Vieta Jumping and Problem 6 | Animated Proof

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.

[ryanm]

Vieta Jumping is my favorite math thing right now. This or the Eigenvectors from eigenvalues proof. Thanks for this!

[hk]

Fantastic proof really hit the intuition

[dieago]

I love how they gave this problem to a 13 year old Tao with little more than 90 mins per question, a scrap of paper and a pencil.

[zal]

You lost me at “consider the equation”

[theevilmathematician]

This is a nice and clever solution for problem 6!

[coldestmutt]

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

[iainfulton]

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

[jharnamandal]

Only vdo that made me understand the proof.Thank you very much

[toastyarmor]

Mr. Penguin (idk your name) this video is absolutely brilliant!

[justsomeboyprobablydressed]

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."

[williejohnson]

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.

[gardenmenuuu]

which software did u use to make this animation?

[aryanandaleebazim]

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?

[vijayaade]

What r the prerequisites to solve this problem?? Pls somebody tell..

[UZPvNUCaaQdF]

@1:53 did you mean "denominator"?

[ivanmarinov]

I get lost in how x2 - (kB)x + (b2 - k) = 0 What does kB stand for? (Please can you explain)

[aryanandaleebazim]

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 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?

[TheProudHeretics]

Technically a=b=0 is another solution in the integers with a=b

:P

[SlurpKing]

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.

[mathsinmo]

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.