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Vieta Jumping and Problem 6 | Animated Proof
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- čas přidán 19. 08. 2019
- Problem 6 of the 1988 International Math Olympiad is notorious for its difficulty to prove. There exists a very elegant way to prove it that lends itself nicely to being visualized.
Play with this graph on Desmos: www.desmos.com/calculator/teu...
Awesome video!
Thank you! 3Blue1Brown is my absolute favorite CZcams channel and was my inspiration for making math animations in the first place! Your comment means the world to me, thanks for stopping by!
@Mathematics and MMA as a matter of fact, they do. They even watch videos from other channels relating to their niche or their particular field. And I don't know why but I was expecting 3b1b to be here, even the thumbnail of this video gave me 3b1b vibes.
Most underrated video ever!!!!!
w rizz
guy got acknowledged by the man himself
amazing proof
Only vdo that made me understand the proof.Thank you very much
Vieta Jumping is my favorite math thing right now. This or the Eigenvectors from eigenvalues proof. Thanks for this!
Fantastic proof really hit the intuition
Great video! It's the only one I've watched so far that has actually explained the problem and solutions in a way I understood
I'm still a little lost. Why does (B²-k)/A contradicting A's minimality mean k must be a perfect square?
This is a nice and clever solution for problem 6!
My thought is to show that k is not a perfect square.
Mr. Penguin (idk your name) this video is absolutely brilliant!
That's so elegant.
beautiful video 🙏🏼
You lost me at “consider the equation”
Ooh...
That was an expression...
Lol , a true mathematician here 🤣
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
Very clean wow
What r the prerequisites to solve this problem?? Pls somebody tell..
which software did u use to make this animation?
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.
kB is kilobytes right
Amazing proof
1)Also, just a curious question, was there any clever trick behind you choosing a=b^2 during the initial geometric proof you made as an example? Or is it just you randomly made an assumption that worked out?
My guess: you chose the value of a accordingly as you picturized the fact at 2:20 in advance, as b by b square can be divisible by b^2 number of 1 by 1 squares....so you just accordingly assumpted a to be equalling b^3 just to match the rest of the problem. Am I correct?
Aryan Azim there isn’t a trick, by assuming that he didn’t prove anything, the whole point of that is an example of what you might come across when attempting to come up with a solution
Supposedly this will be done by considering solutions to said function, and noting a pattern between a few them.
Impressive
i subscribed!
@1:53 did you mean "denominator"?
I get lost in how x2 - (kB)x + (b2 - k) = 0 What does kB stand for? (Please can you explain)
kB = k * B
Ivan Marinov it’s boltzmanns constant
@@chonchjohnch 😂😂😂
kilobytes smh
@@chonchjohnch I'm wheezing
Nice.
I'm lost. At 9:30, how does(x_2+B^2)/(x_2B+1)>0 show that x_2>0? If x_2 simply were sufficiently negative, both the numerator and the denominator would be negative and the expression on the whole would be positive. I must be missing something, because if x_2 would be allowed to be negative, then (A, B) could be the smallest pair of positive integers, since the next solution after (A, B) would be negative and thus not a part of the allowed solutions.
x_2 is squared in the numerator, so the numerator is always nonnegative
@@PenguinMaths Ah. Thanks
Hey you are genius
Okay, so after studying this same proof in a book and then in this video, the only thing that's disturbing me, is that for the PARTICULAR CASE where x2 {i.e. the other root than A} is greater than zero, we just proved that in that case, x2 will be always less than A (initially defined least valued solution).....My question is, won't it still be true, if k was a perfect square? I mean the same algebraic manipulation would have hold (whatever we did with the quadratic equation), isn't it?
One step that he did was b^2-k can't be equal to 0. He could only do this because if b^2-k=0 then b^2=k and k is a perfect square. So in order for this step to work k can't be a perfect square.
@@adenpower249 well i understand that why it can't be zero, but let's talk about the part where we proved is x2 > 0 and x2 < A , which is a contradiction {{{by x2 i mean x subscript 2, not x^2 }}}}.....My question is, just by the setup of the initially given equation, isn't it obvious for this contradiction to happen for x2, regardless of the fact that k is a perfect square or not?
Good question! I actually should have been more clear on this part. The step that we prove x2 > 0, we were actually using the assumption that x2 != 0. That step by itself only shows that x2 >= 0 (since if x2 = 0, we get B^2 / 1 which is positive), but since directly before this we showed x2 != 0 I used this assumption in asserting that x2 > 0. So if k were a perfect square we would not be able to show that x2 != 0, and so all the next step could show is x2 >= 0 which would not lead to a contradiction, since if x2 = 0 it would not meet the original premise.
@@PenguinMaths by x2! Do you mean factorial?
@@aryanandaleebazim823 Nope, I wrote "x2 != 0" to mean "x2 is not equal to 0". The notation is used in some programming languages, although I'm now realizing it looks funny if you haven't seen it before
Shouldn't the jumps be perpendicular each time, I guess after fixing "a" you find next "b", then fixing b you find next "a", and so on ..... This process will make the steps orthogonal to the axes right?
The graphic that you show as a zig zag line should be perpendicular to the axes each time for k = 4, 9, ....
Am I right or am I missing something?
One of the two coordinate jumps is parallel to one of the axes (you choose whether A=B or B=A); the other one is to the other branch of the hyperbola, which is symmetric to a 45 degrees line, so it's not parallel to either axis.
@@dlevi67 By interchanging A & B we get the symmetricity about line (Y=X) but the steps should be perpendicular to axes. There would be 2 (perpendicular) staircases symmetric about (Y=X).
Let (A = x axis, B = y axis)
i.e: If I take A = 8, then I get B = (2 & 30), so I draw a line from (8,2) to (8,30) then I get to (8,30)
Now I take B = 30, then I get A = (8 & 112), so I draw a line from (8,30) to (112,30) then I get to (112,30)
Now I take A = 112, then I get B = (30 & 418), so I draw a line from (112,30) to (112,418) then I get to .....
.....
So this continues and these lines will be 90 degree to the axes, this will be mirrored against the (Y=X) line and those will be staircases too.
As shown in another video: i.ytimg.com/vi/K0hjLDFw4dA/hqdefault.jpg
@@swapnildas4533 Slightly different process, same result following a different path.
@@dlevi67 Yeah, while writing the previous reply I figured out that it totally depends on how you want to find the next solution, I guess I have always seen and heard as orthogonal steps (seems nice as shown in the picture given in the link) so that tends to be more natural to me.
@@swapnildas4533 Staircases are beautiful; lightning bolts are beautiful too. Take care and stay safe - good talking with you!
I wonder where the assumption "(A, B) being minimal" is used in the proof ? What if we did not assume (A, B) to be minimal ?
This is how we obtained a contradiction. What we showed is that, if k is not a perfect square, then any solution (A, B) can generate a smaller solution (x, B). Normally, this isn't a contradiction. But if (A, B) is the minimum solution, then this is a contradiction.
9:30 what if B was negative? Than if x2 is negative we're going to end up having positive denominator and, thus, k is positive as well
B cannot be negative since we're looking at the minimum (A,B), which is in the first quadrant. If it was negative then A has to be negative, which would not be possible since the "pattern" terminates at x=0
a and b have been defined to be positive integers right from the start. So A and B are also positive integers
Is that you FEW?
Technically a=b=0 is another solution in the integers with a=b
:P
0 isn’t a integer lul
@@harshbaliyan5867 0 is for sure an integer. You're thinking of the counting numbers, aka the natural numbers
is this solution correct 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 realize that you only found one part of sets of points
a²+b²=0 implies both a and b to be zero. (0,0) point
And ab = 0 gives (0,k) and (k,0) for k positive integers
Ummm ok , let's say I follow the proof for the most part, but when we take a = 8 and b = 4, we end up with: 8^2+4^2/(8*4) + 1 = 80/33, which clearly is not an integer, much less a perfect square. a and b are positive integers and a is larger than b. So you haven't proven that k will [i]always[/i] be a perfect square. I mean, have you?
I mean if the question was posed a bit more specifically, as in: "Prove that if the result of the division (k) is an integer, it must also be a perfect square", then we'd be fine and dandy
We are supposing that (ab + 1) divides (a^2 + b^2) which by definition means k is an integer, so if it doesn't, as in your example, it does not meet our premise and so the statement is vacuously true.
@@PenguinMaths Nice, I didn't think that an assumption that something is a perfect square also includedthe assumption that it is an integer by default, nor that the phrasing "a divides b" implies an integer. Then again, I'm not too well versed in math lingo yet. Thanks!
@@SaturnineXTS He actually explicitly made the assumption at the beginning of the video.
@@adenpower249 I didn't know that "a divides b" means it divides it with no remainder
@@SaturnineXTS It does.
10:49 This is all an illusion. The problem strictly proscribes one from using an a or b equal to zero. This proof then goes off, creates its own world of Vieta jumping, which allows, indeed requires, that a or b must eventually reach zero which therefore, by contradiction requires k to be a perfect square. Great. No problem perfectly sound. But you can't then transfer the results of this contradiction proof to one where a=0 or b=0 is proscribed. Such logic then says given the proscription one may assume K is not a perfect square and therefore and b may indeed equal zero (this contradicts the original proscription). You go through your Vieta jumping and lo and behold, k is a perfect square but that also creates the contradiction that a and b can be zero. So even though I get a perfect square I can't use the result because a and b are prohibited from reaching zero given the stated contradiction.
In short given Vieta jumping world rules k can be a perfect square if and only if a or b at some point is equal to zero. The original world the rules proscribes this. If this rule is not valid in the Vieta jumping world then the Vieta jumping world proof is invalid in the original rules arena.
You're using the word "equation" incorrectly. An equation has an equals sign (as the name suggests) and two sides. What you keep referring to as an equation is called an "expression."