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Recall Question 6 of the 1988 Math Olympiad

Question 6 is as follows:

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.

My proof:

Proof:

Let us denote the square of an integer by $k^2$. We have to show that $$\frac{a^2+b^2}{ab+1}=k^2$$ for some $a > 0$ and $b > 0$ where $a, b, > k \in \Bbb Z$.

Rewriting the equation:

$$a^2+b^2=k^2(ab+1)=k^2ab+k^2$$

Let $k = b$:

$$a^2+b^2=b^3a+b^2 \Rightarrow a^2=b^3a \Rightarrow a=b^3$$

To check, let's substitute these relations back in the original expression:

$$\frac{a^2+b^2}{ab+1}=\frac{(k^3)^2+(k)^2}{(k^3)(k)+1}=\frac{k^6+k^2}{k^4+1}=\frac{k^2(k^4+1)}{k^4+1}=k^2$$

Two of things I would like to know about this Olympiad Question:

1. Is this proof correct?
2. Is there a reliable way to check if a certain solution to a proof already exist?

Background

Every now and then I visit Youtube for educational purposes. I've been a great fan of Kurzgesagt, Numberphile, Computerphile, and other educational channels. For anyone not having heard of these channels I would greatly recommended to check them out!

Today I was watching this video on the channel of Numberphile: The World's Best Mathematician (*) - Numberphile. In this video a great mathematician Professor Tao tells about his passion for mathematics from a young age. At some point in this video it is mentioned that Tao only scored 1 out of 7 points for the infamous Question 6 from the 1988 Math Olympiad. In the next couple of frames this video shows the details of the question. At that moment I paused the video to try and solve this question by myself, with only pen and paper.

After 10 minutes or so I figured out the solution to this problem and questioned myself why the question is called infamous. I came across some solutions at this site that are, in my opinion, mindblowing in terms of length and mathematical notation, for example this solution. Then I read something about Vieta-jumping that I'm not familiar with and how it's used to solve this question (or not). A rather simple solution I came across is this one. I think the elegance and quality of a solution is greatly influenced by its length and level of simplicity in terms of notation, terminology, and construction.

As far as I did my silly online research I couldn't find a solution that was equivalent to mine. Therefore I thought it would be a good idea to share it and let other people, with a much broader understandig of existing mathematical proofs, decide if this proof is correct (and a known solution).

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2 Answers

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$a=8,b=30$ gives $k=2$. So you haven't gotten all answers when you've assumed $k=b$.

You are trying to prove it for all cases. You've only proven it for some cases.

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Dividing by the first principle$$ ab+1)a^2+b^2(\frac{a}{b} $$$$ \ \ \ \ \ \ a^2+\frac{a}{b} $$

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$$ \ \ \ \ \ \ b^2-\frac{a}{b} $$Since ab+1 completely divides $$a^2+b^2$$$$b^2-\frac{a}{b}=0\\ \therefore a=b^3 $$The quotient is $$ a\over{b}$$Putting the value of a$$\frac{a}{b}=b^2$$Hence it is a perfect square

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