The Legend of Question Six - Numberphile

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Recommends Rutger Moody · 2016-08 · Mathematics Stack Exchange · on “Math olympiad 1988 problem 6: canonical solution (2) without Vieta jumping”

There is a recent question about this famous problem from 1988 on this forum, but I'm unable to respond to this because the subject is closed for me (insufficient reputation). Therefore this new post on the subject. here's the link to the earlier subject [LINK] The problem: Let a and b be positive integers. Let $$ k={{a^2+b^2}\over{1+ab}} $$ Show that if $k$ is an integer then $k$ is a perfect square. The question was: Is there a more direct and intuitive way to arrive at the solution (instead of the usual proof using Vieta jumping and proof by contradiction)? After seeing this hilarious Numberphile video [this video] on Youtube i decided i had to give it a go myself. The solution i came up with

Recommends user383527 · 2017-12 · Mathematics Stack Exchange · on “How to prove if $\frac{20b}{19b-20}=a$ is possible or not, where $a$ and $b$ are positive integers?”

Here it is: [this video] I will watch it, but I don't think the problem I've posted will be that hard, as it is for high school level proofs.

Cites cactus314 · 2016-11 · Mathematics Stack Exchange · on “Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square.”

Here is a bit of a hacky answer to your question in the affirmative You observe that $k = \mathrm{gcd}(a,b)^2$. After plodding through various resources, it is because one can Viete Jump: $$ (a,b) \mapsto \big(a_1, b_1\big) \mapsto \dots \mapsto (k,0) $$ and the gcd is [LINK] conserved. Once I agree with you, let $a = \text{gcd}(a,b) \, c$ and $b = \text{gcd}(a,b) \, d$ so that \begin{eqnarray} ab+1 &=& \frac{a^2+b^2}{\mathrm{gcd}(a,b)^2}= c^2 + d^2 \\ &=& \,\text{gcd}(a,b)^2 \, cd + 1 \end{eqnarray} In this way there are two conditions $c, d$ might solve (where the two $\square$ are different) and $\text{gcd}(c,d)=1$: \begin{eqnarray*} c^2 - \square \, cd + d^2 &=& 1 \\ c^2 + d^2 &=& \squar

Cites Arthur · 2017-11 · Mathematics Stack Exchange · on “does this proof for "question 6" have a flaw?”

Given that this was problem 6 of the 1988 IMO in Australia [LINK] , a problem that was long considered the hardest one to ever appear in the olympiad [this video] , I don't think the solution is that simple.

Cites Omri Dor · 2017-12 · Mathematics Stack Exchange · on “SOOO close to a really clean solution to Question 6 from the 1988 Math Olympiad”

I recently came across the "infamous" Question 6 after watching this youtube video by Numberphile [this video] . Just to refresh everyone's memory, here it is: Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $$\frac{a^2+b^2}{ab+1}$$ is the square of an integer. Being ever the optimist, I took out a pen and paper and got to a very simple solution in less than 30 minutes. With all this hoo-hah about this problem, I assumed my solution is probably wrong, so I decided to upload here. As soon as I began to write it in a more orderly fashion, I realized I have a missing step :( I still think this approach is very elegant (a lot more than those Vietta jumps which a

Cites Patrick Danzi · 2020-05 · Mathematics Stack Exchange · on “Simple proof for the Legendary Question 6. International Mathematical Olympiad (IMO) 1988”

Question number 6 posed at the 1988 International Mathematical Olympiad (IMO) has become famous for its relative complexity. It consisted in showing that: Let $n_{i}$ and $n_{i+1}$ be positive integers such that $(1+n_{i} × n_{i+1})$ divides $(n_{i}^2+n_{i+1}^2)$ . Show that $$ \frac{n_{i}^2+n_{i+1}^2}{1+n_{i} n_{i+1}} \tag{1}$$ is the square of an integer. This problem has been made famous also through the work [this video] of popularizers, therefore I think that a simple proof of it could be of interest even for those who are not particularly skilled in mathematics. So I tried to formulate an accessible one for anyone who has minimal confidence with the equations of the second degree. $ \ $ Any