тому +324

    The critical part is A1 is not 0 because B^2 - k can't vanish. Blink and you miss it, still a great job laying out the proof!

    • @MarioRossi-sh4uk
      @MarioRossi-sh4uk 3 роки тому +16

      Yes, sure. Well said.
      If you omit that detail, you may start the demonstration by stating °let's assume k is a perfect square°, and then conclude that by contradiction k must not be a perfect square.

    • @Guillaume_Paczek
      @Guillaume_Paczek 3 роки тому +1

      ah yes thanks, i was wondering the link with the fact we forbid k to be a square

    • @GothicKin
      @GothicKin 3 роки тому

      @@MarioRossi-sh4uk Probably both elegant, I wonder if that would complicate the proof or make it simpler

    • @GauravPandeyIISc
      @GauravPandeyIISc 3 роки тому +12

      This proves a stronger claim: Not only k is a square, it is the square of the smaller of A and B. Effectively, this proves that (a^2 + b^2)/(ab + 1) = b^2 if a>=b and a,b are both positive integers. Solving the above, we get that (b^3, b) are the only solutions to the above equation.

    • @werdo9638
      @werdo9638 2 роки тому +7

      @@GauravPandeyIISc thats actually not true. (a,b)=(30,8) would be a counterexample

  • @PranavGarg_
    @PranavGarg_ 3 роки тому +7

    This was the second video I watched from this channel and it was a good understandable solution. Just subscribed.

  • @kephalopod3054
    @kephalopod3054 3 роки тому +28

    I saw somewhere that this problem is a particular case of the nice generalization (much harder to prove):
    Let a, b positive integers. Prove that
    if (ab)^(n-1) + 1 | a^n + b^n, then
    (a^n + b^n) / ((ab)^(n-1) + 1) is a perfect n-th power.

    • @dimitrismelas508
      @dimitrismelas508 3 роки тому +1

      This looks interesting, where did you find it?

    • @mattiascardecchia799
      @mattiascardecchia799 3 роки тому +36

      I have a wonderful proof of this, but I'm afraid it doesn't fit within the margin of this UA-cam comment section...

    • @keescanalfp5143
      @keescanalfp5143 2 роки тому +1

      @@mattiascardecchia799, quite interesting. could you give an indication for the starting direction of that proof.

    • @themathsgeek8528
      @themathsgeek8528 2 роки тому +1

      @@mattiascardecchia799 lol

    • @imauz1127
      @imauz1127 2 місяці тому +2

      @keescanalfp5143 it’s a fermat reference

  • @buxeessingh2571
    @buxeessingh2571 3 роки тому +394

    Remember: even Terry Tao did not find a complete proof to this question.

    • @cr1216
      @cr1216 3 роки тому +110

      In the limited time of the exam though. Remember in that amount of time he had to do two other questions as well, which he did well.

    • @timothy6955
      @timothy6955 3 роки тому +129

      and he was 12 or 13 years old

    • @kevinm1317
      @kevinm1317 3 роки тому +72

      Also remember that by today's standards, this is a relatively easy problem compared to then. Its about as standard as Vieta Jumping gets, but Vieta Jumping was nearly unheard of back then, which is why this problem is so famous

    • @zawadulhoque4511
      @zawadulhoque4511 3 роки тому +9

      @@kevinm1317 yeah today it would make a hard p1/easy p2

    • @vaibhavsingh1113
      @vaibhavsingh1113 3 роки тому +19

      Terrance Tao won Bronze medal in IMO at age of 11
      and I failed to even qualify for National team at age of 15

  • @ignaciobenjamingarridoboba2071
    @ignaciobenjamingarridoboba2071 2 роки тому +56

    This solve is brilliant, always you assume there is a minimum (a,b) when the equation is not a perfect square, you'll always find out a smaller number than the minimum

  • @msk4246
    @msk4246 3 роки тому +32

    Elegance at its peak...... 🙏🙏🙏🙏🙏

  • @ratulee
    @ratulee 5 місяців тому +7

    It can be solved with elementary number theory without Vieta jumping. If you modify the rest of the theorem a little bit.
    The size of b is between ak-a and ak.
    Set b = ak - r (0

  • @pbj4184
    @pbj4184 3 роки тому +193

    5:47 I would like to add a few more steps here so that the jump may become clearer for those who didn't get it. A1B+1>0
    => A1B > -1
    As A1 was proved to be an integer and we know B is an integer as well (natural to be more specific), A1*B must also be integral. Hence A1B can be 0 at minimum as that is the next integer after 0
    => A1B >= 0
    A1 was shown not to be 0 and B is natural so it can't be 0. Hence since both A1 and B are non-zero, their product must also be non-zero and therefore it can only be >0
    => A1*B > 0
    Now since B is positive by virtue of being natural, A1 must also be positive. QED

    • @nicholasroberts2933
      @nicholasroberts2933 3 роки тому +7

      Other viewers will appreciate this comment. Thank you

    • @lewischeung868
      @lewischeung868 3 роки тому +7

      This comment clearly saves the proof :)

    • @tioa.p.1058
      @tioa.p.1058 3 роки тому +1

      thanks

    • @pedrojose392
      @pedrojose392 3 роки тому +6

      @@lewischeung868 , I do not agree, that it saves the proof. The proof itselve is clean and safety. It was just a step so easy, that he did not waste time explaining. The proof it is so pretty, excellent. Nothing to repair.

    • @lewischeung868
      @lewischeung868 3 роки тому +3

      @@pedrojose392 i am sorry to tell you that i don't agree with your point. The logic behind proof by infinite descent is to show "we can generate a smaller counter example from a given counter example". However, natural number has a lower bound, which is "1" accordingly. We can't accept an infinite descent algorithm for such problem. Then, a contradiction arises.
      Why this step is necessary? If we can't make sure A1 is positive, we cannot say (A1,B) is a possible candidate for a smaller counter example. Our infinite descent algorithm cannot bee carried out and eventually the proof is meaningless.
      I hope my terrible english can persuade you the reason behind. :)

  • @aua6330
    @aua6330 Рік тому +1

    Perfectly done, thank you.

  • @alexandergolys2087
    @alexandergolys2087 3 роки тому +4

    Such a cool proof, thanks!

  • @littlefermat
    @littlefermat 3 роки тому +25

    The problem that you'll see in every NT book for math Olympiad.

  • @danielontheedge
    @danielontheedge 3 роки тому +153

    It's quite hard to find examples of a and b that satisfy this... One example is (1, 1)

    • @lucacastenetto1230
      @lucacastenetto1230 3 роки тому +45

      If you take a=b³ those are all the soluzions i think

    • @patrickng8974
      @patrickng8974 3 роки тому +13

      There are endless number of examples: 1,1;2,8;3,27;4,64;5,125......

    • @Qermaq
      @Qermaq 3 роки тому +4

      Look at the thread I started a few weeks ago. People have posted a lot of insights.

    • @patrickng8974
      @patrickng8974 3 роки тому +10

      other than setting a = b^3, another set of solutions can be found by setting a = n^2 and k = n^3 where k(ab+1) = a^2+b^2

    • @grammairiennase624
      @grammairiennase624 3 роки тому +1

      @Luca Castenetto
      Wrong, (0, k) or (k, 0) for all k != 0 is good, too.
      Infinity of trivial examples, not following the a = b^3 or b = a^3 rule.

  • @Luarhackererreape
    @Luarhackererreape 3 роки тому +12

    Excelente bro! me gusta que pongas los subtítulos en español! Ganaste un suscriptor :)

  • @mohamedazizghorbel6413
    @mohamedazizghorbel6413 2 роки тому +3

    Thank you for such a nice work , all my Support ❤️

  • @user-ip4bm4xp2q
    @user-ip4bm4xp2q 3 роки тому +32

    Thanks,author. Please make more content like that(I'm the Russian olympiad participant)

    • @mehjabin5571
      @mehjabin5571 Рік тому +2

      did you participate in imo?

    • @guptahaha
      @guptahaha 2 місяці тому +1

      Did you win any medals?

  • @pahularora9642
    @pahularora9642 3 роки тому +7

    Amazing solution....Loved it..

  • @kayson971
    @kayson971 3 роки тому +223

    I remember seeing this problem in one of my math sessions disguised as a harmless question
    And the whole class was struggling to solve it

    • @4ltrz555
      @4ltrz555 3 роки тому +66

      Does your math teacher hate u guys lmao

    • @kayson971
      @kayson971 3 роки тому +59

      @@4ltrz555 I believe it was the coordinator that gave the question, so the funnier thing is that the teacher didn't know that this was an imo question either

    • @4ltrz555
      @4ltrz555 3 роки тому +6

      @@kayson971 haha

    • @yatharthsingh5349
      @yatharthsingh5349 3 роки тому +4

      Same, lmao.

    • @yat_ii
      @yat_ii 2 роки тому +11

      we do a little trolling

  • @MrCarlosmario22
    @MrCarlosmario22 Рік тому +1

    Exelente razonamiento. Muchas Gracias.

  • @ren200758
    @ren200758 9 місяців тому +7

    struggled with the contradiction a bit in the end. the trick is in order for this not to contradict, B^2 must equal k. nice trick!

  • @ary480
    @ary480 3 роки тому +1

    This channel will get 1 million by December 2021

  • @theevilmathematician
    @theevilmathematician 3 роки тому +7

    Very interesting number theory problem.

  • @jasonleelawlight
    @jasonleelawlight 4 місяці тому

    This method is very interesting! I didn’t expect this setup could lead to a solution. I need to watch again to better evaluate what role “being a perfect square” plays in solving this problem.

    • @jasonleelawlight
      @jasonleelawlight 4 місяці тому

      I think I figured it out, basically if we twist the problem a bit and assume we are only given that “k is a natural number” and nothing is said about perfect square, we can still find that the solution is actually limited to a specific structure, i.e. k must be B^2 and A must be B^3, as this is the only way this whole thing can hold up.

    • @unemployed5373
      @unemployed5373 2 місяці тому +1

      Thank you for clearing things up, I had no idea how this solution explains k being a perfect square.

  • @webtoon1121
    @webtoon1121 3 роки тому +19

    Thank you very much. this idea is very helpful for my problem that ask me to prove for positve x,y if (x²+y²+6)/xy integer then it must be perfect cubes

  • @mahxylim7983
    @mahxylim7983 9 місяців тому

    Nice explaination!

  • @prithujsarkar2010
    @prithujsarkar2010 3 роки тому +11

    That's soooo cool

  • @HaotianWu-bm2fx
    @HaotianWu-bm2fx 2 місяці тому

    A very clear explanation👍

  • @KJ-zs7pi
    @KJ-zs7pi 3 роки тому +3

    Awesome😃

  • @abhinavpj1729
    @abhinavpj1729 3 роки тому +1

    Awesome🥰

  • @sparkaks-gr8647
    @sparkaks-gr8647 3 роки тому +1

    Woah its great

  • @MegaRainnyday
    @MegaRainnyday 3 роки тому +3

    From your proof, we can strengthen the statement by replacing a perfect square with b^2, right?
    Edit: It also need to add an assumption b

    • @zerosumgame9071
      @zerosumgame9071 3 роки тому +2

      No it’s not b^2. For example (8,30) is a solution which equals 4, which is not the square of either input

    • @BrunoVisnadi1
      @BrunoVisnadi1 3 роки тому

      We only know that for sure if a+b is mimimal

  • @SuperYoonHo
    @SuperYoonHo Рік тому +2

    Thank you!!!!!!!!!!!!!!!!!!!!

  • @allaboutcommands4984
    @allaboutcommands4984 3 роки тому

    you make it look so easy lol ^^'

  • @trungnhanpham7694
    @trungnhanpham7694 Місяць тому +1

    Me, a 14 years old, suck at math, watching this, having no idea what he's talking about, but its very interesting

  • @brendanchamberlain9388
    @brendanchamberlain9388 3 роки тому +5

    really good

  • @iainfulton3781
    @iainfulton3781 Рік тому +16

    There's only one negative integer solution to the equation which is -5. The 8 non reducible sets of a and b are (-1,2) (-1,3) (2,-1) (3,-1) (1,-2) (1,-3) (-2,1) and (-3,1) and with these you can Vieta jump to larger absolute values. Like -5(3) - (-1) yields -14,3

  • @user-lr8od4uz1n
    @user-lr8od4uz1n 9 місяців тому

    Beautiful

  • @biscuitnerd7243
    @biscuitnerd7243 3 місяці тому +2

    So yay we are done :D

  • @math_qz_2
    @math_qz_2 5 місяців тому

    Perfect

  • @padraiggluck2980
    @padraiggluck2980 2 роки тому

    Very nice. 👍

  • @mkj1887
    @mkj1887 2 роки тому +2

    My preferred wording of the problem is that the given expression is not a prime. Then: case 1 is that the expression is not an integer, in which case it certainly is not a prime. case 2 is where we show that it must be a square, and a square is never a prime. QED.

  • @pauselab5569
    @pauselab5569 3 місяці тому

    A nice trick is to quickly abuse symmetry and transform this into symmetric polynomials form. Then it becomes a lot easier but still hard to solve without the hell a lot of ring theory

  • @mathisnotforthefaintofheart
    @mathisnotforthefaintofheart 2 роки тому +2

    Sometimes I think it is even harder to come up with a theorem like this...

  • @quantumgaming9180
    @quantumgaming9180 5 місяців тому +2

    Dude, my college professor posted on his facebook page your video here. Glad I was able to find a simpler proof of this problem

  • @iainfulton3781
    @iainfulton3781 Рік тому +3

    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.

    • @spiderjerusalem4009
      @spiderjerusalem4009 6 місяців тому

      from where did u get all these?

    • @victory6468
      @victory6468 4 місяці тому +1

      @@spiderjerusalem4009 proof by intimidation, write a whole bunch of mathematical jargon no one can read, and no one will doubt your proof

    • @spiderjerusalem4009
      @spiderjerusalem4009 4 місяці тому

      @@victory6468 the jargons are comprehensible. It's just the derivations, where it came from were utter vague

  • @anshugupta793
    @anshugupta793 Рік тому

    Awesome

  • @aaryan8104
    @aaryan8104 4 місяці тому +1

    i dont know a lot on how to solve these type of questions or how these even work rather but heres how i solved,please just tell me if im wrong anywhere(i certainly will be)
    let us assume
    a^2+b^2/ab+1=p where p is a natural number is not a square ----(1)
    ab+1/a^2+b^2=y which is a natural number
    ab+1=(a^2+b^2)(y)
    (a^2+b^2)(y)/(a^2+b^2)=p
    1/y=p
    y=1/p
    but according to (1) p is a natural number but i/natural number is not a natural number
    therefore our assumption is false and p is a square number

  • @andreadevescovi4166
    @andreadevescovi4166 Рік тому

    If ab+1 divides a^2+b^2 then b^2=a/b (it is simple: divide (a^2+b^2) by ab+1 and to make the rest=0 it is necessary b^2=a/b)
    Then b^2=a/b --> b^3=a. ----> substituting in (a^2+b^2)/(ab+1) --> (a^2+a^6)/(a^4+1)=(a^2(a^4+1))/(a^4+1)=a^2.

    • @notmymain2256
      @notmymain2256 9 місяців тому

      Long division "divisibility" works on polynomials, you're confusing divisibility on every a, b and divisibility on specific a, b

    • @notmymain2256
      @notmymain2256 9 місяців тому

      Also, don't you think it's a problem if you get a result like that since, by symmetry, you could conclude b=a^3 and so a=b=1 only solution? (btw you can easily see (2, 8) is another solution)

    • @antonioorlando5246
      @antonioorlando5246 8 місяців тому

      I am not able to find the condition b^2=a/b. By dividing a^b+b^2 by ab+1 the rest is a^2-a^2b-a+b^2. Now how do you elaborate on a^2-a^2b-a+b^2=0 to get that there must be b^2=a/b. Thanks

    • @BossDropbear
      @BossDropbear 5 місяців тому +1

      Just to express differently ... (a^2+b^2)/(ab+1)=k, where a,b,k all pos integers.
      So need k*(ab+1)=kab+k to equal a^2+b^2.
      Hence need (1) kab=a^2 i.e. kb=a and (2) k=b^2.
      Substituting in for k in eq1, then kb=(b^2)*b=b^3=a.
      With a=b^3 we substitute and simplify:
      a^2+b^2 =b^6+b^2 =(b^2)*(b^4+1)
      ab+1 = b^4+1
      So ratio = b^2 = k.
      Done.

  • @NakSrea84
    @NakSrea84 3 роки тому +1

    Wow so good teacher I will teach my students the same to you
    Because your skill is very nice

  • @muhendisgenc8216
    @muhendisgenc8216 2 роки тому

    Wow nice one

  • @Dinosaur-xj3kx
    @Dinosaur-xj3kx 5 місяців тому

    What does ab+1 | a²+b² mean ? Why we are using vertical line between two equations?

  • @lewischeung868
    @lewischeung868 3 роки тому +3

    May I ask how to make sure A1 is a positive number?

    • @pbj4184
      @pbj4184 3 роки тому +2

      I posted a comment about this. Hope it helps

  • @ankitkumar-pw6pu
    @ankitkumar-pw6pu 3 роки тому

    Sir I don't understand any thing what should I do to understand this solution I mean any basic available

  • @subramanyakarthik5843
    @subramanyakarthik5843 2 місяці тому

    This equation satisfies only if A and B are perfect squares when substituting to that equation will result to a perfect solution😊

  • @mustydustard
    @mustydustard 3 місяці тому

    how do you know its an integer

  • @LongNguyen-lg4zi
    @LongNguyen-lg4zi 3 роки тому +4

    why can you conclude k is a perfect square? you just proved that for every k there is only one satisfying set

    • @bwobkjobrien2508
      @bwobkjobrien2508 2 роки тому

      The case of the repeated root would require A^2 = B^2 - k since it would be when A1 = A (the same equation used in the proof in the video), but k = B^2 - A^2 is only positive when b>a, which is false by assumption.

  • @peponi3456
    @peponi3456 2 роки тому +3

    7:29
    Why does this cobtradiction arises because of k not being perfectly squared?
    If k was a perfect square then it would be
    A1> or =0 so A1

    • @anshumanagrawal346
      @anshumanagrawal346 2 роки тому

      Then B^2 - k =0

    • @peponi3456
      @peponi3456 2 роки тому +3

      @@anshumanagrawal346 if k not being a perfect square leads to a contradiction then k being a perfect square must not lead to a contradiction. The contradiction is a2

    • @ericzhu6620
      @ericzhu6620 2 роки тому +1

      @@peponi3456 4:50 from this term we can see if B^2-k = 0 then A1=0, which is not a natural number, which does not lead to a contradiction at the end since A1 is never valid as a solution, in the actual solution A1 leads to contradiction because A1 > 0, which contradicts to the assumption "(A+B) is minimal"

    • @123integration9
      @123integration9 Рік тому

      Yaa man you can assume k as not a trangular number and with the contradiction you can prove that k is a trangula number.

  • @bilkishchowdhury8318
    @bilkishchowdhury8318 Рік тому

    4:36 How are A,B (the minimum roots of the equation) known to be integers?

    • @jilow
      @jilow 9 місяців тому

      It's not like that.
      The problem is claiming that ALL natural solutions also happen to produce a perfect square.
      So the guy says let's say we find a solution that meets all the criteria a,b are naturals and that those two expressions divide. Suppose we find a solution and not just any solution we find the smallest solution. Which of course there will be.
      Assume we have the smallest solution that is NOT a perfect square then this proofs shows if that were the case you could always make a smaller one..which is a contradiction. Therefore, it must be a perfect square.

  • @ranjitprasad2155
    @ranjitprasad2155 3 роки тому +2

    Those 11 students , 🤯🤯

  • @suuujuuus
    @suuujuuus 8 місяців тому

    How do we know that A, the root, is an Integer, i.e. a non floating point number in proof that A1 is in Z?
    Also, that A1 is >0 comes from A1B+1>0 A1>(-1)/B which gets us A1>(-1) since B is an Integer. Since we just showed that A1 is a whole number and we assumed for our proof by contradiction that A1 /= 0, otherwise k would be an Integer square, A1 has to be in IN/0. Therefore A1>0.
    Feel like you not only skipped a lot of steps there, but also presented them in a wrong order.

  • @Uknowwhois
    @Uknowwhois 11 місяців тому

    Bro has proved hardest imo problem by contradiction

  • @marlongrau246
    @marlongrau246 Рік тому

    I'm sorry is this related to phytarean triples? It doesn't seemed to be.

  • @jmart474
    @jmart474 2 роки тому +3

    I found an easy solution, but of course there must be something wrong with my assumption.
    a^2+b^2 = k (ab+1)
    a^2+b^2 = kab + k
    Then I consider !!!
    a^2 = kab
    b^2 = k
    So k=a/b and k=b^2 and thus a = b^3
    Substituting (b^6+b^2)/(b^4+1) = b^2
    Which is a perfect square
    Hope that you can comment on this solution.

    • @sinistergaming1418
      @sinistergaming1418 2 роки тому +3

      How is that possible as k cannot be equated to b^2 as we didnt prove k is a perfect square ,the main motive is to prove k is a perfect square so we cannot assume it

    • @aaykat6078
      @aaykat6078 2 роки тому

      @@sinistergaming1418 it doesn't really assume that k is a perfect square
      a+b
      ------ = n
      c+d
      if a/c =n
      Then b/d also equal n
      9+18
      -------- = 3
      3+6
      9/3=3,18/6=3
      27/9=3
      This is how division and ratio works, since we have unknowns, it's safe to say a²/ab = k, and same with b²/1= k
      Although there will be times where the solution isn't like this, so i guess this is just possible answers

  • @represent409
    @represent409 3 роки тому +2

    nice

  • @babulalyogi1952
    @babulalyogi1952 2 роки тому +1

    Well I solved it in few minutes and astonishingly my solution was also correct...
    Can I send it to someone to verify it????

  • @omaralvarezzaleta4728
    @omaralvarezzaleta4728 3 роки тому

    Soy asesor de olimpiadas de matemáticas en prepa, nivel regional,Chiapas México

  • @chanderkumar7061
    @chanderkumar7061 2 місяці тому

    Sir my answer firstly distributed ab +1 in a^2+b^2 take a>=b so a^2 greater than ab +1 so if we divide than remainder will be -a/b and if we divide b^2/ab+1 remainder will be b^2 net remainder will be zero -a/b+b^2=0 so a=b^3 if we put this value in expression we got b^2 which is perfect square ... Thank you I am from india

    • @meowplays2151
      @meowplays2151 Місяць тому

      really...??
      remainder is not -a/b......or how?

    • @mickerson3979
      @mickerson3979 Місяць тому

      Your solution is not correct

  • @garydetlefs6095
    @garydetlefs6095 3 місяці тому

    I enjoy your videos but I am very curious about seeing things in the flesh so I was curious as to what numbers actually satisfy this condition. It took me about 10 seconds to write a line of maple code to produce the results and it is interesting to see that any two numbers x and x cubed will satisfy the conditions for a and b
    The only pairs less than a thousand which also satisfy this condition are
    (30,8)...(112,30)...(240,27)...(418,112)

  • @omaralvarezzaleta4728
    @omaralvarezzaleta4728 3 роки тому

    Muy bueno

  • @gauranshbansal
    @gauranshbansal 9 місяців тому

    I really like your accent, that stereotypical Asian accent (I mean it in a good way, I'm not being racist, I'm Asian too) makes me much more comfortable dunno, if I'm the only one

  • @sawyersmith5373
    @sawyersmith5373 2 роки тому

    Why can't the contradiction arise for perfect square k?

  • @fernandoalmer3312
    @fernandoalmer3312 Рік тому +1

    Can anyone explain what is the relation between the assumption that k is not a perfect square and the minimality of the roots?

    • @florentinmunch6769
      @florentinmunch6769 10 місяців тому +3

      k not square was used to deduce that A1 is not zero, and hence positive by a later argument. Minimality of roots is a fancy formulation of induction. Having (A,B) a solution, it is shown that (A1,B) is a smaller solution which is a contradiction assuming that (A,B) is a minimal solution. Here, positivity of A1 is needed so that (A1,B) is a proper solution. In other words, the Vieta jumping produces smaller and smaller roots, hitting zero at some point. But hitting zero is only possible if k is square.

    • @spiderjerusalem4009
      @spiderjerusalem4009 6 місяців тому

      the root A_1 = (B²-k)/A. k not being square means that can't vanish

  • @Qermaq
    @Qermaq 3 роки тому +1

    What i find interesting are the answers I find: a and/or b = 0, or a = b^3, or a^3 = b, and that seems to be it.

    • @petersievert6830
      @petersievert6830 3 роки тому +1

      Would be sweet to proof this thing by showing, that these are the only solutions possible, because then it easily breaks down to k=b^2 (respectively k=a^2)
      There might be a way to show this in a way, that any prime factor that is in a must also be in b and vice versa and once you are there, then conclude that the exponent must be exactly 3.

    • @vindex7
      @vindex7 3 роки тому +3

      Unfortunately only some of the solutions are of this form. Take for example a=30, b=8.

    • @petersievert6830
      @petersievert6830 3 роки тому

      @@vindex7 thanks for pointing this out.

    • @Qermaq
      @Qermaq 3 роки тому

      @@vindex7 Yep, I'm finding 30, 112 and 27, 240 as well. As 8 and 27 are both cubes I suspect 8, 30 and 27, 240 are related. But 30,112 is a mystery.

    • @bsmith6276
      @bsmith6276 3 роки тому +7

      I found this: Let (a, b) be any solution pair with a>b and let s = (a^2+b^2)/(ab+1). Then another solution can be derived by creating solution pair (s*a-b, a). So if we start with a trivial (a,0) solution then that generates (a^3, a). Then from (a^3,a) we can generate (a^5-a, a^3) as another solution. And of course we can keep going to generate larger solutions.

  • @adithya3642
    @adithya3642 3 місяці тому

    6:01 im confused, what if A1 and B are both negative? also how does it being positive tell us its an integer?
    not sure how you got A1+B > 0

    • @benkahtan6802
      @benkahtan6802 29 днів тому

      Since A1 = kB - A, where k, B, and A are all integers, we know A1 is an integer.
      We know A1 = (B^2 - k) / A by Vieta's formulas. Since B is an integer, and we are supposing k is not a square, then B^2 - k ≠ 0, so A1 ≠ 0.
      Combining the above two results, we know that A1 is a non-zero integer.
      We know (A1^2 + B^2) / (A1 * B + 1) = k > 0. Since the numerator A1^2 + B^2 > 0, then this quotient is only positive if the denominator A1 * B + 1 is also positive.
      A1 * B + 1 > 0 implies A1 * B > -1. We know B > 0 since it was defined that way when setting up the problem. We know from above that A1 ≠ 0. Since A1 and B are integers, their product can't be between -1 and 0. So A1 * B can't be less than 0 (-1, -2, -3, ...) and it can't be 0, so it must be greater than 0.

  • @daemonturk
    @daemonturk Рік тому

    Why does the proof by contradiction imply that the assumption about k not being a perfect square is false? It could also imply the assumption about k being a natural number is false. Why is the proof sound?

  • @TanvirSami-jo4tx
    @TanvirSami-jo4tx Місяць тому

    I did it(vieta jumping),Andromida and milkiway,cassiopeia

  • @Miguel-xd7xp
    @Miguel-xd7xp 3 роки тому +4

    Vieta jumping is the elegant solution, but the others guys who solved this problem with which solution did it? 🤔

    • @prithujsarkar2010
      @prithujsarkar2010 3 роки тому

      most probably all of the people who got a 7 did vieta jump

    • @wayneyam1262
      @wayneyam1262 3 роки тому +1

      @@prithujsarkar2010 nah, numberphile said only one solved that problem perfectly

    • @Miguel-xd7xp
      @Miguel-xd7xp 3 роки тому +2

      @@wayneyam1262 I don't think so, if you search in the IMO web site, there were people who got 42 but only one guy got a special prize :p

    • @pbj4184
      @pbj4184 3 роки тому

      @@Miguel-xd7xp And that guy did it this way :)

  • @Alberto-nz6er
    @Alberto-nz6er 3 місяці тому

    One of the students who solved the problem, is now the mayor of Bucharest, the city I’m living in

  • @navjotsayal
    @navjotsayal 3 роки тому

    ab+1|a²+b² or (ab+1)|(a²+b²)

  • @10names55
    @10names55 2 роки тому

    Why you got that A >= B

  • @mathsinmo4372
    @mathsinmo4372 6 місяців тому +2

    hey 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.

    • @francescogennaro5873
      @francescogennaro5873 6 місяців тому +1

      you just showed that it works if a = 0 or b = 0, not for any case

    • @ostdog9385
      @ostdog9385 5 місяців тому +1

      Your first step is wrong. You can only say ab(a^2+b^2) is divisible by ab+1, not that it is zero. For example 2|8, but 8=(8)(2)-1(8), but 8isnt 0.

    • @mathsinmo4372
      @mathsinmo4372 5 місяців тому

      @@ostdog9385 i am already wrong just fun see the divisor must be greater then the remainder that is 1 + ab > -ab(a²+b²)

  • @glitser2021
    @glitser2021 9 місяців тому +1

    Can this be done via Mathematical Induction?

  • @marlongrau246
    @marlongrau246 Рік тому

    Okay, there must be some values of a and b when divides by ab+1 gives you a 0 remainder. Okay. Please provide some samples.

  • @dannamilenamedranoquintero5866

    Why the contradicción say that k has to be a perfect square?

  • @PracticeMakePerfectMuslim93
    @PracticeMakePerfectMuslim93 9 місяців тому

    why did not predict the that a perfect square is positive number like 0 greater rather than just tell it a perfect squareroot

  • @srinidhikarthikbs981
    @srinidhikarthikbs981 9 місяців тому

    How did you assume that A1=(B^2-k)/A belongs to N (natural numbers)? Without proving any sort of relation between B^2 and k, we cannot plug in A1 in the original equation. Just because A1 is not 0 and it is an Integer, we cannot plug it into the original equation. We have to prove A1 is a natural number.

  • @iainfulton3781
    @iainfulton3781 Рік тому

    Turn on postifications

  • @s.w.148
    @s.w.148 3 роки тому

    It is a somewhat confused solution. Where did you use 'not perfect square condition'?

    • @s.w.148
      @s.w.148 3 роки тому

      Okay, I answered myself (B^2-k)
      eq 0...

  • @akirakato1293
    @akirakato1293 2 роки тому

    Why can he just state A,B are minimal, does he not gave to prove they exist with example?

  • @eimisahil7255
    @eimisahil7255 3 роки тому

    Bro..i . Do all process same and assume that , k is a perfect square instead of assuming k is not a perfect square.. Still the contradiction occurs... Please explain this. Where do i make a mistake

    • @zac5658
      @zac5658 3 роки тому

      4:50

    • @jimallysonnevado3973
      @jimallysonnevado3973 5 місяців тому

      A_1= (B^2-k)/A is not equal to 0 comes from the assumption that k is not perfect square. (b^2 is a perfect square, k is not hence k cannot equal to B^2 hence B^2-k is not equal to 0 hence the entire fraction is not equal to 0.) In contrast, if k is a perfect square, then you cannot proceed from here because there is always the possibility that k=B^2 and thus A_1=0 which makes (A_1,B) not a solution to the problem which does not lead to a contradiction.

  • @sageofsixpack226
    @sageofsixpack226 3 роки тому +9

    5:41 what if B was negative? Than if A1 is negative we're going to end up having positive denominator and, thus, k is positive as well

    • @earthlington2
      @earthlington2 3 роки тому +35

      A and B are both natural numbers, so B can't be negative

    • @anshumanagrawal346
      @anshumanagrawal346 2 роки тому

      The question states that a and b are strictly positive integers

  • @Alan-dg6io
    @Alan-dg6io 2 роки тому +1

    Why people made this so complicated?
    Obvious (ab+1) must be greater or equal to (a^2+b^).
    If (a^2+b^2) is greater than (ab+1), the result of (ab+1)/(a^2+b^2) is less than one.
    (ab+1)>=(a^2+b^2), both side minus 2ab, then we have
    (1-ab)>=(a^+b^2-2ab)=(a-b)^2, which is greater or equal to zero.
    Then, we have (1-ab)>=0, it implies 1>=ab,
    since both a and b are positive integer, the only solution is a and b equal to 1.
    (a^2+b^2)/(ab+1)=(1+1)/(1+1)=2/2 = 1, which is perfect square.
    Is it a primary school mathematics?

    • @alainsavard8147
      @alainsavard8147 2 роки тому +1

      if q | r and q and r are integer, then q

    • @Alan-dg6io
      @Alan-dg6io 2 роки тому

      @@alainsavard8147 if q | r, q & r are integer, then q >= r. according to wikepedia.org, "|" is divisible. If q is divisible by r, q should be greater than or equal to r. Otherwise, if q < r, q/r is a fractional number.

    • @alainsavard8147
      @alainsavard8147 2 роки тому +2

      @@Alan-dg6io q | r means that "q divides r".

  • @AmazingVideoGaming
    @AmazingVideoGaming 3 роки тому +2

    *And I thought my handwriting was bad!*

  • @nikolavasic1947
    @nikolavasic1947 3 роки тому +2

    How is this so easier than the Numberphile solution?

    • @dp121273
      @dp121273 3 роки тому +1

      Do you have the link to the video?

    • @TtTt-ur5hd
      @TtTt-ur5hd 3 роки тому +1

      It is less rigorous...

    • @sadkritx6200
      @sadkritx6200 3 роки тому +2

      @@dp121273 yes bro I just watched it a few days ago. Will edit the link here...
      Edit: ua-cam.com/video/Y30VF3cSIYQ/v-deo.html
      Here's the question
      ua-cam.com/video/L0Vj_7Y2-xY/v-deo.html
      Here's the answer.

    • @pbj4184
      @pbj4184 3 роки тому

      @@TtTt-ur5hd How so? It uses some facts he didn't prove but those facts are very easily provable themselves

  • @migry
    @migry 2 місяці тому

    I thought that the vertical line was the C computer language “or” operator 😅

  • @Red-Brick-Dream
    @Red-Brick-Dream 9 місяців тому +1

    I hate how these "Olympiad" problems rely so much on niche knowledge and parlour tricks that half the professors don't even know. It's like playing football on a minefield and then blaming the players for their random bad luck.

  • @user-oy3fd1yg7e
    @user-oy3fd1yg7e 3 місяці тому

    I have proved:
    Let a≤b
    a is any positive integer
    If ab+1 | a²+b² and a is not a perfect cube, then b=a³.
    If ab+1 | a²+b² and a is a perfect cube, then b=x⁵-x where a=x³ and x is a positive integer.

  • @peterkiedron8949
    @peterkiedron8949 2 роки тому

    You did not show that A1 is not negative and nowhere you used the assumption of ab+1|a^2+b^2

  • @dr.merlot1532
    @dr.merlot1532 3 роки тому

    where are these 11 kids who solved this?

  • @asgharali4649
    @asgharali4649 2 роки тому

    Is this class 12th problem. Or this imo problem came in the test of class 12 students? Plz answer who knows

    • @SaturnineXTS
      @SaturnineXTS 2 роки тому

      no, you don't have to worry - it's the hardest question from a math olympiad

  • @michaelaristidou2605
    @michaelaristidou2605 Місяць тому

    Couldn't A1 = B ? Because if it could, then we don't have a contradiction.