Solving the Legendary IMO Problem 6 in 8 minutes | International Mathematical Olympiad 1988
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- เผยแพร่เมื่อ 4 ก.ย. 2024
- #IMO #IMO1988 #MathOlympiad
Here is the solution to the Legendary Problem 6 of IMO 1988!!
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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!
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.
ah yes thanks, i was wondering the link with the fact we forbid k to be a square
@@MarioRossi-sh4uk Probably both elegant, I wonder if that would complicate the proof or make it simpler
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.
@@GauravPandeyIISc thats actually not true. (a,b)=(30,8) would be a counterexample
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.
This looks interesting, where did you find it?
I have a wonderful proof of this, but I'm afraid it doesn't fit within the margin of this TH-cam comment section...
@@mattiascardecchia799, quite interesting. could you give an indication for the starting direction of that proof.
@@mattiascardecchia799 lol
@keescanalfp5143 it’s a fermat reference
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
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
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
Other viewers will appreciate this comment. Thank you
This comment clearly saves the proof :)
thanks
@@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.
@@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. :)
Elegance at its peak...... 🙏🙏🙏🙏🙏
Remember: even Terry Tao did not find a complete proof to this question.
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.
and he was 12 or 13 years old
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
@@kevinm1317 yeah today it would make a hard p1/easy p2
Terrance Tao won Bronze medal in IMO at age of 11
and I failed to even qualify for National team at age of 15
Thanks,author. Please make more content like that(I'm the Russian olympiad participant)
did you participate in imo?
Did you win any medals?
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!
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
Does your math teacher hate u guys lmao
@@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
@@kayson971 haha
Same, lmao.
we do a little trolling
It's quite hard to find examples of a and b that satisfy this... One example is (1, 1)
If you take a=b³ those are all the soluzions i think
There are endless number of examples: 1,1;2,8;3,27;4,64;5,125......
Look at the thread I started a few weeks ago. People have posted a lot of insights.
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
@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.
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
The problem that you'll see in every NT book for math Olympiad.
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.
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.
Thank you for clearing things up, I had no idea how this solution explains k being a perfect square.
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
Me, a 14 years old, suck at math, watching this, having no idea what he's talking about, but its very interesting
Very interesting number theory problem.
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.
you just showed that it works if a = 0 or b = 0, not for any case
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.
@@ostdog9385 i am already wrong just fun see the divisor must be greater then the remainder that is 1 + ab > -ab(a²+b²)
This was the second video I watched from this channel and it was a good understandable solution. Just subscribed.
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
Amazing solution....Loved it..
By the same solution we can prove that the all couples (a,b) satisfaying this property are ( a , a^3 ) and ( a^3 , a ) for all integer a . It is more general solution, on particular when we calculate the ratio for this only case we find a^2 wich is perfect square.
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.
Yep, even harder to find now
This channel will get 1 million by December 2021
Dude, my college professor posted on his facebook page your video here. Glad I was able to find a simpler proof of this problem
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.
from where did u get all these?
@@spiderjerusalem4009 proof by intimidation, write a whole bunch of mathematical jargon no one can read, and no one will doubt your proof
@@victory6468 the jargons are comprehensible. It's just the derivations, where it came from were utter vague
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
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.
Long division "divisibility" works on polynomials, you're confusing divisibility on every a, b and divisibility on specific a, b
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)
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
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.
Excelente bro! me gusta que pongas los subtítulos en español! Ganaste un suscriptor :)
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.
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
@@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
This equation satisfies only if A and B are perfect squares when substituting to that equation will result to a perfect solution😊
So yay we are done :D
One of the students who solved the problem, is now the mayor of Bucharest, the city I’m living in
Why would we assume A+B is the minimum in the proof? Without the assumption that A+B is minimal, shouldn’t the Vieta formula still hold true? Then you just found another solution, A1, B to the original equation, but you have nothing to contradict with. (Just for my understanding)
Exactly my doubt
In the end contradiction may not hav been due to taking k is not perfect square
It may have been due to A+B Not being minimal ....
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
really...??
remainder is not -a/b......or how?
Your solution is not correct
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)
Such a cool proof, thanks!
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
Then B^2 - k =0
@@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
@@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"
Yaa man you can assume k as not a trangular number and with the contradiction you can prove that k is a trangula number.
Thank you for such a nice work , all my Support ❤️
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.
I did it(vieta jumping),Andromida and milkiway,cassiopeia
Perfectly done, thank you.
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?
if q | r and q and r are integer, then q
@@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.
@@Alan-dg6io q | r means that "q divides r".
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
That's soooo cool
Thank you!!
Hi prithuj :)
A very clear explanation👍
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
No it’s not b^2. For example (8,30) is a solution which equals 4, which is not the square of either input
We only know that for sure if a+b is mimimal
if a+b is minimal then we can strengthen the statement by replacing a perfect square with b^2
Well I solved it in few minutes and astonishingly my solution was also correct...
Can I send it to someone to verify it????
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.
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.
Unfortunately only some of the solutions are of this form. Take for example a=30, b=8.
@@vindex7 thanks for pointing this out.
@@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.
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.
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.
a^2 + b^2 = d^2 is granted by pythagoras.
P.S.: That, of course, is nonsense. But i have a biological brain, so this trigger went off, and the anti-trigger "but only if in a right triangle" didn't.
Thus, the rest is sadly nonsense as well (but it looked so nice, you know :D)...
d^2 = c^2 * f has to have both factors on the RHS to be either equal or each square (else the product ends up having one prime exponent uneven, thus being not square).
Equal they cannot be, since f = (a*b + 1) is given, and its square is not equal to a^2 + b^2.
Done.
If we assume that A=B then we have
k=(2A^2)/(A^2+1) this is less than 2, which forces our k to be a positive integer less than 2, this is k=1 which is a perfect square. So it is better to assume either A
Sometimes I think it is even harder to come up with a theorem like this...
Yes. It is
Vieta jumping is the elegant solution, but the others guys who solved this problem with which solution did it? 🤔
most probably all of the people who got a 7 did vieta jump
@@prithujsarkar2010 nah, numberphile said only one solved that problem perfectly
@@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
@@Miguel-xd7xp And that guy did it this way :)
Exelente razonamiento. Muchas Gracias.
It is a very good problem
why can you conclude k is a perfect square? you just proved that for every k there is only one satisfying set
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.
May I ask how to make sure A1 is a positive number?
I posted a comment about this. Hope it helps
Those 11 students , 🤯🤯
Thank you!!!!!!!!!!!!!!!!!!!!
What does ab+1 | a²+b² mean ? Why we are using vertical line between two equations?
ab+1 divides a²+b²
really good
Thank you!!
I thought that the vertical line was the C computer language “or” operator 😅
Can anyone explain what is the relation between the assumption that k is not a perfect square and the minimality of the roots?
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.
the root A_1 = (B²-k)/A. k not being square means that can't vanish
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?
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
A and B are both natural numbers, so B can't be negative
The question states that a and b are strictly positive integers
HEY BRO ITS VERY EASY QUESTION EASIEST QUESTION EVER EXIST SO FIRST OF ALL WE ASSUME A AND B BE 1 WHICH IS NATURAL NO SO PUT A EQUALS TO 1 AND B EQUALS TO 1 SO ITS SOLVED IN 5SECONDS EASIEST QUESTION
I think you never read the problem,sat down and look back the problem
Turn on postifications
Wow so good teacher I will teach my students the same to you
Because your skill is very nice
Awesome😃
Nice explaination!
Bro has proved hardest imo problem by contradiction
Why the contradicción say that k has to be a perfect square?
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.
The proof was based on the case where a + b is the minimal being assumed. What about the rest of the cases where a + b is not the minimal ?
This uses a proof by contradiction, where you assume that (a²+b²)/(ab+1) is not a perfect square, then you draw a contradiction from the initial assumption. In this case, we proved that there is no minimal solution that equals a non-square. Since a and b are in the natural numbers, this effectively proves that there are no solutions at all which equal a non-square. This doesn't only prove a minimal case, it proves that all the cases must lead to a perfect square.
We didn't _assume_ a+b was minimal. We chose se (A,B) such that it gave minimal A+B as we know _a_ pair giving minimal sum must exist because there is always a minimum value in a list of integers (the list of A+B here) Then we showed that there exists a pair that gives an even smaller sum which is impossible since we chose the pair which already
had the lowest possible sum. Therefore a logical contradiction happened and so some assumption must have been wrong. There was only one that k wasn't a perfect square. So k must be a perfect square.
The existence of a case where a+b is minimal isn't assumed as there always exist A,B satisfying that.
Since only the existence of (A,B) is necessary here, we are fine
@@pbj4184 Really good explanation, the key here is that choosing the minimum solution is what leads to the contradiction later. The minimum solution isn't one case that was checked, but it proves every case.
@@kenthchen Something true must work for all cases. So if it doesn't work for the case where A+B is minimal, it isn't true. And since k can only either be perfect or non-perfect square and we showed it cannot be a non-perfect square (as that leads to a contradiction), it must be a perfect square.
@@pbj4184 sir can you please explain me the basis of assumption @4:58
"That since k is not a perfect square
Surely A1 is not equal to 0
I don't know why the TH-cam algorithm tossed me this tonight, but:
It needs to be established that B is not 0 in order to say that A_1 B + 1 > 0 gives you A_1 > 0. This is not hard to do, but it relies on k not being a perfect square, which is not actually ever used in the video and is critically important. If k is a perfect square, then you Vieta jump from (sqrt(k),0) to (-sqrt(k),0) and there are no issues; your minimal example is indeed minimal because the smaller solution you Vieta jump to has a negative value. Only when k is not a perfect square do you get stuck in infinite descent within the positive solutions
Edit: A slight tweak to the argument establishes that *every* solution to the equation is a Vieta jump of (n,0) for some non-negative integer n
i dont get it, isnt the entire proof built on the assumption that k isn't a "perfect square"?
Yea but he missed the important part that you know A_1 is not equal to zero at 5:49 because k is not a perfect square and the other thing on numerator is@@namannamish3343
why did not predict the that a perfect square is positive number like 0 greater rather than just tell it a perfect squareroot
Why can't the contradiction arise for perfect square k?
I'm sorry is this related to phytarean triples? It doesn't seemed to be.
how do you know its an integer
It's better version is to prove it =(gcd(a,b))^2
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.
a²+b²=k(ab+1)
b²-k = a(bk-1)
*And I thought my handwriting was bad!*
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
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.
My own proof ( might not be original, so for entertainment purposed only)
1. for a = b. it's (0,0) and (1,1), the last one is also the only solution for c^2 = 1, for everything else c^2> 1 and without loss of generality a < b
2. if (a^2+b^2)/(a*b+1) = c^2 where a < b and c^2> 1 then c^2 - 1 < b/a < c^2 + 1
3. if c^2-1 < b/a < c^2 +1 and c^2> 1 then Always : a*c^2 - b < a and b*c^2 - a > b
4. if exists pair (a,b) such as (a*b+1) | (a^2+b^2) and (a^2+b^2)/(a*b+1) = c^2 Then pairs (a*c^2-b, a) and (b, b*c^2 - a) also satisfy the condition and
((a*c^2-b)^2 + a^2)/(a*(a*c^2-b)+1) = c^2 and also ((b*c^2-a)^2 + b^2)/(b*(b*c^2-a)+1) = c^2 , easy to check, but tldr; so I'll skip the proof.
5. Hence for each c^2 there is an infinite number of pairs (a,b) and a and b are always two consecutive numbers in the sequence A(n) = A(n-1)*c^2 - A(n-2) or the same works in backward direction : A(n) = A(n+1)*c^2 - A(n+2) and this is the most interesting thing about this sequence
6. sequence is growing to both plus and minus infinity. But from (a^2+b^2)/(a*b +1) we can see that there can't be a pair (a,b) when a< 0 and b >0 so there are only exist sequences that are going through 0. Or pair (a,b) satisfying conditions (a^2+b^2)/(a*b+1) = c^2 exists only if in the sequence A(n) = A(n-1)*c^2 - A(n-2) this pair belongs to, also exists pair (0, d) when (0^2+d^2)/(0*d+1) = c^2. Or d^2 = c^2 d is an integer, so c is an integer too and c^2 is a square.
Nice and elegant, congratulations! To my personal taste the statement that all the (a,b)-s are contructed by this very algorithm is even more interesting than just the squareness proof of the c^2. It gives a complete picture of what's happening.
Why can he just state A,B are minimal, does he not gave to prove they exist with example?
Didn't get How U have done for perfect Square u shows it is positive
A perfect square is always positive.
There is no perfect square that is negative
Because negative numbers don't have roots
+ Into + = +
- into - = +
@@RogerSmith-ee4zi engineers would beg to differ. Take a trip to the complex plane yo, 😎😎😎
@@benyseus6325 yes in the complex plane we have the imaginary number calculations
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.
ab+1|a²+b² or (ab+1)|(a²+b²)
Couldn't A1 = B ? Because if it could, then we don't have a contradiction.
4:36 How are A,B (the minimum roots of the equation) known to be integers?
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.
Soy asesor de olimpiadas de matemáticas en prepa, nivel regional,Chiapas México
Sir I don't understand any thing what should I do to understand this solution I mean any basic available
No wonder this was double starred and put last on the test.
Unless you are aware of "Vieta jumping" the chances of finding the solution in a short time are small.
There's even a Wikipedia entry which solve this.
Very nice. 👍
where is the fact that k is not perfect square be applied??? can also the proof in the video be applied when k is perfect square so that k cannot be an integer???
I wonder where the assumption "(A + B) is minimal" is used in the proof ? What if we did not assume (A + B) to be minimal ?
"(A + B) is minimal" is used in the very last part to create contradiction, in this solution whenever we assume a non-perfect square k which has a minimal solution (A + B), we can always find an even smaller pair (A1 + B), which is supposed to be invalid, which proves that the assumption was invalid at start, therefore proved the actual problem
I must please ask for help on this problem !
the sequence a(n) is defined like this:
a(n) = a(n-1) / 5 , if a(n-1) is divisible with 5 or
a(n) = [a(n-1) * sqrt(5)] , otherwise. ( [x] is the floor function ) , with a(0) being a natural number , not equal with 0.
Prove that this sequence contains only a finite number of terms divisible with 5 .
If both a(n) and a(n+1) are not divisible by 5, then a(n+2) is roughly 5a(n), but a little bit smaller. More precisely,
0
Can u prove that k=gcd(a,b)² pls. This one was given to us by teacher and till this day I still have no clue how to do it. Only thing I know is this use strong induction.
You can google "IMO 1988 Problem 6 Induction". The first or second result is a pdf containing the solution you're talking about
@@pbj4184 thx
This must be the case for special case of perfect numbers like phytagorean triples but I think k must be written as k^2.
Beautiful