The hardest mathematics problem on the hardest exam…

Fun fact: the current president of Romania participated in that Olympiad and solved that particular problem. He actually presented his solution to it on national TV at some point.

Let a and b be positive integers and assume ab + 1 divides a^2 + b^2. So we can write a^2 + b^2 = k(ab + 1) for some positive integer k. We must prove that k is a perfect square. 1.Fix k and choose a minimal solution Fix this k. Look at all positive integer pairs (x, y) satisfying x^2 + y^2 = k(xy + 1) … (1) Among these, choose a pair for which min(x, y) is as small as possible. Call this pair (A, B), and assume B ≤ A. Then A^2 + B^2 = k(AB + 1). … (2) 2.Regard (2) as a quadratic in A Rewrite (2) as a quadratic equation in A: A^2 − kBA + (B^2 − k) = 0. Let C be the other root of this quadratic. By Vieta’s formulas, A + C = kB, AC = B^2 − k. … (3) The coefficients of the quadratic are integers and A is an integer root, so the other root C = (B^2 − k) / A is also an integer. that C cannot be positive Because B ≤ A, we have C = (B^2 − k) / A < B^2 / A ≤ B. So if C > 0, then (C, B) is another positive integer solution of (1), and min(C, B) = C < B = min(A, B), which contradicts our choice of (A, B) as having minimal min(x, y). Therefore C is not a positive integer, so C ≤ 0. … (4) 4.But C is also greater than −1 From (3) we compute: (A + 1)(C + 1) = AC + A + C + 1 = (B^2 − k) + kB + 1 = B^2 + k(B − 1) + 1. Since B ≥ 1 and k ≥ 1, the right-hand side is strictly positive, so (A + 1)(C + 1) > 0. Because A + 1 > 0, it follows that C + 1 > 0, i.e. C > −1. Combining this with (4) and using that C is an integer, we must have C = 0. 5.Determine k Putting C = 0 into AC = B^2 − k from (3) gives 0 = B^2 − k ⟹ k = B^2. Thus k is the square of the integer B. Recall that k = (a^2 + b^2) / (ab + 1), so we have shown that this quotient is always a perfect square whenever it is an integer: (a^2 + b^2) / (ab + 1) is the square of an integer.

Why would you post this and not link the solution video in the replies?

Full video here!

Clever solution, Ellie. Explained beautifully.

Damn it was a video trap, now I’ll have to watch this other video and probably end up buying something

Fuck your engagement farming. Which method did you use? There's an integral derivative approach or substitution or elimination at first look.

(In)Famous problem that can be solved with Vieta jumping

a = 1 and b = 1 shows that that's the square of an integer. Specifically, 1. That wasn't so hard. Were they asking for a general solution? I guess I have to watch the video to understand what the question is asking. And I hope the math olympiad people learned to ask questions better

Thanks Ellie

If you want people to head to your YouTube channel, you should put the link to that video here.

You really speak in soothing voice :)

Indeed, that was a very elegant proof by contradiction. Great job with that video, instant follow :D

Hello Ellie, nice TL ! I would like to know what are your thought about the result of 1 to the power of infinite Thank You 😉✨

there's a lot of talk. but i do not think 4 hours show us true value

a=1, b=1 (1²+1²)/(1×1+1) = 1 √1=1

Gonna solve that problem, I think I have seen that before

Great, now I have to learn the impossible problem… didn’t have that in my cards before my coffee 🤣

@grok can you solve this?

As I always said Tao is not that good 😎

on test i write... can't .....won't...... not gonnna

Fiction.

Hell yea Ellie 🔥

Doesn't a=3 and b=3 kind of just immediately prove the premise wrong though? (3^2+3^2)/(3*3)+1 = 9+9/9+1 = 18/10 = 9/5 That is rational, and thus cannot be the square of any integer.

Bullshit 2 and 3 give (4+9)/(2*3+1) = 13/7 IS NO INTEGER SQUARE.