1988 Question 6.

In 1988's International Mathematical Olympiad following question was number six (little bit edited): Let $a,b\in {\mathbb{Z}}_{+}$ such that $ab+1$ divides $a^2+b^2$. Show that $$\frac{a^2+b^2}{ab+1}=x^2$$ such that $x\in \mathbb{Z}$.

I come a cross this probelm in the ending of Numberphile's video. Video was about proof by induction and the ending claimed Professor Zvezdelina Stankova was able to proof question 6 via induction. This problem inquired me. Since her explanation of the proof was avaible, I decided not to watch the video and try my own hand on the proof first. After that, I probably contrast it with her explanation.

First, $x$ been in $\mathbb{Z}$ is rather large. Since squares of negative numbers is same as the square of negative of that negative number, we can just consider ${\mathbb{Z}}_{+}$.

To determent variables for induction and the base case, let's play around. Following table summarizes the tests:

NEEDS TO BE RETHOUGHT FROM HERE.

It appears that statement is true because pair $❬1,1❭$ is only valid pair. Base case to verify this is our last test in the table. It proofs that every pair $❬n+1,1❭$ and additionally with symmetry every pair , where $n\in {\mathbb{Z}}_{+}$, does not have $ab+1$ divading $a^2+b^2$.

The induction hypothesis is: $$\forall n\in {\mathbb{Z}}_{+}\exists m\in {\mathbb{Z}}_{+}\left(\right(n+1)m+1\nmid {(n+1)}^2+m^2)$$

The induction step for $b=m+1$ is: $$\begin{array}{ccc}\frac{{(n+1)}^2+{(m+1)}^2}{(n+1)(m+1)+1}& =& \frac{n^2+2n+1+m^2+2m+1}{nm+n+m+2}\\ & =& \frac{n^2+m^2+n+m+n+m+2}{nm+n+m+2}\\ & =& \frac{n^2+m^2+n+m}{nm+n+m+2}+\frac{n+m+2}{nm+n+m+2}\\ & =& \frac{n^2+m^2+n+m}{nm+n+m+2}+\frac{nm+n+m+2}{nm+n+m+2}-\frac{nm}{nm+n+m+2}\\ & =& \frac{n^2+m^2+n+m}{nm+n+m+2}+1-\frac{nm}{nm+n+m+2}\end{array}$$ SECOND TRY!!!! $$\begin{array}{ccc}\frac{{(n+1)}^2+{(m+1)}^2}{(n+1)(m+1)+1}& =& \frac{n^2+2n+1+m^2+2m+1}{nm+n+m+2}\\ & =& \frac{n^2+m^2+2n+2m+2}{nm+n+m+2}\end{array}$$